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Students can Download 1st PUC Biology Model Question Paper 6 with Answers, Karnataka 1st PUC Biology Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Biology Model Question Paper 6 with Answers

Time: 3.15 Hours
Max Marks: 70

General Instructions:

1. The question paper consists of four parts A, B, C and D.
2. All the parts are compulsory.
3. Draw diagrams wherever necessary. Unlabelled diagrams or illustration do not attract any marks.

Part – A

Answer the following questions in one word / one sentence each: ( 10 × 1 = 10 )

Question 1.
What is a Monograph?
Answer:
Monographs give a comprehensive account of a complete compilation of available information of anyone’s family or genus at a time.

Question 2.
What are viroids?
Answer:
These are the small infectious self-replicating RNA particles lacking protein coats.

Question 3.
Define Species?
Answer:
It is a group of individuals with similar morphological characters and who can freely interbreed among themselves, to produce fertile offsprings of their own kind.

Question 4.
What is phyllotaxy?
Answer:
The mode of arrangement of leaves on the stem axis is known as phyllotaxy.

Question 5.
Name the Chemical matrix of bone?
Answer:
Ossein.

Question 6.
Name the main component of the secondary cell wall?
Answer:
Lignin or pectin.

Question 7.
Define Imbibition?
Answer:
Imbibition is a special kind of diffusion that involves the movement of water molecules along a diffusion gradient from a region of higher concentration onto a suitable solid matrix (adsorption).

Question 8.
Name the hormone responsible for fruit ripening?
Answer:
Ethylene.

Question 9.
What is Gout?
Answer:
It is a disorder caused due to the accumulation of uric acid crystals in joints.

Question 10.
Name the hyperglycemic hormone?
Answer:
Glucagon.

PART-B

Answer any FIVE of the following questions in 3-5 sentences each. (5 × 2 = 10)

Question 11.
List any two uses of Bacteria?
Answer:
As natural Scavengers: They decompose dead parts of animals, plants, and other organic matters into simpler inorganic matters. e.g.: Proteus, Clostridium, Bacillus, etc.

As fermenting agents:
eg: Vinegar (acetic acid) is obtained from alcohol by fermentation in the presence of the bacterium Acetobacter acetic.

Question 12.
Draw a neatly labeled diagram of Chloroplast?
Answer:

Electron microscopic structure of the chloroplast

Question 13.
Write the Linnaenean Hierarchy of Mango?
Answer:

Kingdom	Plantae
Phylum	Tracheophyta
Class	Dicotyledonae
Order	Sapindales
Family	Anacardiaceae
Genus	Mangifera
Species	Indica

Question 14.
Write any four characters of meristematic Tissues?
Answer:

1. Cells of the meristem may be spherical, oval, polygonal, or rectangular in shape.
2. The cell wall is thin and made up of cellulose only.
3. Intercellular spaces are absent because of compact arrangements.
4. Each cell has a large nucleus and a dense cytoplasm usually without vacuoles.
5. Cells are highly active and exhibit rapid divisions.
6. Plastids are in proplastid state.
7. Mitochondria and endoplasmic reticulum are less complex in organization.
8. Cells do not store food materials.
9. The cells lack orgastic substances.

Question 15.
What are the functions of the Liver?
Answer:

1. It is a storehouse for glycogen and some vitamins.
2. It secretes bile juice, which helps in the emulsification of fats.
3. It is responsible for the detoxification of the digested food.
4. It provides proteins like fibrinogen and prothrombin, that are involved in the clotting of blood.

Question 16.
Distinguish between Vital capacity and Total lung capacity?
Answer:
Vital Capacity (VC): It is the total volume of air that can be expelled from the lungs during maximum exhalation.
VC = TV + IRV + ERV
= 500 + 2500+ 1100
= 4100 ml approx.

Total lung capacity (TLC): It is the maximum volume of air that can enter the lungs during forceful inspiration,
TLC = VC + RV
= 4100 + 1600
= 5200 approx.

Question 17.
Which hormonal deficiency is responsible for the following?
(a) Diabetes mellitus
(b) Goitre
Answer:
(a) Insulin
(b) Thyroxine.

Question 18.
Write the excretory organs of (a) Arthropoda (b) Mollusca.
Answer:
(a) Malpighian tubules
(b) Metanephridia.

PART-C

Answer any FIVE of the following questions in 40-80 words each: (5 × 3 = 15)

Question 19.
List the General characters of Gymnosperms.
Answer:
General features:
1. The life cycle has a distinct, dominant, diploid, asexual phase represented by the well-differentiated evergreen woody plant, which is known as the sporophyte.

2. The sporophyte is heterosporot4s bearing .microspores and megaspores within microsporangia and megasporangia respectively. These structures occur on leaf-like microsporophylls and megasporophylls. These are further organized into fertile structures called Strobili or cones.

3. Sporophyte shows the presence of a taproot system which is well developed. The stem possesses branches that are dichotomies. Leaves are well developed and are dimorphic (two types of leaves): viz,

• Green photosynthetic leaves (Foliage).
• Brown-colored scale leaves.

4. Microspore develops into male gametophyte arid megaspore produces female gametophyte.
These gametophytes represent the haploid phase and are highly inconspicuous in comparison with sporophytic generation.

5. Female gametophyte is enclosed within a megasporangium that in turn is covered by an integument. Such a tegument megasporangium possessing the female gametophyte is called an ovule.

6. Endosperm or female gametophyte is regarded as per the fertilized product and is haploid.

7. Semigerminated pollen grains (microspores) are liberated from the microsporangium and are pollinated onto the ovule. In the ovule, they sit in the pollen chamber and complete germination by producing a pollen tube which enters into nucellus, female gametophyte endosperm and ultimately enters archegonia and reaches the egg located in the center region. The tip of the pollen tube bursts releasing sperm cells which fuse with the egg to form a zygote. (Simultaneously 2-3 archegonia in an ovule fertilized).

8. Embryogeny takes place where only one functional embryo is formed along with seed development.

9. The seed germinates to form a new sporophyte.

Question 20.
List the functions of the endoplasmic reticulum.
Answer:
(Serves as secretary, storage, circulatory and nervous system for the cell).
1. ER forms the ultrastructural skeletal framework (cytoskeleton) to the coli and gives mechanical support to the colloidal cytoplasmic matrix.

2. It takes part in the transportation of substances. It is an efficient intracellular transportation channel that connects nuclear membrane and cell membrane and permits the exchange of molecules through its membranes by the process of osmosis, diffusion, and active transport.

3. It provides an increased surface area of various enzymatic reactions like protein synthesis. Also, ¡t contains many enzymes, and thus takes part in synthetic and metabolic activities.

4. It is useful in intracellular impulse conduction, and thus nucleus is able to respond quickly to change in the cell’s immediate environment.

5. It forms the new nuclear envelop, after each nuclear division.

6. It contributes to the formation of the cell plates. during telophase of cell division in plants.

7. Smooth endoplasmic reticulum takes part in lipid synthesis. break down of glycogen and detoxification of drugs.

8. Rough endoplasmic reticulum takes part in protein synthesis and secretion of enzymes.

Question 21.
Mention the role of Gibberellins in plants.
Answer:

1. Discovered by Kurasawa in paddy seedlings, while investigating the cause for Bakane disease.
2. Yabuta, and Suzuki isolated gibberellins from the fungi GibberellafuzikorL
3. Gibberellins break genetic dwarfism ¡n dwarf varieties of pea and beans.
4. Can induce internode elongation and early flowering in rosette plants like cabbage and cauliflower. This is called bolting.
5. Can break seed dormancy and bud dormancy.
6. Can induce parthenocarpic fruits.
7. Gibberellins can substitute cold treatment i.e., can replace ‘vernalization.
8. Can be used in inducing flowers even in off-seasons in long-day plants.

Question 22.
Define the following (a) Epigynous flower (b) phyllotaxy (c) venation.
Answer:
(a) a flower possessing inferior ovary
(b) The mode of arrangement of leaves on the stem axis is known as phyllotaxy.
(c) The mode of arrangement of veins and veinlets is called venation.

Question 23.
Explain the transport of carbon dioxide by blood.
Answer:
CO₂ is produced within the body cell as a metabolic waste product of oxidative metabolism.
CO₂ is transported from cells to the alveoli through the following forms.

1. In the form of carbonic acid (7%).
2. In the form of carbamino hemoglobin (23%).
3. In the form of bicarbonates of Sodium and Potassium (70%).

Question 24.
Briefly explain the digestion of carbohydrates in the small intestine.
Answer:
a. Digestion in the mouth: The salivary amylase or ptyalin converts starch and glycogen into maltose units.

b. Digestion in the intestine :
Pancreatic juice: Carbohydrase secreted by the pancreas is called pancreatic amylase. It converts starch and glycogen into maltose units.

Intestinal juice: It consists of three types of carbohydrates.

Maltase: Maltase catalyzes the hydrolysis of maltose (disaccharide) into glucose units (monosaccharides).

Sucrase: Sucrase catalyzes the splitting of sucrose (disaccharide) into glucose, and fructose units (monosaccharides).

Lactase: Lactase acts on milk sugar lactose (disaccharide), and splits lactose into galactose, and glucose units (monosaccharides).

Question 25.
Briefly explain any two disorders of the respiratory system.
Answer:
Asthma (Bronchial asthma):
It is a common disorder of the lower parts of the respiratory system involving smaller bronchi and bronchioles. It is caused due to an allergic reaction to foreign substances in the respiratory system. The substances which are responsible for asthma are called allergic substances or allergic agents.

In allergic asthma, the individuals become allergic to certain allergens like platinum salts, sulfur dioxide, and aeroallergic agents like pollen grains, dust, and animal dandruff. Asthma is also caused due to bacterial infection to bronchi, and bronchioles.

Symptoms [Effect]:

1. Difficulty in breathing and suffocation.
2. Continuous coughing and wheezing.
3. Tightness in the chest (Lungs).
4. The swelling of bronchial walls takes place and secretes large quantities of thick mucus Allergic asthma can be prevented by avoiding exposure to the particular allergic agent, which causes it.

Question 26.
What is binomial nomenclature? Write any two rules of binomial nomenclature.
Answer:
This method was introduced by Carolus Linnaeus.

In this method, every organism is given a scientific name, which has two parts, the first is the name of the genus (generic name), and the second ¡s the name of the species (specific epithet).

e.g. Mangifera indica for mangoes and Homo sapiens for human beings. In the above, Mangifera and Homo are generic names, while India and sapiens are the names of the species belonging to Man gear and Homo respectively.

The classification consists of a hierarchy of steps where each step represents a range or category. The various ranges or categories used in classification are called Taxonomic categories.

PART-D (Section I)

Answer any FOUR of the following questions in 200-250 words each wherever applicable: (4 × 5 = 20)

Question 27.
Explain the different steps in Mitosis?
Answer:
Mitotic division: Mitotic division has two processes occurring simultaneously.
They are:
(A) Karyokinesis and
(B) Cytokinesis.
A. Karyoldnesis: (Gr: Karyon-nucleus; kinesis-movement): It is a continuous process and for convenience divided into four stages. They are prophase, metaphase, anaphase, and telophase.

I. Prophase: (Gr: pro-first, phases-stage): It follows the interphaše and is characterized by the following features :
The already duplicated chromosomes become progressively shorter and thicker due to condensation. The spindle fibers appear between the centrioles present at the poles.

• The nucleolus diminishes in size and finally disappears.
• The nuclear envelope ¡s partly disintegrated.

II. Metaphase: (Gr. Meta-after or second: phases-Stage) It is characterized by the following features.

• Spindle fibres are completely formed.
• The spindle fibres that are attached to chromosomes arç called chromosomal fibres.
• The chromosomes aggregate in the center of the cell. They come to lie in the equatorial places to form an equatorial plate.

III. Anaphase (Gr:ana-up; phases-stage):
Anaphase ¡s is characterized by the following features.
The two chromatids of each chromosome separate complétely, to become daughter chromosomes. The two daughter chromosomes move away from the equator towards the opposite poles.

During the anaphasic movement of chromosomes, the centromeres lead the path, and the arms trail behind. As a result, the anaphasic chromosomes app&r V, L, J, and I shaped.

IV. Telophase (Gr:telos-end; phases-stage):
It is characterized by the following features. ,.

• Here, reversal of the prophasic events occur. The daughter chromosomes move and reach the opposite poles where they become thin and thread-like again. These threads overlap one another to form a fine chromatin network.
• The spindle fibers disintegrate and disappear.
• Reconstitution of a new nuclear envelope occurs. Nucleolus reappear.

B. Cytokinesis: (Gr:kytes-hollow; kinesis-movement): The division of the cytoplasm is called cytokinesis. A furrow develops in the middle of the cell. They deepened centripetally, till the cytoplasm is divided into two equal parts. It forms two daughter cells.

Question 28.
Draw a neatly labeled diagram of L.S. of phloem and explain briefly.
Answer:

Phloem is a food conducting tissue and is also called bast. Phloem is composed of four elements:
(a) Sieve elements
(b) Companion cells
(c) Phloem parenchyma
(d) Phloem fibers.

(a) Sieve elements: Sieve elements in lower vascular plants (Pteridophytes and gymnosperms) are referred to as ‘sieve cells’ whereas in higher plants (Angiosperms) they are referred to as ‘sieve tubes’.

Sieve cells are elongated, thin-walled, and arranged one above the other, and show a thin protoplasmic layer within. Sieve cells possess perforated lateral walls and they are not associated with companion cells, but with the cells called albuminous cells.

Sieve tubes are much longer than sieve cells and are arranged one above the other longitudinally with perforated transverse end walls, called ‘sieve plates’. Protoplasmic strands maintain the continuity through these perforations with the adjoining sieve tubes, sieve tubes are living cells but they are enucleated.

At the end of the growing season, sieve pores are covered by deposits of callose.

(b) Companion cells: These are thin-walled living and narrow cells running parallel to sieve tubes in angiosperms. Sieve tubes and companion cells are sister cells derived from the same mother cell. These are known to regulate the functions of sieve tubes.

(c) Phloem parenchyma: Parenchyma cells associated with phloem are referred to as phloem parenchyma. They are stored in function. Phloem parenchyma is absent ¡n monocot stem.

(d) Phloem fibers: These are sclerenchyma fibers associated with them. They are also called hast fibers. They provide mechanical support to the plant body.

Question 29.
Explain the structure of Neuron with a neatly labeled diagram?
Answer:
Structure of Neurons:

1. Each neuron has a cell body, dendrons (dendrites), and an axon.
2. The cell body contains cytoplasm with a nucleus, and certain granular bodies, called Nissil granules.
3. A number of processes (outgrowths) arise from the cell body. The longest among them is called axons while the others are called dendrons and their branches dendrites.
4. The axon is a long fiber and is branched at its distal end. Each branch terminates as a bulb-like structure called synaptic knob, which contains synaptic vesicles with neurotransmitters.
5. The axon transmits the impulses away from the cell body, while the dendrites! dendrons conduct it to the cell body.
6. Axons are of two types depending on the presence or absence of a myelin sheath around them. They are called myelinated axons/nerve fibers and non-myelinated axons.
7. The myeÌinated nerve fibers are enveloped with Schwann cells, which form a myelin sheath around the axon.
8. The myelin sheath is not continuous and the gaps are called nodes of Ranvier.
9. Myelinated nerve fibers are found in the brain and spinal cord, while nonmyelinated fibers are commonly found in autonomous and somatic neural systems.

Question 30.
Explain the factors affecting photosynthesis.
Answer:
The following factors influence photosynthesis:
1. Light: Sunlight is used as a primary source of energy. Its three aspects affecting the process are:
(a) Light Intensity: Optimum light intensity for photosynthesis is 2000-2500 foot candles/ Higher light intensity bleaches chlorophyll and is called solarisation.
(b) Light Quality: Photosynthetic rate ¡s maximum in red light, next maximum in blue light, and least in green light.
(c) Duration: Photósynthetic rate is more efficient in intermittent light than in continuous light supply.

2. CO₂: Used as a raw material in photosynthesis. Its concentration in air is 0.03%. Its increase up to 0.5% increases the rate of the process but above 0.5% inhibits photosynthesis.

3. O2: Liberated as a by-product in photosynthesis. its increase above the normal 21% in air, decreases the rate of the process. It is called the Warburg effect.

4. Temperature: It affects photosynthesis through its influence on the enzyme-controlled dark reaction. The optimum temperature is 18 – 40°C for photosynthesis.

5. Water: Used as a raw material in photosynthesis. In the dehydrated state of cells, photosynthesis is inhibited.

Question 31.
(a) What is photoperiodism? Classify plants based on photoperiodism.
Answer:
Plants, in order to flower, require a particular day length or light period called photoperiod, and the response of the plants to photoperiod in terms of flowering is called photoperiodism.

Photoperiodism was first studied by W.W. Garner, and HA. Allard.

Based on their photoperiodic responses, plants are classified into the following groups:

1. Long Day Plants: These flower in photoperiod more than critical day length. eg: Wheat, oats, etc.
2. Short Day Plants: These flower in photoperiod less than critical day length. e.g: Tobacco, Chrysanthemums, etc.
3. Day Neutral Plants: These are the plants that are not influenced by the duration of light for their flowering.
   e.g: Tomato, cucumb&, cotton, etc.

(b) List the uses of Transpiration?
Answer:
The important external factors that influence the rate of transpiration are:
1. Light: Light directly increases the rate of transpiration by inducing the opening of stomata.
Starch is formed in the guard cells in the presence of light. Starch gets converted into sugars which absorb water into the guard cells. Guard cells become turgid, and stoma opens, and the rate of transpiration increases. When there is no light the stomata gets closed, and the rate of transpiration decreases.

2. Temperature: When temperature increases, the atmospheric humidity decreases, and the rate of transpiration increases, and when temperature decreases, transpiration decreases.

3. Atmospheric humidity: It is the amount of water vapor present in the atmospheric air. When the atmospheric humidity is more, the rate of transpiration is less, and when atmospheric humidity is less the rate of transpiration is more.

4. Wind velocity: Plants produce a cloud of water vapor just above the leaves which reduces the rate of transpiration. When wind velocity removes the water vapor around the plant, then the rate of transpiration increases. But a very high wind velocity tends to close the stomata, and the rate of transpiration decreases.

5. AvaiLability of soil water: When there is less soil water, the rate of water absorption Asics, and rate of transpiration is also less.

Question 31.
Explain Transpiration -pull theory of Ascent of Sap?
Answer:
It is one of the most successful physical theories of the ascent of sap. Dixon and Jolly proposed this theory. According to this theory, the ascent of sap is due to three factors namely,
(a) Cohesive force of water
(b) Adhesion of water molecules to the walls of the xylem vessel
(c) Transpiration pull

A strong intermolecular force of attraction exists between water molecules. Thus water molecules are bound to each other forming a continuous column of water.

There is a force of attraction between water molecules on the inner walls of xylem elements.
Therefore water molecules are attached to the wall of the xylem. It is called adhesion.

Due to the adhesive and cohesive properties of water, a continuous column of water is presold in the xylem.

Transpiration at the leaf surface forces the mesophyll cells of the leaves to draw water from the neighboring xylem elements. This creates a suction force known as Transpiration pull.

This suction force pulls the water column in the xylem upwards. This ascent of sap is due to the combined effect of adhesive, and cohesive properties of water and transpiration pull.

Question 32.
List the salient features of Phylum Annelida.
Answer:
Annelida and the remaining phyla under kingdom Animaba possess a true body cavity, hence are included under the grade Eucoelomata.

A principal feature of this phylum is the division of the body into similar parts or segments or metameres, marked externally by transverse grooves and internally by septa. Each segment or metamere posts the representatives of all the organs of the body namely blood vessels, ‘muscles, nerves, excretory and reproductive organs. This phenomenon is known as segmentation or metamerism.

Annelids show cephalization union of a few of the anterior body segments to form ahead which is specialized by the development of nervous, sensory, and feeding structures.

Section II

Answer any THREE of the following questions in 200-250 words each. (3 × 5 = 15)

Question 33.
Draw a neat labeled diagram of the female reproductive system of cockroaches.
Answer:

Question 34.
Write the sequential steps of Glycolysis.
Answer:

• It occurs ¡n the cytoplasm of the cell.
• It is an enzymatic reaction, thus temperature sensitive.
• It is a common reaction for both aerobic and anaerobic respiration.

Question 35.
Explain the structure of Mitochondria with a neatly labeled diagram.
Answer:

Section through mitochondrion (diagrammatic)

The elementary particles have been shown only on one crista.
It is present in eukaryotes except for mammalian RBC, and absent in prokaryotes. Its shape is oval, or sausage. Its number per cell depends on the metabolic state of the cell.

Structure: The mitochondria are bounded by two lipoprotein unit membranes namely the outer membrane and inner membrane. In between, their lies per mitochondrial matrix containing water, minerals, and enzymes.

The outer membrane is smooth and unfolded while the inner membrane ¡s folded and it produces an inward finger-like process called Cristae. Along the inner surface of the inner membrane, there is numerous tiny tadpole-like structures called elementary particles or F₁ particles, Racker’s particles, or exosomes. F₁ particle contains basal piece. stalk and head.

Sites of ATPase are between adjacent elementary particles. The inner membrane contains an electron transport system. it is made up of a chain of co-enzymes in the order of NAD, FAD, cytochrome B, cyt C, cyt A, cyt A₃. The inner space of mitochondria is filled with a dense fluid called mitochondria matrix containing water, proteins, lipids, all enzymes of Kreb’s cycle, circular DNA, and 80s ribosomes.

Functions:

1. Mitochondria are the centers of aerobic respiration.
2. They are the sites of synthesis and storage of energy as ATP. Hence called the powerhouse of the cell.
3. As they have circular DNA & ribosomes, they synthesize a few proteins for their own requirement. Hence they are called ‘Semi-autonomous cell organelles’.

Question 36.
Mention any five hormones of the pituitary with one function of each.
Answer:
The pituitary gland weighs about 0.5 gm and is a pea-sized endocrine gland that lies on the ventral surface of the brain attached to the hypothalamus by a nervous stalk called the infundibulum.

It is called the master gland or conductor of the endocrine orchestra as several of its hormones control other endocrine glands directly. The hormones of the pituitary that influence other endocrine glands are called tropins or trophic hormones. However, the pituitary gland itself works under the influence of the hypothalamus through releasing factors. The pituitary gland is called the master gland because it controls the functions of other endocrine glands by secreting hormones.

Based on origin, it is divided into two parts namely Adenohypophysis and Neurohypophysis.
1. Adenohypophysis (Anterior Pituitary) :
It accounts for nearly 75% of the total weight of the gland. It is derived from the buccal cavity in the form of a projection called Rathke’s pouch.

Adenohypophysis is further divided into three regions: Pars distalis, Pars intermedia, and Pars tubular. Pars intermedia degenerates during fetal development and occurs only as a small strip in adults. Pars tubular has no functional significance. Adenohypophysis cells secrete seven major hormones.

1. Somatotrophic hormone or Human growth hormone (STH or HGH): Promotes the growth of muscles. It stimulates the uptake of amino acids by tissues and their synthesis into proteins.

• Hypo secretion in childhood causes Pituitary dwarfism. Such an individual will be abnormally dwarf and is called a midget.
• Hypersecretion in childhood causes Pituitary gigantism. Such an individual will be abnormally tall.
• Hypersecretion in adolescence causes acromegaly which is characterized by the formation of disproportionately large hands, feet, cheekbones, jaws, etc.,

2. Thyroid-stimulating Honnone (TSH): Controls secretion of thyroid hormones by the thyroid gland and also regulates iodine intake by the thyroid gland.

• Hypo secretion leads to goiter, cretinism, and myxoedema.
• Hypo secretion leads to hyperthyroidism.

3. Adrenocorticotrophic hormone (ACTH): It controls the( secretion of hormones (Cortisone) by the adrenal cortex.

• Hypersecretion of ACTH causes Cushing’s syndrome.
• Hypo secretion causes Addison’s disease.

4. Follicle Stimulating Hormone (FSH): In females the induces growth and maturation of Graffian follicle and stimulates follicular secretion of estrogen. In males, it stimulates the testis to produce sperm.

5. Luteinizing Hormone or Interstitial Cell Stimulating Hormone (EH or ICSH): In females, it stimulates ovulation and the formation of the corpus luteum. In males, it stimulates interstitial cells in the testis to secrete testosterone.

6. Prolactin or Lactogenic or Luteotropic Hormone (LTH): In females, it causes growth and development of breasts during pregnancy. It stimulates milk production and secretion after childbirth, it inhibits ovulation during pregnancy and breastfeeding.

7. Melanocyte Stimulating Hormone (MSH): Increases skin pigmentation by stimulating dispersion of melanin in amphibians but its exact role in humans is unknown.

Question 37.
List the salient features of the phylum Arthropoda.
Answer:
The phylum Arthropoda is the• largest phylum of the animal kingdom. It includes more than 80% of all the known animals. They are found in marine, terrestrial and aerial habitats.

They are the only major invertebrates adapted to live on dry land.
Insects among arthropods are the only invertebrates capable of flight. More than about 10,00,000 species of arthropods have been described so far.
1. Arthropods are bilaterally symmetrical, triploblastic, metamerically segmented, and coelomates.

2. Segmentation of the body is less apparent and the number of segments is fewer in number.

3. They show the highest degree of cephalization, that is the formation of a well-developed head. The Head is usually formed by the fusion of six body segments. Following the head, the region is the trunk which is usually divisible into an anterior thorax and a posterior abdomen.

4.Their body segment usually bears paired lateral and jointed appendages. A segment of the appendage is known as podomere. The appendages are variously modified in different groups and even in different parts of the same animal.

• A few appendages close to the mouth are modified into jaws and are helpful in feeding.
• A few appendages are sensory in function.

5. Some appendages help, in locomotion. Due to the presence of jointed appendages or legs for locomotion, the phylum gets the name Arthropoda, which means jointed feet.

6. The exoskeleton ¡s periodically cast off and regrown at regular interv1s. This is redundant. This periodic casting off of the exoskeleton is known as ecdysis or mouthing. The presence of a chitinous exoskeleton does not permit the free growth of the animal. to overcome this problem, the exoskeleton is periodically cast off.

The stages between the series of molts are called instars. The animal grows during intervals between successive molts.

7. Muscles are present, which are segmentally arranged in bundles.

8. The body cavity is filled with blood and such a cavity or coelom is known as a hemocoel. The hemocoel doesn’t bear the lining of the coelomic epithelium. The true coelom is reduced and is confused to the spaces of excretory and genital organs.

Students can Download 1st PUC Biology Model Question Paper 5 with Answers, Karnataka 1st PUC Biology Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Biology Model Question Paper 5 with Answers

Time: 3.15 Hours
Max Marks: 70

General Instructions:

1. The question paper consists of four parts A, B, C and D.
2. All the parts are compulsory.
3. Draw diagrams wherever necessary. Unlabelled diagrams or illustration do not attract any marks.

Part – A

Answer the following questions in one word / one sentence each: ( 10 × 1 = 10 )

Question 1.
What is a botanical garden?
Answer:
These are collections of living plants maintained for reference, i.e., they are meant for identification and classification. Both local and exotic plants are grown.

Question 2.
What are heterocysts?
Answer:
Heterocysts are enlarged thick-walled cells present in cyanobacteria which help in nitrogen fixation and asexual reproduction.

Question 3.
Name the chemical present in the cell wall of a diatom.
Answer:
Silica and Cellulose.

Question 4.
What is a phylloclade?
Answer:
A stem modification to carry out photosynthesis and often without leaves.

Question 5.
What is a ligament?
Answer:
Fibers connect bone to bone.

Question 6.
What is bolting?
Answer:
Internode elongation and early flowering induced by gibberellins are called bolting.

Question 7.
Define Imbibition?
Answer:
Imbibition is a special kind of diffusion that involves the movement of water molecules along a diffusion gradient from a region of higher concentration to a suitable solid matrix (adsorption).

Question 8.
Name any two hormones produced in the Gastro-Intestinal tract.
Answer:
Gastrin and Secretin.

Question 9.
Name the chemical present in the secondary cell wall?
Answer:
Lignin.

Question 10.
What is adelphy?
Answer:
When stamens are partially or completely fused among themselves, it is called adelphy.

PART-B

Answer any FIVE of the following questions in 3-5 sentences each. (5 × 2 = 10)

Question 11.
List the main differences between mitosis and meiosis. .
Answer:
(i) Mitosis consists of single nuclear divisions.
Meiosis consists of two nuclear divisions.
(ii) In Mitosis, crossing over is absent.
In Meiosis, crossing over is present.

Question 12.
What is a transamination reaction? What is its significance?
Answer:
Transamination:
In this process, the amino group is transferred from one amino acid to the keto group of a keto-acid.
Glutamic acid is the main amino acid that transfers its NH₂ group to form seventeen other amino acids.

The reaction is catalyzed by the enzyme transaminase.

It results in the formation of amino acids.

Question 13.
What are nephridia? Mention the types of nephridia in earthworms?
Answer:
It is the excretory organ of the members belonging to Annelida.

There are 3 types of nephridia:
(a) Septal nephridia – Being is the segment.
(b) Integumentary nephridia – Inner side of the body wall in all segment expects.
(c) Pharyngeal nephridia – 4th, 5th, and 6th segments.

Question 14.
What are Ammonotelic animals? Give an example?
Answer:
Animals that excrete ammonia as nitrogenous waste products are called ammonotelic animals.
e.g: Aquatic invertebrates and bony fish, larvae of frog (Tadpole).

Question 15.
With reference to growth, define the following terms:
(a) Differentiation.
(b) Dedifferentiation.
Answer:
Differentiation: It is the conversion (maturation) of meristematic cells, to regain the power to divide under certain conditions.

Dedifferentiation: Permanent cells like parenchyma can revert back to meristematic activities (cell division) to form secondary meristem. This conversion is called Dedifferentiation.

Question 16.
What are the characteristic features of Euglenoids?
Answer:
Members of this phylum are freshwater, flagellate, and motile forms.
A single flagellum is found at the anterior end.

The flagellum develops from basal granule and an eyespot or stigma is present near to it. The cells have an elastic pellicle that gives a definite shape to the cell.

Euglena is photosynthetic when optimum light is available and is non-photosynthetic in the absence of light and engulfs the food materials. Hence it is described as mixotrophic. Contractile vacuoles help in osmoregulation.

Reserve food occurs in the form of paramylum starch.
Asexual reproduction is by longitudinal binary fission.

Question 17.
How are viroids different from viruses?
Answer:
Viroids: These are the small infectious self-replicating RNAparticles lacking protein coats.
Viruses: Viruses are regarded as ‘living entities’ since they exhibit living character when they occur within a host cell as intracellular parasites.
A typical virus consists of a central core of nucleic acid either DNA or RNA, surrounded by a protective protein coat called a capsid.

Question 18.
Name the cells of cartilage and bone.
Answer:
Chondrocytes and osteocytes.

PART-C

Answer any FIVE of the following questions in 40-80 words each: (5 × 3 = 15)

Question 19.
What is aestivation? Mention any two types of aestivation with one example of each.
Answer:
The mode of arrangement of sepals, petals, or even tepals in a flower bud is called aestivation.

The different kinds of aestivation are as follows:
1. Valvate aestivation: When sepals, petals, or tepals are not overlapping.

2. Imbricate aestivation: When out of the total number of sepals, petals, or tepals, one is completely out, one is completely in and the rest are in and out.

3. Descending imbricate aestivation: When the standard petals are large and overlap the two wing petals, which in turn overlap the keel petals. It is technically known as vexillary aestivation.

Note: It is characteristic of the members of the subfamily Papilionoideae (Papilionaceae).
e.g: Pea, Bean, Indigofera, Tephrosia, etc.

4. Ascending imbricate aestivation: When the small standard petal is completely ¡n and is overlapped by the lateral wing petals which in turn are overlapped, by the keel petals.

Note: It is characteristic of the sub-family Caesalpinioideae (Caesalpinae).
e.g: Caesalpinia pulcherhima, Delonix regia, etc.

5. Quincuncial aestivation: When out of the total number of sepals or petals, two are completely out, two are completely in and the rest are in and out.

6. Contorted or twisted aestivation: When all sepals or petals are both in and out.

Question 20.
Explain the digestion of Carbohydrates in the Human Small Intestine.
Answer:
a. Digestion in the mouth: The salivary amylase or ptyalin converts starch and glycogen into maltose units.

Intestinal juice: It consists of three types of carbohydrates.

Maltase: Maltase catalyzes the hydrolysis of maltose (disaccharide) into glucose units (monosaccharides).

Intestinal juice: It consists of three types of carbohydrates.

Maltase: Maltase catalyzes the hydrolysis of maltose (disaccharide) into glucose units (monosaccharides).

Sucrase: Sucrase catalyzes the splitting of súcrose (disaccharide) into glucose, and fructose units (monosaccharides).

Lactase: Lactase acts on milk sugar lactose (disaccharide), and splits lactose into galactose, and glucose units (monosaccharides).

Question 21.
Write the schematic representation of a Nitrogen cycle.
Answer:
Nitrogen Cycle:

• Nitrogen is a limiting nutrient for both natural, and agricultural ecosystems.
• It exists as two nitrogen atoms held together by strong triple covalent bonds (N = N).

The nitrogen cycle involves the following steps:

1. Nitrogen fixation
2. Ammonification
3. Nitrification and
4. Denitrification.

1. Nitrogen Fixation:

• The process of conversion of nitrogen into ammonia and/or other nitrogen compounds is known as nitrogen fixation.
• In nature, lighting and ultraviolet radiation provide energy, to convert nitrogen into nitrogen compounds like N₂O, NO, and NO₂.
• The atmosphere also gets some amount of nitrogen oxides from forest fires, automobile exhausts, industrial combustion, and power generating stations.

2. Ammonification:

• The process by which organic nitrogenous compounds arc decomposed to produce ammonia is known as ammonification.
• Some ammonia volatilizes and re-enters the atmosphere.
• Most of the ammonia is converted into nitrites and nitrates.

3. Nitrification:
Nitrification is the process of converting ammonia first into nitrite, and then into nitrate.
2NH₃ + 3O₂ → 2NO₂ + 2H⁺ + 2H₂O
2NO₂ + O₂ → 2NO₃

• These processes are carried out by soil bacteria that are chemoautotrophs.
• Ammonia is oxidized into nitrite by Nitrosomonas and Nitrococcus.
• Nitrite ¡s oxidized to nitrate by Nitrobacter,
• The nitrates are absorbed by the plants. and reduced to nitrites.
• The nitrites are transported to the leaves and reduced to ammonia, which forms the amino group of amino acids.

4. Denitnfication:

• It is the process of conversion or reduction of nitrates into free nitrogen.
• It is carried out by bacteria like Pseudomonas and Thiobacillus.

Question 22.
What is meristem? Write any two kinds of meristems.
Answer:
Meristems are classified on the basis of their:
(A) Position.
(B) Function and
(C) Origin and development.

(A) Based on position, there are mainly three types:
(a) Apical Meristems: They are found at the tip of the stem and root. The activity of this, meristem leads to an increase in length.
(b) Intercalary Meristems: These are found between permanent tissues, at the base of the internode and leaf. The activity of this meristem leads to an increase in the length of the internode.
(c) Lateral Menstems: These are meristem is situated parallel to the sides of the organs and permanent tissues. Their activity leads to secondary growth.

(B) Based on function, there are three types:
(a) Protoderm: Gives rise to the epidermis.
(b) Procambium: Gives rise to primary vascular tissues (xylem, phloem, and intrafascicular cambium).
(c) Ground meristem: Gives rise to the cortex, hypodermis, pith, and medullary rays.

(C) Based on Origin, there are three types:
(a) Primordial meristems: Small groups of young embryonic cells existing in early stages of development. They give rise to Primary meristem. They are present in the stem tip and root tip.

(b) Primary meristems: They are directly derived from the Promeristems and divide rapidly. They are responsible for primary growth.

(c) Secondary meristems: These arise secondarily from permanent tissues by a process of dedifferentiation. Hence they appear much later in the life of a plant. They result in the later growth of plants.

Question 23.
List the living and nonliving characters of viruses.
Answer:
The living characters are:

1. They occur as obligate parasites and can grow and reproduce within a host cell.
2. They occur indefinite forms and contain genetic material -either DNA or RNA.
3. They are pathogenic and cause infectious diseases like other pathogenic organisms.
4. They show sensitiveness to heat and cold, like other organisms.

The non-living characters are:

1. They exhibit a cellular nature i.e., they lack protoplasmic bodies such as cell wall, cell membrane, cytoplasm, nucleus, mitochondria, Golgi bodiès, E.R, ribosomes, etc.
2. They cannot grow and multiply when they are outside the host cell., They look inert.
3. They do not show cellular metabolism like respiration.
4. They can be isolated, crystallized, and stored like any other chemicals.

Note:

• Shape: Viruses may be rod-shaped, spherical, tadpole, tetrahedral, octahedral, or icosahedral.
• Size: Viruses are measured ¡n nanometer units. The size of viruses arises from 30 to 350 nm. eg: Pox vines – 300 nm, foot and mouth virus – 30 nm, psittacines -300 mn.
• Structure: A typical virus consists of a central core of nucleic acid either DNA or RNA, surrounded by a protective protein coat called a capsid. The nucleic acid constitutes the viral genome and may be either

Question 24.
(a) What is vernalization? How is the process of vernalization advantageous to plants?
Answer:
Vernalisation:
Besides correct photoperiod, some plants require low-temperature treatment for flowering. This treatment is known as vernalization.

Vernalization prevents precocious reproductive development late in the growing season and enables the plant to have sufficient time to reach maturity.

Certain food plants like wheat, barley, and rye have two varieties called,
(i) Spring variety, and
(ii) Winter variety.

The spring variety is planted in spring, and it completes the lifecycle before the growing season.
The winter variety is normally planted in autumn or spring and is harvested by mid-summer.

(b) What is the growth rate? Write the mathematical expression of arithmetic growth rates?
Answer:
Growth Rate:
The increase in growth per unit time is termed as growth rate.
The rate of growth may be arithmetic or geometric. In arithmetic growth, one of the daughter cells formed by mitosis continues to divide, while the other undergoes differentiation and maturation.

In Geometrical growth, both the daughter cells formed after mitosis, retain the inability to divide.

Question 25.
Mention the role of any five macronutrients in plants.
Answer:
Nitrogen :
It is absorbed as NO₃^–, NO₂^– and NH₄ ions.

Functions:
It is a major constituent of amino acids, proteins, nucleic acids, vitamins, etc.

Deficiency symptoms:

1. Chlorosis, observed first in older leaves.
2. Stunted growth in plants.
3. Purple coloration due to anthocyanin development in the shoot axis surface.
4. Dormancy of lateral buds.
5. Delayed flowering.
6. Wrinkling of cereal grains.

Question 26.
Explain the mechanism of breathing.
Answer:
It is a process of intake of oxygen and leaving out carbon dioxide from the lungs. The mechanism of breathing involves.
1. Inspiration (Inhalation: It is the process of drawing air into the lungs from the outside atmosphere (intake of air). During this process, muscles of the diaphragm contract, which increases the length of the thorax. In meanwhile intercostal muscles also contract, pulling the ribs outward. This phenomenon increases the width of the thorax, resulting in the expansion of the lungs. As a consequence pulmonary pressure falls. So oxygen-rich air rushes into the lungs and fills alveoli.

2. Exchange of gases: The alveoli are closely surrounded by a thin wallet epithelium, having a capillary network. The oxygen is under high pressure in alveoli, because of its higher concentration. Similarly, carbon dioxide concentration and pressure will be more in the capillaries containing impure blood.

So, the exchange occurs by diffusion through the capillary walls. The oxygen is drawn into the blood, and carbon dioxide is pushed into the alveoli.

3. Expiration (exhalation): lt is the process of throwing out carbon dioxide from the lungs to the outside. During this process, muscles of the diaphragm and intercostal muscles relax causing the collapse of the rib cage. This decreases the volume of the thorax and lungs.

As a consequence pulmonary pressure increases. So, the carbon dioxide-rich air of the alveoli is drawn out through the respiratory passage.

PART-D (Section I)

Answer any FOUR of the following questions in 200-250 words each wherever applicable: (4 × 5 = 20)

Question 27.
With schematic representation, explain the diplontic life cycle.
Answer:

• The dominant phase in the life cycle is the independent sporophytic plant.
• Meiosis occurs in the spore mother cells during spore formation and the gametophyte is short- and is dependent on the sporophyte.
  e.g. Fucus (brown alga), gymnosperms and angiosperms.

Question 28.
Write the General characters of the phylum Echinodermata?
Answer:
Echinodermata comprises animals with a spiny skin. About 6000 species of echinoderms have been described.
General characters:
1. They are exclusively marine.

2. Echinoderms are rounded, cylindrical, or star-shaped organisms.

3. The body is triploblastic, coelomate, and unsegmented with radial symmetry. (But larvae are bilaterally symmetrical).

4. Body is generally covered by skin or exoskeleton supported by an endoskeleton made up of calcareous plates/ossicles. From these plates, spines may arise which project through the epidermis (skin).

5. Head is absent, and the body is distinguishable into oral, and aboral surfaces.

6. Coelom is extensive, enterococcus filled with coelomic fluid, and amoebocytes.

7. They possess a characteristic Ambulacra) or water vascular system. It consists of a system of tubes that emerge as a series of pouch-like projections from the body surface as body appendages 1 podia/tube feet, and help in locomotion, food capture, and respiration.

8. Alimentary canal (Digestive system) is complete with a mouth on the oral surface and anus on an aboral surface.

9. Circulatory system is primitive and reduced. It s haemal, macular, or coelomic channels. (Thu it is an open type of circulatory system).

10. Respiration may body minute gills called dermal branchiae or papillae (star-fishes) or by tube feet, or by. genital bursae (brittle star) or peristomial gills (sea urchin) or by cloacal respiratorý trees (cucumaria).

11. Excretory system is absent. The excretory product is ammonia, which is eliminated by amoebocytes of the coelomic fluid.

12. Nervous system is primitive, and consists of 3 nerve rings, and radial nerves.

13. Sensory system ¡s poorly developed. It comprises simple eyes and chemoreceptors.

14. Sexes are separate with large gonads, and simple ducts.

15. Fertilisation is external.

16. Development is indirect with free-swimming ciliated larval stages.

17. Regeneration of lost parts takes place.

18. They exhibit autotomy i.e., self-amputation of bodý parts to escape from the enemy.

Question 29.
Describe the major types of simple epithelial tissue, with the help of a labeled diagram?
Answer:
Epithelial tissues are broadly classified into two types namely:
Simple epithelium: Single layered epithelial tissue is known as the simple epithelium.

Stratified epithelium: Multilayered epithelial tissue is known as the stratified epithelium.

Simple epithelium: Simple epithelium is further classified into:

1. Squamous epithelium.
2. Columnar epithelium.
3. Cuboidal epithelium.
4. Ciliated epithelium.
5. Glandular epithelium.

Question 30.
Describe the structure of chloroplast with a neatly labeled diagram?
Answer:

Electron microscopic structure of the chloroplast

Chlorophyll-containing plastids are called Chloroplasts. They are present in the cells of all green plants and abundant in the leaf mesophyll cells. They are elliptical or oval in shape. Chloroplast is bounded by two membranes with the intermembrane space called peri plastidial space, containing peri plastidial fluid. It is made up of H₂O, mineral ions, proteins, etc.

It is a lubricating fluid that avoids friction between the two membranes. The inner chamber is filled with a colorless proteinaceous fluid matrix called the stroma. Besides abundant proteins, stroma contains 70s ribosomes, circular DNA, and all the enzymes of the Calvin cycle.

Embedded in the stroma, there are green-colored bodies called grana, which are inter-connected by frets. Grana is the site of light reaction, each granum consists of a stack of lipoprotein membrane discs called Thylakoids. Each thylakoid contains several photosynthetic centers called quanta somes. Each quantasome contains about 250 chlorophyll pigments and a few xanthophylls and carotenes.

Only chlorophyll- is capable of harvesting light energy into photosynthesis and hence it is called the primary photosynthetic pigment. All the other pigments, merely absorb lightener’ and pass it on to chlorophyll, and hence are called accessory photosynthetic pigments.

Note: As chloroplasts contain circular DNA 70s ribosomes and as they go for protein synthesis, they are regarded as semi-autonomous cell organelles.

Question 31.
Draw a neatly labeled diagram of the nephron.
Answer:

Question 32.
Discuss the role of pancreatic hormones in the regulation of blood glucose levels.
Answer:
The pancreas is both an exocrine and an endocrine gland. For its endocrine function, it has islets of Langerhans. These islets have β (beta) or β cells, α (alpha) or A cells, and (delta) or D cells. The β cells secrete insulin and A cells secrete glucagon. Both are proteinaceous hormones regulating blood glucose levels.

The D cells secrete a hormone called somatostatin. It inhibits the digestion and absorption of nutrients. It also inhibits the secretion of insulin and glucagon.

1. Glucagon:

• It is a peptide hormone.
• It is a hyperglycemic hormone, i.e., it increases the level of glucose.
• It acts on the liver cells and hepatocytes ¡n the blood.
• It stimulates glycogenolysis and increases the level of glucose in the blood.
• It stimulates gluconeogenesis and the synthesis of glucose from amino acids.

2. Insulin:

• It is a peptide hormone.
• It is a hypoglycemic hormone, Le., it reduces the level of glucose in the blood by stimulating the uptake and utilization of glucose by tissue cells.
• It acts on the hepatocytes and stimulates the conversion of glucose into fats.
• It also acts on adipocytes and stimulates the conversion of glucose into fats.
• Deficiency of insulin causes hyperglycemia and prolonged hyperglycemia results in diabetes mellitus

Section II

Answer any THREE of the following questions in 200-250 words each. (3 × 5 = 15)

Question 33.
With a neat labeled diagram, explain phloem tissue.
Answer:

Phloem is a food conducting tissue and is also called bast. Phloem is composed of four elements:
(a) Sieve elements
(b) Companion cells
(c) Phloem parenchyma
(d) Phloem fibers.

(a) Sieve elements: Sieve elements in lower vascular plants (Pteridophytes and gymnosperms) are referred to as ‘sieve cells’ whereas in higher plants (Angiosperms) they are referred to as ‘sieve tubes’.

Sieve cells are elongated, thin-walled, and arranged one above the other, and show a thin protoplasmic layer within. Sieve cells possess perforated lateral walls and they are not associated with companion cells, but with the cells called albuminous cells.

Sieve tubes are much longer than sieve cells and are arranged one above the other longitudinally with perforated transverse end walls, called ‘sieve plates’. Protoplasmic strands maintain the continuity through these perforations with the adjoining sieve tubes, sieve tubes are living cells but they are enucleated.

At the end of the growing season, sieve pores are covered by deposits of callose.

(b) Companion cells: These are thin-walled living and narrow cells running parallel to sieve tubes in angiosperms. Sieve tubes and companion cells are sister cells derived from the same mother cell. These are known to regulate the functions of sieve tubes.

(c) Phloem parenchyma: Parenchyma cells associated with phloem are referred to as phloem parenchyma. They are stored in function. Phloem parenchyma is absent in the monocot stem.

(d) Phloem fibers: These are sclerenchyma fibers associated with phloem. They are also called bast fibers. They provide mechanical support to the plant body.

Question 34.
Draw a neatly labeled diagram of the sagittal section of the brain.
Answer:

Question 35.
List the physiological effects of Auxins on plants.
Answer:

1. Auxins were the first hormones to be discovered in plants.
2. F.W. Went confirmed and isolated the auxins from coleoptiles of Avena saliva.
3. Chemically it is called Indole-3-Acetic Acid and is derived from amino acid tryptophan.
4. Auxins can stimulate cell division, cell elongation, and cell maturation.
5. It promotes Apical dominance.
6. It initiates Root formation.
7. Can promote parthenocarpy.
8. Prevent premature fall of flowers, buds, and leaves.
9. 2, 4D, and 2, 4, 5T are used as selective weedicides.
10. Promote phototrophic and hydrotropic movements.
11. Promote xylem differentiation.

Question 36.
Describe the various stages of prophase I of meiosis I.
Answer:
During this process the diploid parent cell divides into two daughter cells which are haploid, hence it ¡s a reduction division.

A. Karyokinesis I:
1. Prophase I: It lasts for quite a long period and is studied under five substages.

1. Leptotene
2. Zygotene
3. Pachytene
4. Diplotene
5. Diakinesis.

1. Leptotene: (Gr leptons slender; tene – thread),

• The chromosomes are thin, long, and uncoiled. Each is a double chromosome consisting of two chromatids. These chromatids are held firmly.
• Each chromosome appears as a string of beads, the beads are the chromomeres.

2. Zygotene: (Gr: Zygon-male; tene- thread)
During this stage, the pairing of homologous chromosomes (half of the maternal and half of them paternal). This pairing is called synapsis. The pairs so formed are called bivalents.

Each bivalent consists of four chromatids and is therefore called a tetrad. The two chromatids of the same chromosomes are called sister chromatids and the belonging to two different chromosomes of a homologous pair are termed non-sister chromatids.

3. Pachytene: (Gr: Pachus – thick tene – thread).
Crossing over takes place by breakage and reunion of chromatid segments. After, crossing over, the two chromatids of a chromosome become dissimilar.
The points of interchange are X-shaped and are called chiasmata. (smg.chiasma)

4. Diplotene: (Gr: Diplo – double; one – thread)

• Repulsion between homologous chromosomes begins.
• Each tetrad now appears in different shapes j.c, X-shaped, ‘8’ shaped, or ‘O’ shaped.
• Repulsion results in the criminalization of chiasmata (sliding of chiasmata towards the ends of chromosomes.

5. Diakinesis: (Gr: dia-cross; kinesis- movement).

• The nucleolus breaks down and disappears.
• Spindle fibers appear.
• The nuclear membrane breaks down.
• Chromosomes are released into the cytoplasm.

1. Metaphase I: It is characterized by the completion of spindle formation. In animals cells, asters are formed around the centrioles at two poles. The paired chromosomes arrange themselves on the equator of the spindle fibers.

2. Anaphase I: Ñere, the spindle fibers contract and this makes the bivalents move towards the opposite poles of the spindle without the division of centromere. The separated chromosomes or univalents are also called dyads because each of them consists of two chromatics that remains attached to each other at the centromere.

3. Telophase I: Chromosomes undergo uncoiling. A nuclear envelope is formed around each group of chromosomes. The nuclei do not reappear now.

Question 37.
Draw a neat labeled diagram of the digestive system of cockroaches.
Answer:

Students can Download 1st PUC Basic Maths Model Question Paper 8 with Answers, Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Model Question Paper 8 with Answers

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any TEN questions. (10 × 1 = 10)

Question 1.
Write the real part of 2 + 3i.
Answer:
2.

Question 2.
If A = {a, b, c, d}, B = {a, b, c} find A n B.
Answer:
A∩B = {a, b, c}.

Question 3.
Simplify : (3°)² + (3²)°.
Answer:
(1)² + (9)°= 1 + 1= 2.

Question 4.
Find the value of log₁₀ 1000
Answer:
log₁₀ 1000 = log₁₀ 10³ = 3 log₁₀ 10 = 3(1) = 3

Question 5.
Find the 8th term of the A.P. 2, 4, 6………….
Answer:
a = 2, d = 2 ,n = 8, T8 = ?
T_n = a + (n – 1)d
= 2 + (8 – 1) (2) = 2 + 7 (2) = 2 + 14 = 16

Question 6.
Solve for x : 2 (- 8 + x) = 28 – 2x
Answer:
– 16 + 2x = 28 – 2x
2x + 2x = 28 + 16
4x = 44 ⇒ x = 11

Question 7.
Find the S. I. on Rs. 500 at 5% for 5 years.
Answer:
SI = \(\frac{500 \times 5 \times 5}{100}\) = 25 × 5 = 125

Question 8.
Define perpetuity.
Answer:
If the installments (annuities) are made for an infinite period them it is called a perpetuity.

Question 9.
Convert 0.25 into persentage.
Answer:
0.25 = 0.25 × 100% = \(\frac{25}{100}\) × 100% = 25%.

Question 10.
Convent 45° into radians.
Answer:
45° =45 × \(\frac{\pi}{180}=\frac{\pi}{4}\)

Question 11.
Find the value of Sin²60° + Cos²60°.
Answer:
Sin² 60 +Cos² 60 = 1.

Question 12.
Find the locus of a point that moves such that its distance x-axis is 5.
Answer:
y = 5

PART-B

II. Answer any TEN of the following questions: (10 x 2 = 20)

Question 13.
Find the number of positive divisors of 360
Answer:
360 = 2 × 3² × 5¹
= p₁^α₁ × p₂^α₂ × p₃^α₃
T(a) = (1 + α₁)(1 + α₂)(1 + α₃)
T(360) =(1 + 3)(1 + 2)(1 + 1)
= (4)(3)(2) = 12(2) = 24.

Question 14.
Find the L.C.M. of 9, 15 and 24.
Answer:
9 = 32
15 = 3 × 5
24 = 2³ × 3¹
LCM = 2³ × 3² × 5¹ = 8 × 9 × 5 = 360.

Question 15.
If u= {1, 2, 3, ……….. 9}, A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6, 7} then verify (A ∪ B)’ = A’ ∩ B’
Answer:
A∪B = {1, 2, 3, 4, 5, 6,7} (A∪B)’ = {8,9}
A’ = {6, 7, 8, 9} B’ = {0, 2, 8, 9} A’∩B’ = {8,9}
∴ A (A∪B)’= A’∩B’

Question 16.
If K + 9, – 6 and 4 are in GP., then find the value of K.
Answer:
(- 6)2 = (K + 9)4
36 = 4K + 36
⇒ 4K = 0
K = 0.

Question 17.
If α and β are the roots of the equation x² + 5x + 6 = 0. Then find the value of \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\)
Answer:
α + β = 5 αβ = 6
\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}=\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha \beta}=\frac{25-12}{6}=\frac{13}{6}\)

Question 18.
Find two numbers whose sum is 64 and the difference is 16.
Answer:
x + y = 64
x – y = 16
2x + 80
⇒ x = 40
x + y – 64
40 – y = 64
y = 64 – 40 = 24
y = 24

Question 19.
Sove the following linear inequality: 5x – 3 < 3x + 1, x ∈ real numbers. Also represent it on a number line.
Answer:
5x – 3 < 3x + 1
5 x – 3x < 3 + 1
2x < 4 ⇒ x < 2 or x ∈ (-∞, 2)

Question 20.
The average score of 40 boys is 60% and an average of 60 girls is 70%. Find the combined average of boys and girls.
Answer:
N₁ = 40 X₁ = 60% N₂ = 60 X₂ = 70%
X̄₁₂ = \(\frac{\mathrm{X}_{1} \mathrm{~N}_{1}+\mathrm{X}_{2} \mathrm{~N}_{2}}{\mathrm{~N}_{1}+\mathrm{N}_{2}}=\frac{60(40)+70(60)}{40+60}\)
\(\frac{2400+4200}{100}=\frac{6600}{100}\)
= 66

Question 21.
Find the value of: Sin² \(\frac{5 \pi}{6}\) + cos² \(\frac{5 \pi}{6}\) – tan²\(\frac{ \pi}{4}\)
Answer:
Sin 5π/6 = Sin 150 = Sin (180-30) = Sin 30 = 1/2
Cos 5π/6 = Cos 150 = Cos (180-30) = -Cos 30 = –\(\)
tan π/4 = tan 45 = 1
LHS = (1/2)² – (-3/2)² – 1²
= 1/4 + 1/3 – 1 = \(\frac{1+3-4}{4}=\frac{0}{4}\) = 0

Question 22.
Find the value of x, if the distance between (x, 3) and (4, 5) is 5 units.
Answer:
\(\sqrt{(x-4)^{2}+(3-5)^{2}}\) = 5
(x – 4)² + 4 = 25
x² – 8x+ 16 + 4-25 = 0
x² – 8x – 5 = 0
= \(\frac{8 \pm \sqrt{64-4(1)(-5)}}{2(1)}=\frac{8 \pm \sqrt{84}}{2}\)
= \(\frac{8 \pm 2 \sqrt{21}}{2}\)
x = 4 ± \(\sqrt{2}\).

Question 23.
Find the locus of a point that moves such that it is equidistant from (5, 0) and (-5, 0).
Answer:
P = (x, y) A = (5, 0) B = (- 5. 0)
PA = PB
PA² = PB²
(x – 5)² + y² = (x + 5)² + y²

20 x = 0 ⇒ x = i.e., y – axis

Question 24.
Find the equation of a straight line passing through (3, 4) and having a slope of 3.
Answer:
y – 4 = 3 (x – 3)
y – 4 = 3x – 9 ⇒ 3x – y – 5 = 0.

Question 25.
Derive the equation of the straight line passing through two points (x₁, y₁) and (x₂, y₂).
Answer:
Let P (x, y) be any point an the line joining points A (x₁, y₁) and B(x₂, y₂)
∴ APB are collinear
The slope of AP = Slope of AB

\(\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) (Two Point form)

PART – C

III. Answer any ten questions : (10 × 3 = 30)

Question 26.
Prove that \(\sqrt{2}\) is not a rational number.
Answer:
We shall prove it by the method ofcontradiction.
If possible Let \(\sqrt{2}\) be a rational number
Let \(\sqrt{2}\) = \(\frac{p}{q}\) where p and q are Integers and q ≠ 0.
Further let p and q are coprime i.e. H.C.F. of p and q = 1.
\(\sqrt{2}\) = \(\frac{p}{q}\) ⇒ \(\sqrt{2}\)q = p
⇒ 2q² = p²
⇒ 2 divides p² ⇒ 2 divides p
⇒ p is even
Let p = 2k where k is an integer p² = 4k²
2q² = 4k²
q² = 2k² ⇒ q² is even
⇒ q is even.
Now p is even and q is even which implies p and q have a common factor 2. which is a
contradiction of the fact that p and q are co-prime.
∴ our assumption that \(\sqrt{2}\) is rational is wrong and hence \(\sqrt{2}\) is irrational.

Question 27.
If R-1 = {(2, 4), (1, 2), (3, 1), (3, 2)}, find R. Also find its domain and range.
Answer:
R = {(4, 2), (2, 1), (1, 3), (2, 3)}
Domain of R= {1, 2, 4}
Range of R= {1, 2, 3}.

Question 28.
If a^x = b^y = c^x and b² = ac show that \(\frac{1}{x}+\frac{1}{z}=\frac{2}{y}\)
Answer:
Let a^x = b^y = c^z = k (say)
∴ a^x = k ⇒ a = k^1/x, b^y = k ⇒ b = k^1/y, c^z = k ⇒ c = k^1/z
Now, b² = ac
∴ (k^1/y)² = k^1/x. k^1/z
∴ k^2/y = k^1/x+1/z
Bases are the same ∴ Equating powers on both sides, we get.
\(\frac{1}{x}+\frac{1}{z}=\frac{2}{y}\)

Question 29.
Prove that \(\frac{1}{\log _{2} 4}+\frac{1}{\log _{8} 4}+\frac{1}{\log _{16} 4}\) = 4.
Answer:
LHS = \(\frac{1}{\log _{2} 4}+\frac{1}{\log _{8} 4}+\frac{1}{\log _{16} 4}\) = log 42+ log4 8+ log4 16
= log4 2. 8 .16 = log4 256 = log4 44 = 4 log4 4 = 4(1) = 4 = RHS.

Question 30.
The sum of three numbers in an A.P. is 15 and their product is 105. Find the numbers.
Answer:
Let the numbers be a – b, a, a + d
(a-d) + a + (a + d)= 15
3a= 15
a = 15/3 = 5
(a – d) a (a + d) = 105 .
(5 – d)5(5 + d) = 105
(5 – d)(5 + d) = \(\frac{105}{5}\)
25 – d² = 21
25 – 21 = d² ⇒ d² = 4
⇒ d = ±2
Let d = 2

∴ The numbers are
5-2, 5, 5 + 2
3, 5, 7.

Question 31.
If sum of the digits of a two digit numbers is 9 and if 9 is added to the number, then the digits get reversed.
Answer:
Let the number be 10x + y, x + y = 9
10x + y + 9 = 10y + x
9x – 9
y = – 9

x = 4
y = 5
∴ The number in 10 x +y = 10 (4) + 5 = 40 + 5 = 45.

Question 32.
In what time, a sum of Rs. 500 will earn Rs. 975 at the rate of 6% p.a. if C. 1 is payable half yearly
Answer:
P = 500 CI = 975
A = P + CI = 500 + 975 = 1475
R = 6% = 0. 06 (Nominal rate of interst)
9 = 2
r = \(\left(1+\frac{R}{q}\right)^{q}\) – 1
= \(\left(1+\frac{0.06}{2}\right)^{2}\) – 1
= 0. 0609 = 6. 09 %
r= 0. 0609
A + P (1 + r)ⁿ
log A= log P + n log (1 + r)

n = \(\frac{\log \mathrm{A}-\log \mathrm{P}}{\log (1+r)}\)
= \(\frac{\log 1475-\log 500}{\log 1.0609}\)
= \(\frac{3.1688-2.6990}{0.0253}\)
n = 18. 56 years.

Question 33.
By how much should the use of tea be decreased if the Price of tea is increased by 25% so that the expenditure remains unchanged.
Answer:
Let the Price be 100 Let the quantity be 100
Total expenditure = 100 × 100 = 10,000
New Price = 100 + 25 = 125
Let the new quantity be y
Total Expenditure = 125 y
∴ 125 y = 1000
y = 80 ∴ Decrease on quantity = 100 – 80 = 20
% decrease = \(\frac{\text { Decrease }}{\text { Original Value }}\) × 100%
= \(\frac{20}{100}\) x 100%
= 20%.

Question 34.
Find the B.E.P in units if total cost and the revenue functions are C(x) = 450 + 1. 5JC and R(x) = 3x. (x = level of out put).
Answer:
At Break Even point,
Total revenue = Total cost
∴ R (x) = C(x)
3x = 450 + 1.5x
3x – 1.5x = 450
1.5x = 450
∴ x = \(\frac{450}{1.5}\)
x = 300 units.

Question 35.
If cot A = \(\frac{5}{12}\) and A is acute show that 2 cosec A – 4SecA = – \(\frac{247}{30}\)
Answer:
cot A = \(\frac{5}{12}\)
cosec A = \(\frac{13}{12}\)
sec A = \(\frac{13}{5}\)

2 cosec A – 4SecA = 2.\(\frac{13}{12}\) – 4.\(\frac{13}{5}=\frac{26}{12}-\frac{52}{5}=\frac{130-624}{60}=\frac{-494}{60}=-\frac{247}{30}\)

Question 36.
If A, B, C are angles of a triangle then prove that.
(i) Sin (A + B) = SinC
(ii) Cos \(\left(\frac{\mathbf{A}+\mathbf{B}}{2}\right)\) = Sin\(\frac{C}{2}\)
Answer:
(i) Sin (A + B) = Sin (180 – C) = Sinc
(ii) Cos\(\left(\frac{\mathbf{A}+\mathbf{B}}{2}\right)\) = Cos\(\left(\frac{180-C}{2}\right)\) = Cos(90 – \(\frac{C}{2}\))
= Sin\(\frac{C}{2}\)

Question 37.
Show that the points A(2, 2), B(6, 3) and C (4, 11) form a right angled triangle.
Answer:
AB = \(\sqrt{(6-2)^{2}+(3-2)^{2}}=\sqrt{16+1}=\sqrt{17}\)
BC = \(\sqrt{(4-6)^{2}+(11-3)^{2}}=\sqrt{4+64}=\sqrt{68}\)
CA = \(\sqrt{(4-2)^{2}+(11-2)^{2}}=\sqrt{4+81}=\sqrt{85}\)
AB² = 17 BC² =68 CA² =85
∴ AB² + BC² = CA²
∴ ABC form a right angled triangle

Question 38.
Find the distance between the parallel lines : 3x + 4y + 5 = 0 and 6x + 8y + 20 = 0.
Answer:
6x + 8y + 10 = 0
6x + 8y + 20 = 0.
d = \(\left|\frac{\mathrm{C}_{1}-\mathrm{C}_{2}}{\sqrt{a_{2}+6_{2}}}\right|=\left|\frac{10-20}{\sqrt{6^{2}+8^{2}}}\right|=\left|\frac{-10}{10}\right|\) = |-1| = 1 units

PART – D

IV. Answer any six questions : (6 × 5 = 30)

Question 39.
Out of 50 people, 20 drink tea and 10 take both tea and coffee. How many take at least one of the two drinks. Also, represent the above problem by Venn diagram.
Answer:

n(T∪C) = n(T) + n(C) – n(T∩C)
50 = 20 + n(C) – 10
40 = n(C)
Number of people taking atleast one of the two drinks 10 + 10 + 30 = 40 + 10 = 50.

Question 40.
Find the value of the following using logarithmic table. \(\frac{213.781 \times 7.434}{6.321}\)
Answer:
Let x = \(\frac{213.781 \times 7.434}{6.321}\)
logx = log\(\left[\frac{213.781 \times 7.434}{6.321}\right]\)
= log213.781 + log 7.434 – log 6.321
= 2.3298 + 0.8712 – 0.8008.
= 2.4002
x = Antilog 2.4002
x = 251.3.

Question 41.
Find the sum of all the numbers betw een 50 and 200 which are divisible by 11.
Answer:
S = 55 + 66 + 77 +……………….+ 198
55, 66, 77, …………….198
Here a = 55, l = T_n = 198, d= 11
Now, T_n = a + (n – 1) d
198 = 55 + (n – 1) 11
198 – 55 = (n – 1) 11
2\(\frac{143}{11}\) = n – 1
n – 1 = 13
n= 13 + 1
n= 14
Now consider:
S_n = \(\frac{n}{2}\)(a + l)
S₁₄ = \(\frac{14}{2}\)(55 + 198)
= 7(253)
= 1,771.

Question 42.
Find the integral roots between – 3 and + 3 by inspection and then using synthetic division, solve the equation.
x³ – 2x² – 5x + 6 = 0.
Answer:
Let f (x) = x³ – 2x² – 5x + 6
f(1)= 1 – 2 – 5 + 6 = 7 – 7 = 0
∴ x = 1 is a root of the given equation by using Synthetic division.

The resulting equation is x² – x – 6 = 0 is the quotient and remainder = 0
x² – x – 6 = 0
(x – 3) (x + 2) = 0
x = 3 or – 2
Thus x = 1, -2, 3 are the roots of the given equation.

Question 43.
Find the difference between S. 1. and C. 1. on Rs. 18000 invested for four years at 8% p.a. where C. 1. is compounded annually.
Answer:
Simple Interest .
P= 18,000
T = 4
R = 4
SI = \(\frac{\text { PTR }}{100}\)
= \(\frac{18,000 \times 4 \times 8}{100}\)
= ₹ 5,760

Compound Interest
P = 18,000
n = 4
I = 0.08
A = P(1 + i)ⁿ
= 18,000(1 +0.08)⁴
= 24,488.8
CI = A – P
= 24,488.8 – 18,000
CI = 6488.8

Difference between Cl and SI is
CI – SI = 6488.8 – 5760
= 728.8.

Question 44.
Find the present value of an annuity of Rs. 2500 payable at the end of 6 months for 5 years, if money is worth 10% computed semi-annually.
Answer:
a = 2500
i = \(\frac{0.10}{2}\) = 0.05
n = 5 x 2 = 10
P = \(\frac{a\left[(1+i)^{n}-1\right]}{i(1+i)^{n}}\)
= 2500\(\frac{\left[(1+0.05)^{10}-1\right]}{0.05(1+0.05)^{10}}\)
= \(\frac{2500[1.6289-1]}{0.05 \times 1.6289}\)
= ₹ 19304.44.

Question 45.
A watch is sold for Rs. 150 at a profit of 25%. At what price should it be sold in order to have 50% profit.
Answer:

Question 46.
Prove that: \(\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}\) = 1 + Sec A. Cosec A.
Answer:

= sec A . cosec A + 1 = R.H.S.

Question 47.
Show that the following points are the vertices of a square A (1, 1), B (4, 1),C(4, 4) and D (1, 4)
Answer:
AB = \(\sqrt{(4-1)^{2}+(1-1)^{2}}=\sqrt{9+0}\) = 3
BC = \(\sqrt{(4-4)^{2}+(4-1)^{2}}=\sqrt{0+9}\) = 3
CD = \(\sqrt{(1-4)^{2}+(4-4)^{2}}=\sqrt{9+0}\) = 3
DA = \(\sqrt{(1-1)^{2}+(4-1)^{2}}=\sqrt{0+9}\) = 3
AB = BC = CD + AD
Sol.
AC = \(\sqrt{(4-1)^{2}+(4-1)^{2}}=\sqrt{9+9}=\sqrt{18}\)
BD = \(\sqrt{(1-4)^{2}+(4-1)^{2}}=\sqrt{9+9}=\sqrt{18}\)
∴ AC = BD
ABCD ∆ is a Square.

Question 48.
Find the area of a quadrilateral whose vertices are A(1, 1), B (4, 1), C (4, 4), and D (1, 4).
Answer:
The required area ABCD = area of the triangle ABC + area of triangle ACD
Area of the triangle ABC = \(\frac{1}{2}\)[1(-3 – 2) + 7(2 -1) + 12(1 + 3)]
= 25 sq. units

Similarly
Area of the triangle ACD =107 sq.units
Area of the triangle ABCD = 25 + 107 = 132 sq.units.

PART – E

V. Answer any one questions : (1 × 10 = 10)

Question 49.
(a) If f(x) = x + 1 and g(x) = x² + 1.
Find (i) fog (1), (ii) fog (2), (iii) gof (1), (iv) gof (2), (v) fog (3)
Answer:
f(x) = x+ 1 and g (x) = x² + 1
(i) fog (1) = f (g (1))
= f(2)
= 2 + 1 = 3

(ii) fog (2) = f (g(2))
= f(5)
= 5 + 1
= 6

(iii) gof(1)
= g(f(1))
= g(2)
= 4 + 1
= 5

(iv) gof (2)
= g(f(2))
=g(3)
= 9 + 1
= 10

(v) fog (3) f(g(3))
f (10) = 10 + 1
= 11.

(b) Find the sum to n terms of series
7 + 77 + 777 +………………n term.
Answer:
Let S = 7 + 77 + 777 + ………….. n terms
= 7 [1 + 11 + 111 +……………….to nterms] = \(\frac{7}{9}\)[9 + 99 + 999 + …………to n terms]
= \(\frac{7}{9}\)[ [(10 – 1) + (100 – 1) + (1000 – 1) + ………………to n terms]
= \(\frac{7}{9}\)[ [(10 + 10² + 10³ + ……………… to n terms) – (1 + 1 + 1 ………………. to n terms] 9
= \(\frac{7}{9}\left[\frac{10\left(10^{n}-1\right)}{10-1}-n\right]\) [∵ 10 + 10² + 10³ + ……. to n terms in GP where a = 10, r = 10]
S_n = \(\frac{7}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]\)

(c) Find the number of digits in 320.
Answer:
(a) 320
Let x = 320
log x = log 320 = 20 log 3 = 20 (0.4771) = 9.542
As the characterstic is 9, number of digits = 9 + 1 = 10.

Question 50.
(a) Solve the following inequalities graphically
x + 2y ≤ 8
2x + y ≤ 8
x ≥ 0
y ≥ 0
Answer:

(b) Find the future value of an annuity of Rs. 500 at 5% p.a. payable for 5 years.
Answer:
A = 500 r = 5% =0.05 n = 5
F = \(\frac{\mathrm{A}\left[(1+r)^{n}-1\right]}{r}\)
= \(\frac{500\left[(1+0.005)^{5}-1\right]}{0.05}\)
\(\frac{500\left[(1 \cdot 05)^{5}-1\right]}{0 \cdot 05}\) = 2762.81

(c) Find the quadratic equation whose roots are 2 and 5.
Answer:
(x – 2) (x – 5) = 0
x² – 2x – 5x + 10 = 0
x² – 7x + 10 = 0.

Students can Download 1st PUC Basic Maths Model Question Paper 7 with Answers, Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Model Question Paper 7 with Answers

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any TEN questions. (10 × 1 = 10)

Question 1.
Write the canonical representation of 96.
Answer:

Question 2.
Define an equivalence relation.
Answer:
A relation R on a set A is called an equivalence relation if it is reflexive. symmetric and transitive.

Question 3.
Simplify: 5(5°) + (52)°
Answer:
5¹ + (25)° = 5 + 1 = 6.

Question 4.
Find ‘x’ if log 625 = 4.
Answer:
x⁴ = 625 ⇒ x⁴ = 5⁴ ⇒ x = 5.

Question 5.
Find the 30th term of the AP -2, -5, 8,………………..
Answer:
a = -2 d = 5 + 2 – 3
T₃₀ = a + (30 – 1)d
= -2 + (29)(-3) = -2 – 87 = -89.

Question 6.
Form the quadratic equation whose roots are 1, 2.
Answer:
(x – 1)(x – 2) = 0 ⇒ x² – 3x + 2 = 0.

Question 7.
Calculate the S.I. on Rs. 18,000 for 4 months at 12\(\frac{1}{2}\)% per annum.
Answer:
P= 18,000 T = \(\frac{4}{12}=\frac{1}{3}\)R = \(\frac{25}{2}\)%

Question 8.
Write the formula to find the future value of Annuity Due.
Answer:
F = \(\frac{\mathrm{A}\left[(1+r)^{n}-1\right]}{r}\)(1 + r)

Question 9.
A person bought a cycle for Rs. 3000. For what price should he sell it to gain 10%?
Answer:
SP = \(\frac{100+\text { gain } \%}{100}\) × CP = \(\frac{100+10}{100}\) × 3000 = \(\frac{110}{100}\) × 3000 = Rs. 3300

Question 10.
Convert 15° into radian measure.
Answer:
15° = \(\frac{15 \pi}{180}=\frac{\pi^{c}}{12}\)

Question 11.
If A = 45°, then prove that sin 2A = 2 A. Cos A.
Answer:
Sin 2A = Sin 90° = 1
2 sin A cos A = 25 in 45° cos 450 = 2\(\frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}\) = 1
∴ LHS = RHS.

Question 12.
Find the equation to the locus of a point moving in a plane such that the sum of the squares of its distance from the coordinate axes is 2.
Answer:
x² + y² = 2.

PART -B

II. Answer any TEN questions. (10 × 1 = 10)

Question 13.
Find the imaginary part of (1 + i) (4 – 3i).
Answer:
(1 + i) (4 – 3i) = 4 – 3i + 4i + 3 = 7 + i
imaginary’ part = i.

Question 14.
Find the HCF \(\frac{8}{9}, \frac{32}{81}, \frac{16}{27}\)
Answer:
HCF of Numerators 8, 32, 16 = 8
LCM of Denominators 9, 81, 27 = 81
∴ HCF of fractions = \(\frac{\text { HCF of Numerators }}{\text { LCM of Denominators }}=\frac{8}{81}\)

Question 15.
If A = {3, 5, 7},B = {5, 7, 9} and U = {1, 2, 3, 4, 5, 6, 7, 8, 9} then find(A – B) × B’.
Answer:
A – B = {3}
B’ = {1, 2, 3, 4, 6, 7, 8)
(A – B) × B = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 6), (3, 7), (3. 8)}.

Question 16.
If the 5th term of a HP is \(\frac{1}{5}\) and its 7th term is \(\frac{6}{5}\), find the 10th term.
Answer:
5th term of HP = \(\frac{1}{5}\)
5th term of AP = 5
a + 4d = 5 ………………..(1)
Seventh term of HP = \(\frac{6}{5}\)
Seventh term of AP = \(\frac{5}{6}\)
a + 6d = \(\frac{5}{6}\) ………….(2)

= –\(\frac{25}{6}\)
d = –\(\frac{25}{12}\)

a + 4d = 5 ⇒ a + 4\(\left(-\frac{25}{12}\right)\) = 5 ⇒ a – \(\frac{25}{3}\) = 5
a = 5 + \(\frac{25}{3}=\frac{15+25}{3}=\frac{40}{3}\)
Tenth term of AP = a + 9d
= \(\frac{40}{3}+9\left(\frac{-25}{12}\right)=\frac{40}{3}+\frac{3(-25)}{4}=\frac{160-225}{12}=\frac{-65}{12}\)
Tenth term of the corresponding HP = –\(\frac{-12}{65}\) .

Question 17.
Nine tables and eight chairs cost Rs. 456. Eight tables and nine chairs cost Rs. 462. Find the cost of each chair and each table.
Answer:

9x + 8 (30) = 456
9x + 240 = 456
9x = 456 – 240
9x = 216
x = 24.
∴ Cost of 1 table = Rs. 24
Cost of 1 chair = Rs. 30.

Question 18.
Solve: \(\frac{x-5}{10}+\frac{x+5}{5}\) = 5
Answer:
\(\frac{x-5}{10}+\frac{x+5}{5}\) = 5
x – 5 + 2x + 10 = 50
3x + 5 = 50
3x + 45
x= 15

Question 19.
Solve the following system of linear inequalities. 2 (x – 1) < x + 5 and 3 (x + 2) > 2 – x.
Answer:
2x – 2 < x + 5 and
2x – x < 5 + 2
x < 7……………(1)

3 (x + 2) > 2 – x
4x > 2 – 6
4x > – 4
x > \(\frac{-4}{45}\)
∴ x > -1 ……………(2)
From (1) and (2)
-1 < x < 7.

Question 20.
The average age of Ashok and Abdul is 45 years, the average age of Abdul and Anthony is 50years and the average age of Anthony and Ashok is 35 years. Find the ages of Ashok and Anthony.
Answer:
Age of Ashok = x
Age of Abdul =y
Age of Anthony = z
Given,
\(\frac{x+y}{2}\) = 45 ⇒ x + y = 90 ………………….(1)
\(\frac{y+z}{2}\) = 50 ⇒ y + z = 100 …………………(2)
\(\frac{z+x}{2}\) = 35 ⇒ z + x = 70 2 ………………….(3)

Add (1), (2), (3)
2x + 2y + 2z = 260
x + y – z = 130
x+ 100 = 130
x = 30

x + y = 90
30 + y = 90
y = 60
z = 100

60 + z = 100
z = 40

Age of Ashok = 30 years.
Age of Abdul = 60 years.
Age of Anthony = 40 years.

Question 21.
Prove that cos 189° + cos 9° = 0.
Answer:
Cos (180 + 9) + cos 9 = – cos 9 + cos 9 = 0.

Question 22.
Three consecutive vertices of a parallelogram are (3, 0), (5, 2) and (-2, 6). Find the fourth vertex.
Answer:
A = (3, 0) B = (5, 2) C = (-2, 6) D = (x, y)
Mid point of AC = mid point of BD
\(\)
3 – 2 = 5 + x
1 = 5 + x
x = -4

0 + 6 = 2 + y
y = 4
D = (- 4, 4).

Question 23.
Find the equation to the locus of a point moving in a plane that is equidistant from (- 1, – 1) and (4, 2).
Answer:
Let P(x, y) be a point which is equidistant from A (-1, -1) and B (4, 2)
Then |PA| = |PB|
Squaring PA2 = PB2
(x + 1)² + (y + 1)² = (x – 4)² + (y – 2)²
i,e. x² + 2x + 1 + y² + 2y + 1 = x² – 8x + 16 + y² – 4y + 4
i.e. 2x + 2y + 2 = – 8x- 4y + 20
i.e. 5x + 3y = 9
Hence the locus of P is the straight-line 5x + 3y = 9.

Question 24.
Find the equation of a line perpendicular to 3x – 2y + 1 = 0 and passing through the point ( 1, -2).
Answer:
Slope of the given line 3x – 2y + 1 = 0 is \(\frac{3}{2}\) = m₁
Since the two lines are perpendicular m₁ m₂ = – 1
m₂ = \(\frac{-1}{m_{1}}=\frac{-1}{3 / 2}=\frac{-2}{3}\)
Thus the equation of the line with slope m = \(\frac{-2}{3}\) and passing through
(x₁, y₁) = (1, -2) is given by
y – y₁ = m(x – x₁)
y + 2 = \(\frac{-2}{3}\)(x – 1)
⇒ 3y + 6 = -2x + 2
⇒ 2x + 3y + 4 = 0.

Question 25.
If the lines x – 2y + 1 = 0, 2x – 5y + 3 = 0 and 5x – 9y + k = 0 are concurrent, find k.
Answer:

y = 1
2x – 4 + 2 = 0 ⇒ 2x – 2 = 0 ⇒ x = 1 ,
5 (1) – 9 (1) + k= 0
– 4 + k = 0
k = 4
Point of concurrency = (1, 1).

PART-C

III. Answer any TEN questions: (10 × 3 = 20)

Question 26.
Three bells toll at intervals of 30 seconds, 40 seconds, and 50 seconds respectively. They start together. After how many minutes will the next bell toll together?
Answer:
Find the LCM of 30, 40 and 50

LCM = 30 × 20 = 600
Hence the bells will fall together after 600 sec.
Or \(\frac{600}{60}\) = 10 minutes.

Question 27.
If f(x) = x and g (x) = x³ + 1, find fog (1), gof (1) and fof (2) .
Answer:
(fog)(1) = f[g(1)] = f[2] = 2
(gof)(1) = g[f(1)] = g[1] = 1 + 1 = 2
(fof)(2) = f[f(2)] = f(2) = 2.

Question 28.
If 2^1/a = 3^1/b = 54^1/c show that a + 3b = c.
Answer:
∴ 2^1/a = k ⇒ 2 = k^a
3^1/b = k ⇒ 3 = k^b
54^1/c = k ⇒ 54 = k^c
Now 2 × 3³ = 54
k^a.(k^b)³ = k^c ⇒ k^ak^3b = k^c
∴ k^a+3b = k^c ⇒ a + 3b = c.

Question 29.
Prove : log_y x². log_z y³. log_x z⁵ = 30.
Answer:
LHS = 2 log_y x. 3log_z y. 5log_x z

Question 30.
30. Find three numbers in GP. whose sum is 39 and their product is 729.
Let the three numbers be \(\frac{a}{r}\), a, ar
Product = 729
⇒ a³ = 729 ⇒ a³ = 91 ⇒ a = 9
Sum = 39
\(\frac{a}{r}\) + a + ar = 39
\(\frac{a}{r}\) + 9 + 9r = 39
\(\frac{a}{r}\) + 9r = 39 – 9
\(\frac{a}{r}\) + 9r = 30 ⇒ 9 + 9r² = 30r ⇒ 9r² – 30r + 9 = 0
⇒ 9r² – 27r – 3r + 9 = 0
9r(r – 3) – 3(r – 3) = 0
(r – 3)(9r – 3) = 0
r = 3 or 9r = 3
r = 3 or r = \(\frac{3}{9}=\frac{1}{3}\)
r = 3 or r = \(\frac{1}{3}\)

The numbers are \(\frac{a}{r}\),a,ar
\(\frac{9}{3}\), 9, 9(3)
3, 9, 27.

Question 31.
If α and β are the roots of the equation 2x – 5x + 7 = 0, then evaluate \(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}\)
Answer:
\(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{a}=\frac{\alpha^{3}+\beta^{3}}{a \beta}\)

Question 32.
Rs. 16,000 invested at 10% p.a. interest compounded semiannually amounts to Rs. 18,522. Find the period of investment.
Answer:
P = Rs. 16,000
A = Rs. 18,522
R = 10%
q = 2
r = \(\left(1+\frac{\mathrm{R}}{q}\right)^{q}\) – 1 = \(\left(1+\frac{0.10}{2}\right)^{2}\) – 1 = (1 + 0.05)² – 1 = (1.05)² – 1 = 1.1025 – 1
= 0.1025 = 10.25%
A = P (1 + r)ⁿ
18522 = 16000(1 + 0.1025)ⁿ
\(\frac{18522}{16000}\) = (1.10025)ⁿ
1.157625 = (1.1025)ⁿ
log (1.157625) = n log 1.1025
n = \(\frac{\log (1.157625)}{\log 1.1025}=\frac{0.0635}{0.0423}\) = 1.501 ≈ 1.5
∴ n = 1.5 years.

Question 33.
The rate of a movie ticket was Rs. 150. This was reduced by 20%, due to which the revenue increased by 20%.
What was the percentage increase in the number of viewers?
Answer:
Let the no. of initial viewers be 100 Cost of each ticket = Rs. 150
Revenue generated initially = 150 × 100 = Rs. 15000
After the reduction in price cost of each ticket
= 150 – 20% of 150 = 150 – \(\frac{20}{100}\) × 150 = 150 – 30 = Rs.120
New revenue generated = 15000 + 20% of 15000 20
= 15000 + \(\frac{20}{100}\) × 15000 = 15000 + 3000 = Rs. 18000 100
No. of viewers = \(\frac{\text { Total revenue }}{\text { Cost of each ticket }}=\frac{18000}{120}\) = 150
Percentage increase in the no. of viewers = 150 – 100 = 50%.

Question 34.
For the first year of production, the fixed cost for setting up a new company is Rs. 3,00,000 and the variable cost is 70. The company expects revenue from sales to be Rs. 270 per unit of the product. Find the revenue function, cost function, and the break-even output.
(i) Construct the revenue function.
(ii) Construct the cost function.
(iii) Find the Break-Even output
(iv) Find the number of Calculator produced for which the company will suffere loss.
Answer:
C (x) = 70x + 3,00,000,
R(x)=270x
P(x) = R(x) – C(x)
270 x = (70x + 3,00,000) = 200x – 3,00,000

For BEP P(x)= 0
R(x) = C(x)
270x = 70x + 3,00,000, 20x = 3,00,000
x = 1500 BEP
less than x = 1500 units the company will suffer loss.

Question 35.
The angles of a triangle are in the ratio 2: 3: 5. Find them in degrees and radians.
Answer:
2 : 3 : 5 or 2x, 3x, 5x.
2x + 3x + 5x – 180
10x = 180 ⇒ x – 18°
2x = 2 (18) = 36° 36 × \(\frac{\pi}{180}=\frac{\pi^{c}}{5}\)
3x = 3 (18) = 54°
= \(\frac{54 \pi}{180}=\frac{3 \pi}{10}\)
5x = 5(18) = 90° = \(\frac{\pi}{2}\).

Question 36.
Find ‘x’ given : \(\frac{x \cdot \sin ^{2} 300^{\circ} \sec ^{2} 240^{\circ}}{\cos ^{2} 225^{\circ} \cdot \cos e c^{2} 240^{\circ}}\) = cot 315.tan2 300°.
Answer:
sin300 = sin(360 – 60) = -sin60 = \(\frac{-\sqrt{3}}{2}\)
sec 240 = sec (180 + 60) = – sec 60° = – 2
cos225 = cos(180 + 45°) = -cos45° = –\(\frac{1}{\sqrt{2}}\)
cosec 240 = cosec (180 + 600) = cosec 60 = –\(\frac{2}{\sqrt{3}}\)
cot 315° = cot(360 – 45°) – cot45° = -1
tan 300 = tan (360 – 600) = – tan 600 = –\(\sqrt{3}\)
∴ The given equation becomes:
\(\frac{x \cdot \frac{3}{4} \cdot 4}{\frac{1}{2} \cdot \frac{4}{3}}\) = 1.3
⇒ x = \(\frac{6}{9}=\frac{2}{3}\)

Question 37.
Show that the points A(2, 4), B(2, 6) and C(2 + \(\sqrt{3}\), 5) form the vertices of an equilateral triangle.
Answer:
AB = \(\sqrt{(2-2)^{2}+(6-4)^{2}}=\sqrt{0+4}=\sqrt{4}\) = 2
BC = \(\sqrt{(2+\sqrt{3}-2)^{2}+(5-6)^{2}}=\sqrt{3+1}=\sqrt{4}\) = 2
CA = \(\sqrt{(2+\sqrt{3}-2)^{2}+(5-4)^{2}}=\sqrt{3+1}=\sqrt{4}\) = 2
AB = BC = CA
∴ A, B, C form an equilateral triangle.

Question 38.
Derive the equation of a straight line in slope-intercept form.
Answer:

Slope of AP = m = tan θ = \(\frac{P N}{\mathrm{AN}}=\frac{P M-M N}{O M}\)
\(\frac{\mathrm{PM}-\mathrm{OA}}{\mathrm{OM}}=\frac{\mathrm{Y}-\mathrm{C}}{x}\)
∴ y – c = mx ⇒ y = mx + c.

PART -D
III. Answer any SIX questions. (6 × 5 = 30)

Question 39.
In a certain college with 500 students, 300 take the milk and 251 take tea. Assuming that every student takes at least one of the two, find how many take.
(a) both milk and tea
(b) milk only
(c) tea only
Answer:
n (M ∪ T) = 500
n (M) = 300
n (T) = 250
n (M ∪ T) = n (M) + n (T) – n (M ∩ T)
n (M ∩ T) = n (M) + n (T) – n (M ∪ T) = 300 + 250 – 500
n (M ∩ T) = 50
n (M – T) = n (M) – n (M ∩ T) = 300 – 50 = 250
n(T – M) = n(T) – n(M ∩ T) = 250 – 50 = 200.

Question 40.
Evaluate using log tables: \(\frac{25.36 \times 0.4569}{847.5}\)
Answer:
Let x = \(\frac{25.36 \times 0.4569}{847.5}\)
log x = log 25.36 + log 0.4569 – log 847.5
= 1.4041 + 1.6599 – 2.9282
= 1.4041 – 1 + 0.6599 – 2.9282
= – 1.8642
= – 1 – 0.8642 = – 1 + 1 -0. 8642
= -2 + 0.1358
= 2.1358
= 01368

Question 41.
Find the sum to ‘n’ terms of the series 4 + 44 + 444 +……………..
Answer:
S_n = 4 + 44 + 444 + …………………….. to n terms
S_n = 4(1 + 11 + 111 + …………………….. to n terms)
\(\frac{9 \mathrm{~S}_{n}}{4}\) = 9 + 99 + 999 + ……………. to n terms
\(\frac{9 \mathrm{~S}_{n}}{4}\) = (10 – 1) + (10² – 1) + (10³ – 1) +……………………
= (10 + 10² + 10³ + ………………. ) – (1 + 1 + 1 +………………… )
\(\frac{9 S_{n}}{4}=\frac{10\left(10^{n}-1\right)}{10-1}\) – n
S_n = \(\frac{4}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]\)

Question 42.
Solve the cubic equation x³ + 6x² + 9x + 4 = 0 using synthetic division, given that there is an integral root between – 3 and 3.
(1) x³ + 6x² + 9x + 4 = 0
Answer:
Put x = – 1
(-1)³ + 6(-1)² + 9 (-1) + 4
= -1 + 6 – 9 + 4
= 10 – 10 = 0
∴ x = – 1 is a root of the equation
∴ x + 1 is the factor (factor theorem)

Quotient = x² + 5x + 4
remainder = 0
x³ + 6x² + 9x + 4 = 0.
(x + 1)(x² + 5x + 4) = 0
x + 1 = 0, x² + 5x + 4 = 0
x² + 4x + x + 4 = 0
x(x + 4) + 1 (x + 4) = 0
(x + 4)(x + 1) = 0
x = -4, x = -1
The roots are -1, -1, -4.

Question 43.
If the difference between simple interest and compound interest for 3 years at 2.5% p.a. is Rs. 625, find the sum invested.
Answer:
Let the sum invested = P
r = 2.5% p.a. = 0.025 p.a.
n = 3 years
A = P(1 +r)ⁿ = P(1 +0.025)³ = P (1.025)³
CI = A – P = P (1.025)³ – P = P [(1.025)3 – 1] = P (0.07689)
SI
r = 2.5
T = 3
SI = \(\frac{P \times T \times R}{100}=\frac{P \times 3 \times 2.5}{100}\) = 0.075P
∴ CI – SI = 625
0.07689P – 0.075 P = 625
∴ 0.00189 P = 625
P = 3,30687.83
∴ P = Rs. 3,30,687.83.

Question 44.
How much should you invest if you want to receive Rs.5000 at the beginning of each year for the next 5 years and the Cl is 16% p.a. compounded quarterly.
Answer:
A = Rs. 5,000 n = 5 years R = 0.16 q = 4
r = \(\left(1+\frac{\mathrm{R}}{q}\right)^{q}\) – 1 = \(\left(1+\frac{0.16}{4}\right)^{4}\) – 1 = 0.1698 = 16.98
Effective rate of interest = 16.98%
P = \(\frac{A\left[(1+r)^{n}-1\right]}{r(1+r)^{n}}\)(1 + r)
= \(\frac{5000\left[(1+0.1698)^{5}-1\right]}{0: 1698(1.1698)^{5}}\)(1.1698) = \(\frac{6963}{0.3719}\) = 18,722.77

Question 45.
Ram Singh purchased 2 camels for Rs. 18,000 and Rs. 15,000 respectively. He sold them at a loss of 15% and a gain of 19% respectively. Find the selling price of each camel. Also, find the overall loss or gain percent in this transaction.
Answer:
C.P of camel = 18000
S.P while sold at loss of 15% = \(\frac{85}{100}\) × 18000 = 15300
C.P. of second camel = 15000
S.P. while sold at gain of 19% = \(\frac{119}{100}\) × 15000 = 17,850
Total CP = 18000 + 15000 = 33000
Total S.P. = 15,300 + 17,850 = 33150
∴ Profit = 33,150 – 33000 = 150
Profit % = \(\frac{150}{33000}\) × 100 = 0.45%.

Question 46.
Prove: \(\frac{\tan \theta}{\sec \theta-1}+\frac{\tan \theta}{\sec \theta+1}\) = 2 cosec θ
Answer:

Question 47.
Find the area of the quadrilateral whose vertices are A (1, 2), B (6, 2), C (5, 3) and D (3,4).
Answer:
(1,2) (6,2) (5, 3) (3,4)
∆ ABC = 1/2 [1 (2 – 3) + 6(3 – 2) + 5 (2 – 2)]
= 1/2 [1 (-1) + 6 (1) + 5 (0)] = 1/2 [-1+6 + 0] = \(\frac{5}{2}\)
ACD = 1/2 [1 (3 – 4) + 5 (4 – 2) + 3 (2 – 3)]
= 1/2 [1 (-1) + 5 (2) + 3 (-1 )] = 1/2 [-1 + 10 – 3] = 1/2 [10-4 ] = 1/2 (6) = 3
Area of the quadrilateral ABCD
= |Area of ∆ABC| + | Area of ∆ACD| = \(\left|\frac{5}{2}\right|\) + |3| = \(\frac{5}{2}\) + 3 = \(\frac{5+6}{2}\) = \(\frac{11}{2}\)sq. unit.

Question 48.
Find the equations of lines passing through the point (3, 4) and having intercepts on the axes such that their sum is 14.
Answer:
Let the x and y intercepts of the line be a and b respectively so that a + b = 14 or b = 14 – a
Thus the equation of the line in the intercept form is
\(\frac{x}{a}+\frac{y}{14-a}\) = 1
But it passes through the point(3,4).
therefore \(\frac{3}{a}+\frac{4}{14-a}\) – 1
3 (14 – a) + 4a = a (14 – a)
i.e., a² – 13a + 42 = 0
i.e., (a – 6) (a – 7) = 0
or a = 6 or a = 7
If a = 6 then b = 8; If a = 7 then 6 = 7

Therefore equations of the lines are \(\frac{x}{6}+\frac{y}{8}\) = 1 or \(\frac{x}{7}+\frac{y}{7}\) = 1
i.e., 4x + 3y – 24 = 0 or x + y – 7 = 0.

PART-E

III. Answer any ONE question. (1 × 1 = 10)

Question 49.
(a) If A {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8}, then verify:
(i) A ∪ (B ∩ C) = ( A ∪ B) ∩ ( A ∪ C)
(ii) A∩(B∪C) = (A∩B)∪(A∩C)
Answer:
(i)Let A = {1, 2, 3, 4}
B = {3, 4, 5, 6}
C = {4, 5, 6, 7, 8}
B∩C = {4,5,6}

A∪(B∩C) = {1, 2, 3, 4} ∪ {4, 5, 6}
= {1,2, 3, 4, 5, 6}

A∪B = {1, 2, 3, 4, 5, 6}
A∪C = {1, 2, 3, 4, 5, 6, 7, 8}
(A∪B) ∩ (A∪C)
= {1, 2, 3, 4, 5,6} ∩ {1, 2, 3,4, 5, 6, 7, 8}
= {1, 2, 3, 4, 5, 6}
A∪(B∩C) = (A∪B) ∩ (A∪C)

(ii) B∪C = {3, 4, 5, 6, 7, 8}
A∩(B∪C)= {1, 2, 3, 4} ∩ {3, 4, 5, 6, 7, 8}
= {3, 4}
A∩B = {3, 4}
A∩C = {4}
(A∩B)∪(A∩C) = {3, 4}
Thus we have
A∩(B∪C) = (A∩B) ∪ (A∩C).

(b) Find the future value of an annuity of Rs. 5000 at 12% p.a. for 6 years.
Answer:
A = Rs. 5,000 r = 0.12 n = 6
F = \(\frac{\mathrm{A}\left[(1+r)^{n}\right]}{r}=\frac{5000\left[(1+0.12)^{6}-1\right]}{0.12}=\frac{4869.11}{0.12}\) =40,575.9

(c) The average age of 12 boys is 8 years. Another boy of 21 years joins the group. Find the average age of the new group.
Answer:
Average age of 12 boys = 8 years
∴ Total age of boys = 8 × 12 = 96 years
when a boy of 21 yrs joins the group,
the total age of 13 boys = 96 + 21 = 117 years
∴ Average age of the new group = \(\frac{117}{13}\) = 9 years.

50. (a) Solve the linear inequalities graphically :
3x + 3y ≤ 6,
x + 4y ≤ 4,
x ≥ 0,y ≥ 0.
Answer:

(b) A watch manufacturer finds that his total cost function is linear. The total cost for 100 watches is Rs. 20,000 and for 200 watches, the total cost increased to Rs. 30,000. What are the fixed cost and variable cost per unit?
Answer:
Total cost = T.V.C. + T.F.C.
C (x) = ax + b
When x = 100 watches
T.C. = a (100) + b
i.e.20,000 = 100a+ 6 …………………. (1)

Similarly when x = 200 watches
30.0 = 200 a + b ……………….(2)
(2) – (1)30,000 = 200 a + b
-20,000 = 100 a + b
10,000 = 100 a
a = \(\frac{10,000}{100}\) = 100 substitute in (1)
20,000= 100 (100) + b
20,000 = 10,000 + b
b =10,000

∴ When x = 150, TC = (100) (150)+ 10,000
= 15,000+10,000,
TC = ₹ 25,000.

(c) The HCF of 2 numbers is 16 and their LCM is 160. If one of the numbers is 32, then find the other num ber.
Answer:
HCF =16 LCM =160 a = 32 b = ?
a × b = H × L
b = \(\frac{\mathrm{H} \times \mathrm{L}}{a}=\frac{16 \times 160}{32}=\frac{2560}{32}\) = 80
∴ The other number is 80.

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2015 (North), Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2015 (North)

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any TEN questions. (10 × 1 = 10)

Question 1.
Write the canonical representation of 156.
Answer:
2² × 3¹ × 13¹

Question 2.
If A = {1, 2, 3, 4, 5} and relation R on A is defined as R = {(x, y) : x = y}, find R.
Answer:
R = {(1, 1), (2,2), (3, 3), (4,4), (5, 5)}

Question 3.
If f is a function defined by f(x) = 3x + 5, find f(3).
Answer:
f(3) = 3 × 3 + 5 = 14

Question 4.
Simplify : a^x+y a^2x-y
Answer:
a^x+y.a^2x-y = a^x+y+2x-y = a^3x

Question 5.
Express 3^-2 = \(\frac{1}{9}\) in logarithmic form.
Answer:
-2 = log₃(\(\frac{1}{9}\))

Question 6.
Find the 30th term of the A.P. 2, 5, 8,………………….
Answer:
T₃₀ = 2 + 29 × 3 = 2 + 87 = 89

Question 7.
Solve for x : 2(7 + x) – 10 = 16 – 2(x – 24).
Answer:
14 + 2x – 10 = 16 – 2x + 48 = 4x = 68 ⇒ x = 17

Question 8.
Find the present value of a perpetuity of ₹ 3,000 to be received forever at 4% p.a.
Answer:
P₈₀ = \(\frac{A}{r}=\frac{3000}{0.04}\) = ₹ 75,000

Question 9.
The average score of 20 boys is 60% and the average score of 30 girls is 70%. Find their combined average score.
Answer:
X̄₁₂ = \(\frac{20 \times 60+30 \times 70}{20+30}\) = 66%

Question 10.
Convert 0.32 into percentages.
Answer:
0.32 × 100 = 32.00 = 32%

Question 11.
Express \(\frac{3 \pi}{5}\) in degree measure.
Answer:
\(\frac{3 \pi}{5}=\frac{3 \times 180}{5}\) = 108

Question 12.
Find the distance of the point (5,4) from the origin.
Answer:
Distance = \(\sqrt{5^{2}+4^{2}}=\sqrt{25+16}=\sqrt{41}\)

PART-B

II. Answer any TEN questions : (10 × 2 = 20)

Question 13.
Find the sum of all positive divisors of 3⁴ × 5³ × 7².
Answer:
Sum of all positive divisors = T(n) ∴ T(n) = (4+ 1)(3 + 1)(2+ 1) = 60

Question 14.
Find the number which when divided by 16, 20, and 40 leaves the same remainder 4.
Answer:
L.C.M of the given numbers 16,20 & 40 is 80 ∴ the required number = 80 + 4 = 84

Question 15.
If A = (1, 3, 5}, B{5}, C = {7}, find A × (B – C)
Answer:
B – C = {5}
A × (B – C) = {1, 3, 5} × {5} = {(1,5) (3,5) (5,5)}

Question 16.
Simplify: \(\frac{2^{\left(2^{0}\right)}+\left(2^{0}\right)^{3}}{\left(2^{3}\right)^{2}+\left(2^{2}\right)^{0}}\)
Answer:
\(\frac{2^{\left(2^{0}\right)}+\left(2^{0}\right)^{3}}{\left(2^{3}\right)^{2}+\left(2^{2}\right)^{0}}=\frac{2^{1}+1^{3}}{8^{2}+1}=\frac{2+1}{64+1}=\frac{3}{65}\)

Question 17.
Insert 3 geometric means between -4 and 64.
Answer:
Let g₁, g₂, g₃ are G.Ms. ∴ -4. g₁, g₂, g₃, -64 are in G.P.
Here a = -4, ar⁴ = -64 ⇒ r⁴ = \(\frac{-64}{-4}\) = 16 = 24 ⇒ r = 2
the 3 G.Ms are g₁ = ar = -8; g₂ = ar² = -16; g₃ = ar³ = -32

Question 18.
Solve by formula method : 5x² – 7x – 12 = 0
Answer:
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
a = 5, b = -7, c = -12.
x = \(\frac{7 \pm \sqrt{(-7)^{2}-4(5)(-12)}}{10}=\frac{7 \pm \sqrt{49+240}}{10}=\frac{7 \pm \sqrt{289}}{10}=\frac{7 \pm \sqrt{17^{2}}}{10}=\frac{7 \pm 17}{10}\)
∴ x = \(\frac{12}{5}\), -1

Question 19.
Solve 3x – 2 < 2x + 1, x ∈ R and represent on the number line.
Answer:
3x – 2 < 2x + 1
3x – 2x < 1 + 2

x < 3

Question 20.
What sum will amount to ₹ 46,000 in 7 years at 12% p.a. SI?
Answer:
A = P(1 + \(\frac{\mathrm{TR}}{100}\))
P = \(\frac{A}{\left(1+\frac{T R}{100}\right)}=\frac{46,000}{\left(1+\frac{7 \times 12}{100}\right)}\) = ₹ 25,000

Question 21.
The average height of a group of people is 6 ft. 10 more people with an average height of 5 ft. join the group. Find the average height of the new group of 60 people.
Answer:
Total height of 50 people = 50 × 6 = 3 00
Total height of 10 people = 10 × 5 = 50
Total height of 60 people = 300 + 50 = 350
Hence new average = \(\frac{350}{60}\) = 5.83 ft.

Question 22.
While taking measurement, a tailor writes 34 instead of 24. What is the percentage error?
Answer:
Error = 34 – 24 = 10
∴ Error percent = \(\frac{10}{24}\) × 100 = 41.6%

Question 23.
Find (1 + tan²θ)(1 – sin²θ) = 1
Answer:
LHS = (1 + tan²θ)(1 – sin²θ) = sec²θ.cos²θ = \(\frac{1}{\cos ^{2} \theta}\) .cos²θ = 1 = RHS

Question 24.
Find the small value of : cos60°- sin30° – cot³345°
Answer:
cos60 – sin30 – cot³(45°) = \(\frac{1}{2}-\frac{1}{2}\) – (1)³ = -1

Question 25.
Find k if the lines 2x – 3y + 4 = 0 and x + ky = 3 are perpendicular.
Answer:
m₁ = \(\frac{-a}{b}=\frac{2}{3}\)
m₂ = \(\frac{-1}{K}\).
∵ lines are perpendicular .. m1 × m, = -1
⇒ \(\frac{2}{3} \times \frac{-1}{\mathrm{k}}\) = -1
⇒ \(\frac{2}{3} \times \frac{-1}{k}\) = -1
⇒ k = \(\frac{2}{3}\)

PART-C

III. Answer any TEN questions: (10 × 3 = 20)

Question 26.
Find the HCF of \(\frac{8}{9}, \frac{32}{81}, \frac{16}{27}\)
Answer:
H.C.F of 8, 32, 16 = 8
L.C.M of 9, 81, 27 = 81
∴ Required HCF = \(\frac{\text { HCF of N.R }}{\text { LCM of DNR }}=\frac{8}{81}\)

Question 27.
If f(x) = x² g(x) = x + 1, find the value of
(i) fog(1) (ii) gof(1) and (iii) fof(2)
Answer:
(i) fog(1) = f(g(1)) = f(1 + 1) = f(2) = 2² = 4.
(ii) gof(1) = g(f(1)) = g(1²) = g(1) = 1 + 1 = 2
(iii) fof(2) = f(f(2)) f(2²) = f(4) = 4² = 16

Question 28.
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {4, 5, 6, 7, 8}, then verify Au(BnC) = (Au B)n(AuC)
Answer:
A∪ B = {1, 2, 3,4, 5, 6}
A∪ C = {1, 2,3,4, 5, 6, 7, 8}
B∩ C= {4,5, 6}
A∪(B∩C) = [1,2,3,4,5,6] = LHS
(A∪B)∩(A∪C) = {1,2,3,4,5,6} = RHS
∴ LHS = RHS

Question 29.
If x = log, 9, y = log, 7, z = log, 4 then show that xyz = 2.
Answer:
LHS = xyz = \(\frac{\log 9}{\log 2} \times \frac{\log 7}{\log 9} \times \frac{\log 4}{\log 7}=\frac{\log 2^{2}}{\log 2}=2 \frac{\log 2}{\log 2}\) = 2 = RHS

Question 30.
The sum of 3 numbers in an AP is 15 and their product is 105. Find the numbers.
Answer:
Let the 3 numbers be (a – b), a, (a + d)
Given sum =15
⇒ a – d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5
Product =105
⇒ (a-d)(a)(a + d) = 105
⇒ d = ±2
∴ The three numbers are 3, 5, 7

Question 31.
Solve x³ + 6x² + 9x + 4 = 0 using synthetic division, given that it has an integral root between -3 and 3.
Answer:
x = -1 is a root of the equation x³ + 6x² + 9x + 4 = 0

Using synthetic division and getting the reduced equation as x² + 5x + 4 = 0 solving we get
x = -1 & x = -4

Question 32.
Solve the system of linear inequalities graphically : x + y < 6 and x + y > 4.
Answer:
Consider x + y = 6, x + y = 4
x = 0, y = 6 ⇒ (0,6) (0,4)
y = 0, x = 6 ⇒ (6,0) (4,0)

Question 33.
The cost of a refrigerator is ₹ 27,000. If it depreciates at the rate of 8% p.a., find its value after 4 years.
Answer:
A = 27,000(1 -0.08)4 = 27,000(0.02)4= ₹19,342.61
Refrigerators value after 4 years = 19,342.61

Question 34.
10 years ago, the average age of a family of 4 was 24 years. 2 children have been born and the average age remains the same. What are the present ages of the 2 children if their ages differ by 2 years?
Answer:
Let the total age of 4 members = 96 = 4 × 24
After 10 years total age of 4 = 96 + 40 = 136
Let Ages of 2 children x & x + 2.
∴ Total age of 6 = 136 + x + x + 2 = 138 + 2x
Given average age = 24 ⇒ \(\frac{138+2 x}{6}\) = 24 ⇒ x = 3
∴ Ages of 2 children = 3 & 5 years

Question 35.
Ritu’s salary was increased by 10% and then again by 5%. If the present salary is ₹ 9,240, then what was her previous salary?
Answer:
Let the original salary – x
With 10% increase, salary = x + 0.1x = 1.1x
with 5% increase, salary = 1.155x
1.155x = 9240
⇒ x = \(\frac{9240}{1.155}\) = ₹ 8,000

Question 36.
If cosecθ = and 90° < θ < 1800, then prove that \(\frac{4 \sin \theta-7 \cos \theta}{3 \sin \theta+2 \cos \theta}\) = 40
Answer:
sinθ = \(\frac{3}{5}\) cosθ = \(\frac{-4}{5}\)
(∵ in II quadrant cosθ is negative)
LHS = \(\frac{4 \sin \theta-7 \cos \theta}{3 \sin \theta+2 \cos \theta}=\frac{4 \cdot \frac{3}{5}-7\left(-\frac{4}{5}\right)}{3 \cdot \frac{3}{5}+2 \cdot\left(\frac{-4}{5}\right)}=\frac{\frac{12}{2}+\frac{28}{5}}{\frac{9}{5}-\frac{8}{5}}=\frac{\frac{40}{5}}{\frac{1}{5}}\) = 40 = RHS

Question 37.
3 consecutive vertices of a parallelogram are A (3, 0), B(5, 2), and C (-2, 6). Find the fourth vertex.
Answer:
Let D(x, y) be the fourth vertex
w.k. that in a parallelogram diagonal bisect each other
∴ midpoints are equal.

\(\left(\frac{x+5}{2,}, \frac{y+2}{2}\right)=\left(\frac{3-2}{2}, \frac{0+6}{2}\right)\)
⇒ x + 5 = 1, y + 2 = 6
⇒ x = -4, y = 4
∴ the fourth vertex is D(-4, 4)

Question 38.
Derive the slope-intercept form of a straight line.
Answer:
Required equation is y = mx + c

PART-D

IV. Answer any SIX questions : (6 × 5 = 30)

Question 39.
In a class of 50 students, 15 do not participate in any games, 25 play cricket, and 20 play football. Find the number of students who play both. Represent the results using the Venn diagram.
Answer:

n(A∪B) = 50-15 = 35
n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
n(A∩B) = 25 + 20-35 = 10

Question 40.
Evaluate using log tables : \(\frac{5.634 \times 25.64}{12.72}\)
Answer:
logx = log\(\left(\frac{5.634 \times 25.64}{12.72}\right)\)
= log 5.634 + log 25.64 – log12.72
logx = 0.7508+ 1.4089-1.1044
x = A.L (1.0553)
x = 11.36

Question 41.
Find the sum of all integers between 60 and 400 which are divisble by 13.
Answer:
S = 65, 78, 91 ……………………390
a = 65, d = 13, T_n = 390, n = ?, S =?
T_n = a + (n – 1)d
390 = 65 + (n- 1)13 ⇒ n = 26
S_n = \(\frac{n}{2}\)(a + l) = \(\frac{26}{2}\)(65 + 390) ⇒ S_n = 5915

Question 42.
A certain two-digit number is 4 times the sum of the digits. If 18 is added to the number, the digits get interchanged. Find the number.
Answer:
Let x be in 10th place, y in units place
∴ the number = 10x + y
Given 10x + y = 4 (x + y)
⇒ y = 2x…………………(1)
Also 10x + y + 18 = 10y + x
⇒ x + 2 = y……………….(2)
Solving (1) and (2) we get
2x = x + 2 ⇒ x = 2
∴ x = 2 & y = 4
Hence required number = 24.

Question 43.
Find the Cl on ₹ 7500 at 14% for 14 years while interest is calculated half-yearly.
Answer:
Effective rate of interest = r = \(\left(1+\frac{0.14}{2}\right)^{2}\) – 1 = 15%
∴ C.I = A-P = 7500(1 + 0.15)4 – 7500 = ₹5617.5

Question 44.
Find the future value of an annuity of ₹ 5000 12% p.a for 6 years.
Answer:
Future value of an annuity = F = \(\frac{A\left((1+i)^{n}-1\right)}{i}\)
F = \(\frac{5000\left((1+0.12)^{6}-1\right)}{0.12}\) = ₹ 40,575.94

Question 45.
A radio is sold at a profit of 25% cost price and selling price are both increased by 100. If the new profit is at the rate of 20%, find the original cost of the radio, i.
Answer:
Let the original cost price be x
I: Profit = 25% SP = 1.25x
II: C.P. = 100 + x then SP= 1.25x + 100 ………(1)
Profit = 20%
S.P = (100 + x) + 0.2(100 + x) ……………..(2)
From 1 & 2 we get
1.25x + 100 = 100 + x + 20 + 0.2x
Solving we get x = ₹ 400 = original cost of the radio.

Question 46.
Prove that : \(\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}\) = 1 + secA.cosecA
Answer:
\(\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=\frac{\tan A}{1-\frac{1}{\tan A}}+\frac{\frac{1}{\tan A}}{1-\tan A}\)

= RHS

Question 47.
Find the area of a quadrilateral whose vertices are (1,2), (6,2), (5,3), and (3,4) taken in as order.
Answer:
Area of a quadrilateral = Area of ∆ABC + Area of ∆ACD

Area of ∆ABC = \(\frac{1}{2}\)Σx₁ (y₂ – y₃) = \(\frac{5}{2}\) sq. units
Area of ∆ACD = 3 units
∴ Area of ∆BCD = \(\frac{5}{2}\) + 3 = \(\frac{11}{2}\)sq. units

Question 48.
Find the equation of a line through he intersection of the lines x – y + 2 = 0, 3x + 2y = 9 and parallel to the line 2x + y – 3.
Answer:
The point of intersection of the lines x -y + 2 = 0, 3x + 2y = 9 is (1, 3)
Any line parallel to the given line 2x + y = 3 is of the form 2x + y + k – 0
But this line passes through the point (1, 3)
2(1) + 3 + k = 0
∴ k = -5
Required equation of the line is 2x + y – 5 = 0

PART-E

V. Answer any ONE question: (6 × 5 = 30)

Question 49.
(a) Find the equation to the locus of a point moving in a plane such that it is equidistant from (-1, 1) and (4, 2). 4
Answer:
Let P(x, y) be the point on the locus
Given PA = PB where A = (-1, 1) B = (4, 2)
PA² = PB²
(x + 1)² + (y – 1)² = (x – 4)² + (y – 2)²
x² + 2x +1 + y² – 2y + 1 = x² +16 – 8x + y² + 4 – 4y
10x – 6y = 18 or 5x – 3y = 9 is the required equation of the locus.

(b) Solve for x if:
x = sec30°, tan60° + sin45°cosec45° + cos30°.cot60°.
x = \(\frac{2}{\sqrt{3}} \cdot \sqrt{3}+\frac{1}{\sqrt{2}} \sqrt{2}+\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}}\)
x = 2 + 1 + \(\frac{1}{2}\) = 3 + \(\frac{1}{2}\)
x = \(\frac{6+1}{2}=\frac{7}{2}\) ⇒ x = \(\frac{7}{2}\)

(c) Find the number of digits in the integral part of 3²⁰. (2M)
Answer:
Consider log3²⁰ = 20 log 3 = 20 × 0.4771 = 9.542
Since the characteristic is 9, there are 10 digits in 3²⁰.

Question 50.
(a) A firm finds that the production cost directly attributed to each product is 25 and fixed costs are 10,000. If each product can be sold for 35, find the
(i) cost function (ii) revenue function
(iii) profit function (iv) break-even level of output.
Answer:
V.C. = ₹ 125 , F.C. = ₹ 10,000, price = ₹ 35
(i) cost function = C(x) = 10,000 + 25x
(ii) Revenue function = R(x) = P × x = 35x
(iii) Profit function = TR – TC = 35x – 10,000 – 25 = 10x – 10,000.

At B.E. level of output T.R. = T.C.
35x = 10,000 + 25x
10x = 10,000
⇒ x = 10,000 units

(b) Find the sum to n terms of the series 5 + 55 + 555 + ……………………
Answer:
Let S = 5 + 55 + 555 +…………… to n terms
\(\frac{\mathrm{S}}{5}\) = 1+ 11 + 111+………..to n terms
\(\frac{9 \mathrm{~S}}{5}\) = 9 + 99 + 999 +…………… to n terms
=(10 – 1) + (100 – 1) + (1000 – 1) +……….. to n terms
= (10 + 100 + 1000 +………… to n terms) – (1 + 1 + 1………. to n ternis)
= \(\frac{10\left(10^{n}-1\right)}{9}\) – 1 × n
S = \(\frac{5}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]\)

(c) The HCF of two numbers is 16 and their LCM is 160. If each of the numbers is 64, then find the other number.
Answer:
Given H = 16, L = 160, a = 64, b = ?
H × L = a × b
b = \(\frac{H \times L}{a}=\frac{16 \times 160}{64}\) =40
∴ the other number is 40.

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2015 (South), Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2015 (South)

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any TEN questions. (10 × 1 = 10)

Question 1.
Write the canonical representation of 360.
Answer:
360 = 2³ × 3² × 5¹

Question 2.
If A= {1, 2, 3, 4} and B = {3, 4, 6} find B -A.
Answer:
B – A = {6}

Question 3.
If f:R→ R is defined by f(x) = x² + 2x – 5. Find f(-1).
Answer:
f (-1) = (-1)² + 2(-1) – 5 = 1 – 2 – 5 = 1 – 7 = – 6

Question 4.
Which is greater?
\(\frac{1}{\left(5^{2}\right)^{3}}\) Or \(\frac{1}{5^{\left(2^{3}\right)}}\)
Answer:
\(\frac{1}{\left(5^{2}\right)^{3}}=\frac{1}{5^{6}}\); \(\frac{1}{5^{2^{3}}}=\frac{1}{5^{8}}\)
\(\frac{1}{\left(5^{2}\right)^{3}}\) is greater than \(\frac{1}{5^{2^{3}}}\)

Question 5.
Evaluate : log_\(\sqrt{3}\)27.
Answer:
Let x = log\(\sqrt{3}\)27
(\(\sqrt{3}\))x = 27 ⇒ 37 = 33 ⇒ \(\frac{x}{2}\) = 3 ⇒ x = 6

Question 6.
Find the sum to infinity of the series :
3, -1, \(\frac{1}{3}\), –\(\frac{1}{9}\) …………………….
Answer:
Sum to infinity = S_∞ = \(\frac{a}{1-r}=\frac{3}{1+\frac{1}{3}}=\frac{3}{\frac{3+1}{3}}=\frac{9}{4}\)

Question 7.
Solve:
Answer:
2(x + 2) = 5(x – 1)
2x + 4 = 5x – 5
4 + 5 = 5x – 2x ⇒ 3x = 9 ⇒ x = 3

Question 8.
Convert 0.12 into a percentage.
Answer:
0.12 × 100= 12%

Question 9.
Define perpetuity.
Answer:
Annuity for ever is called perpetuity
Ex: Endowment prize.

Question 10.
Convert \(\frac{7 \pi}{12}\) into degrees.
Answer:
\(\frac{7 \pi^{\mathrm{c}}}{12}=\frac{7 \times 180}{12}\) = 105°

Question 11.
The average age of 10 boys in a class is 13 years. What is the sum of their ages?
Answer:
X̄ = \(\frac{\sum \mathrm{x}}{\mathrm{n}}\)
13 = \(\frac{\sum \mathrm{X}}{10}\) ⇒ ΣX = 130 years

Question 12.
If the slope of the line is –\(\frac{2}{3}\)find the slope of the perpendicular line to it.
Answer:
Slope of the perpendicular line is \(\frac{3}{2}\)

PART-B

II. Answer any TEN questions: (10 × 2 = 20)

Question 13.
Find the least number which when divided by 6, 10, 15, and 30 leaves no remainder.
Answer:
The least number is nothing but LCM of 6, 10, 15 & 30.

L.C.M = 6 × 5 × 2 × 3 = 180

Question 14.
If X and Y are two sets such that X has 35 elements, (XuY) has 45 elements. How many elements does Y has?
Answer:
n(X) = 35, n(X∪Y) =45, n(Y) = ?
n(X∩Y) = 10
n(X∪Y) = n(X) + n(Y) – n(X∩Y)
45 = 35 + n(Y) – 10
n(Y) = 45 – 25 = 20

Question 15.
If the product of two numbers is 216 and their L.C.M is 36, find their H.C.F.
Answer:
If a × b = 216, L = 36, H = ?
H = \(\frac{\mathrm{a} \times \mathrm{b}}{\mathrm{L}}=\frac{216}{36}\) = 6
∴ HCF = 6

Question 16.
Prove that (x^b-c)^a.(x^b)^c-a.(x^a-b)^c = 1.
Answer:
LHS = x^ab-bc. x^bc-ab. x^ac-bc
= x^{ab-bc+bc-ab+ac-bc}
= x° = 1 = RHS

Question 17.
Insert 4 Arithmetic means between 5 and 10.
Answer:
Let A₁, A₂, A₃, A₄ be the 4 Arithmetic Means between 5 and 10.
∴ 5, A₁, A₂, A₃, A₄, 10 are in A.P.
∴ a = 5, a + 5d = 10 ⇒ 5 + 5d = 10 ⇒ 5d = 5 =* d = 1
∴ A₁ = a + d = 5 + 1= 6
A₂ = a + 2d = 5 + 2 = 7
A₃ = a + 3d = 5 + 3 = 8
A₄ = a + 4d = 5 + 4 = 9

Question 18.
The age of the father is four times that of the son. 5 years ago, the age of the father was seven times that of his son. Find their present ages.
Answer:
Let the present age of the son = x yrs.
5 years ago, the age of the son was x – 5 and that of the father was 4x – 5.
Given 4x – 5 = 7(x – 5)
4x – 5 = 7x – 35
30 = 3x ⇒ x= 10 and 4x = 40
∴ Father’s age is 40 yrs and the son’s age is 10 years.

Question 19.
What principal will amount to Rs? 46,000 in 7 years at 12% p.a.?
Answer:
A = 46, 000, T = 7 yrs , R = 12% •
A = P(1 + \(\frac{\mathrm{TR}}{100}\))
46000 = P(1 + 7 × 0.12)
P = \(\frac{46000}{1+0.84}=\frac{46000}{1.84}\) = 25000
P = 25,000

Question 20.
Solve for x, if 7x + 3 < 5x + 9, x in R and represent the solution on the number line.
Answer:
7x + 3 < 5x + 9
2x < 6
x < 3

Question 21.
The average score of 35 girls is 80 and the average score of 25 boys is 68. Find the average score of both boys and girls together.
Answer:
X̄₁₂ = \(\frac{35 \times 80+25 \times 68}{35+25}=\frac{2800+1700}{60}=\frac{4500}{60}\) = 75
∴ Average score of both is 75.

Question 22.
By selling 8 erasers a trader gains the selling price of 1 eraser. Calculate the gain percent.
Answer:
Let the S.R of one eraser be x.
∴ the SP of 8 erasers = 8x
profit = x
C.P = S.P – Profit = 8x – x = 7x
Profit % = \(\frac{\text { Profit }}{\text { C.P }}\) × 100 = \(\frac{\mathrm{x}}{7 \mathrm{x}}\) × 100 = 14\(\frac{2}{7}\)%

Question 23.
Prove that : \(\sqrt{\frac{1+\cos A}{1-\cos A}}\) = cosecA + cotA
Answer:
LHS = \(\sqrt{\frac{1+\cos A}{1-\cos A}}=\sqrt{\frac{(1+\cos A)^{2}}{1-\cos ^{2} A}}=\sqrt{\left(\frac{1+\cos A}{\sin A}\right)^{2}}=\frac{1}{\sin A}+\frac{\cos A}{\sin A}\) = cosecA + cosA = RHS

Question 24.
Evaluate: sin2\(\frac{\pi}{6}\) + cos2 \(\frac{\pi}{3}\) – tan2\(\frac{\pi}{4}\) + cot2\(\frac{\pi}{4}\)
Answer:
(sin 30)² + (cos 60)² – (tan 45)² + (cot 45)²
= \(\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\) – 12 + 12
= \(\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\)

Question 25.
Using the concept of the slope, show that the points (-4, -5), (1, -1) and (6, 3) are collinear.
Answer:
Slope of AB = slope of AC
\(\frac{-1+5}{1+4}=\frac{3+5}{6+4}\)
\(\frac{4}{5}=\frac{8}{10}\)
⇒ slopes are equal
∴ the points are collinear.

PART-C

II. Answer any TEN questions: (10 × 3 = 20)

Question 26.
In a survey of 5000 persons, it was found that 2800 read Indian Express and 2300 read Deccan Herald while 400 read both. How many read either Indian Express nor Deccan Herald?
Answer:
n(I) = 2800, n(D) = 2300, n(ID) = 400

n(I∪D) = n(I) + n(D) + n(I∩D)
= 2800 + 2300 – 400
= 4700
∴ 5000 – 4700 = 300 will read either Indian Express nor Deccan Herald.

Question 27.
Show that the relation “is congruent to” is an equivalence relation on a set T of triangles.
Answer:
Any ∆ T₁ is congruent to itself ∴ R is reflexive.
If a ∆T₁ is congruent to ∆ T₂ then T₂ is congruent to T₁. ∴ R is symmetric.
If a ∆T₁ is congruent to ∆ T₂ & ∆T₂ is congruent to ∆ T₃ then ∆T₁ is congruent to ∆ T₃ ∴ R is transitive.
Since R is reflexive, symmetric, and transitive. Hence it is an equivalence relation.

Question 28.
Prove that : log₄8.log₂32.log₁₆4 = \(\frac{15}{4}\)
Answer:
L.H.S = \(\frac{\log 8}{\log 4} \times \frac{\log 32}{\log 2} \times \frac{\log 4}{\log 16}=3 \log 2 \times \frac{5 \log 2}{\log 2} \times \frac{1}{4 \log 2}=\frac{15}{4}\)

Question 29.
Solve : 2^2x – 6.2^x + 8 = 0
Answer:
(2x)² – 6.2^x + 8 = 0
Put y = 2x
y² – 6y + 8 = 0
(y – 2)(y – 4) = 0 ⇒ y = 2 or y = 4
2^x = 2¹ or 2^x = 2² ⇒ x = 1 or x = 2

Question 30.
Find the sum to n terms of the GP. 7 + 77 + 777 +………………..
Answer:
Let S = 7 + 77 + 777 +………………+ n terms
\(\frac{\mathrm{S}}{7}\) = 1 + 11 + 111 +…………..n terms
\(\frac{9 \mathrm{~S}}{7}\) = 9 + 99 + 999 +……………n terms
\(\frac{9 \mathrm{~S}}{7}\) = (10 – 1) + (100 – 1) + (1000 – 1) +………………… n terms
\(\frac{9 \mathrm{~S}}{7}\) = (10 +100 + 1000 + …………….n terms) – (1 + 1 + ………………n terms)
\(\frac{9 \mathrm{~S}}{7}\) = (10+100 + 1000 + …………….n terms) – (1 + 1 +………………. n terms)
= \(\frac{10\left(10^{n}-1\right)}{9}\) – n
S = \(\frac{7}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]\)

Question 31.
Divide ₹ 110 into two parts so that 5 times of one part together with 6 times of other parts will be equal to Rs. 610.
Answer:
Let x & y be the two parts.
Then the number will be x + y
Given: x + y = 110 …………(1)
Also given 5x + 6y = 610 ……………….(2)
Solving (1) and (2) we get x = 50 & y = 60
Hence the two parts of 110 are ₹ 50 and ₹ 60.

Question 32.
Solve the following inequalities graphically :
x + 3y > 3
2x + y > 2
x > 0, y > 0
Answer:
Consider x + 3y = 3
Put x = 0, y = 1 (0,1)
put y = 0, x = 3(3, 0)

Let 2x + y = 2
Put x = 0, y = 2 (0, 2)
Put y = 0, x = 1 (1,0)

Question 33.
Find the present value of an annuity due of Rs. 8,000 for 5 years at 5% p.a.
Answer:
The present value of an annuity due is.
P = \(\frac{\mathrm{A}\left((1+\mathrm{i})^{\mathrm{n}}-1\right)(1+\mathrm{i})}{\mathrm{i}(1+\mathrm{i})^{n}}=\frac{8000\left[(1+0.05)^{2}-1\right](1+0.05)}{(0.05)(1+0.05)^{5}}\) = ₹ 36,369.50

Question 34.
Find the ratio in which the line joining the points (3, 5) and (-7, 9) is divided by the point.
Answer:
A = (3, 5) B = (-7, 9)
P(x, y) = (\(\frac{1}{2}\), 6)
The ratio m : n on which P divides AB is given by
\(\frac{\mathrm{m}}{\mathrm{n}}=\frac{\mathrm{x}-\mathrm{x}_{1}}{\mathrm{x}_{2}-\mathrm{x}}=\frac{\frac{1}{2}-3}{-7-\frac{1}{2}}=\frac{1-6}{-14-1}=\frac{-5}{-15}=\frac{1}{3}\)
∴ \(\frac{\mathrm{m}}{\mathrm{n}}=\frac{1}{3}\) ⇒ m: n = 1: 3
∴ P divides AB in the ratio 1 : 3 internally.

Question 35.
A dealer buys 200 quintals of wheat at Rs. 1,200 a quintal. He spends Rs. 10,000 on transportation and storage. He sells the wheat at Rs. 13 per kg. Find his profit or loss. Also, calculate it as a percentage.
Answer:
Cost Price of 200 quintals of wheat = 1200 × 200 = 2,40, 000
Transportation and storage cost = Rs. 10,000
Total C. P = 2,40,000 + 10,000 = 2,50,000
Total S. P = 13 × 20000 = 2,60,000
Profit = 2,60,000 – 2,50,000 = 10,000
Profit% = \(\frac{\text { Profit }}{\mathrm{CP}}\) × 100 = \(\frac{10,000}{2,50,000}\) × 100 = 4%

Question 36.
If sin θ = \(\frac{-3}{5}\) and lies in IV quadrant, then prove that \(\frac{3 \tan \theta-4 \cos \theta}{4 \tan \theta+3 \cos \theta}=\frac{109}{12}\)
Answer:
sin θ = \(\frac{-3}{5}\), cos θ = \(\frac{4}{5}\), tan θ = \(\frac{-3}{4}\)
LHS = \(\frac{3 \tan \theta-4 \cos \theta}{4 \tan \theta+3 \cos \theta}=\frac{3 \cdot\left(-\frac{3}{4}\right)-4\left(\frac{4}{5}\right)}{4\left(-\frac{3}{4}\right)+3\left(\frac{4}{5}\right)}=\frac{-\frac{9}{4}-\frac{16}{5}}{-\frac{12}{4}+\frac{12}{5}}=\frac{-\frac{45-64}{20}}{\frac{-60+48}{20}}=\frac{-109}{-12}=\frac{109}{12}\) = RHS

Question 37.
Show that the points (4,4) (3,5) and (-1, -1) are the vertices of a right-angled triangle.
Answer:
A = (4, 4) B = (3, 5) C = (-1,-1)
AB = \(\sqrt{(3-4)^{2}+(5-4)^{2}}=\sqrt{(-1)^{2}+1^{2}}=\sqrt{2}\)
AB² = 2 …………….(1)
BC = \(\sqrt{(-1-3)^{2}+(-1-5)^{2}}=\sqrt{(-4)^{2}+(-6)^{2}}=\sqrt{16+36}=\sqrt{52}\)
BC² = 52 ……………..(2)
CA = \(\sqrt{(-1-4)^{2}+(-1-4)^{2}}=\sqrt{(-5)^{2}+(-5)^{2}}=\sqrt{25+25}=\sqrt{50}\)
CA² = 50 …………………(3)
From 1, 2 and 3 we get
AB² + CA² = BC²
∴ A, B & C form a right angled triangle.

Question 38.
Find the equation of the straight line which passes through the point of intersection of 2x-3y = 4 and 2x + 2y = 1 and perpendicular to the line x + 4y = 8.
Answer:
Solving 2x – 3y = 4………………(1)
2x + 2y = 1 ……………(2)
We get x = \(\frac{11}{10}\) & y = –\(\frac{3}{5}\)
Any line perpendicular to x + 4y = 8 is of the form 4x -y + k = 0. But this line passes through \(\)
4.\(\left(\frac{11}{10},-\frac{3}{5}\right)\) + k = 0
\(\frac{11}{10}-\left(-\frac{3}{5}\right)\) = -K
-K = \(\frac{25}{5}\) = 5 ⇒ k = -5
The required equation of the line is 4x – y – 5 = 0

PART-D

IV. Answer any SIX questions : (6 × 5 = 30)

Question 39.
In a survey of 100 persons, it was found that 28 read magazine A, 30 read magazine B, 42 read magazine C, 5 read magazine B and C, while 3 read all three magazines. Find
(i) How many read one of the three magazines?
(ii) How many read-only magazines C?
Answer:
Given U = 100, n(A) = 28, n(B) = 30, n(C) = 42,
n(A∩B) = 8, n(A∩C) = 10, n(B∩C) = 5

n(A∩B∩C) = 3
(i) Number of people who read some of the magazines
= 100 – (13 + 5 + 20 + 7 + 3 + 2 + 30)
= 100 – 80 = 20
(ii) Number of people who read the only magazine C = 30.

Question 40.
Find the sum of all numbers between 50 and 200 which are divisble by 11.
Answer:
AP is given by
55, 66, 77…………………198
Here a = 55, l = T = 198 d= 11
Tn = a + (n – 1)d
198 = 55 + (n – 1) 11
198 – 55 = (n – 1)11
143 = (n – 1)11
(n – 1) = \(\frac{143}{11}\) = 13
n= 13 + 1 = 14
∴ n = 14
S_n= \(\frac{\mathrm{n}}{2}\)(a + l) = \(\frac{14}{2}\)(55 + 198) = 7(253) = 1771
∴ Sum = 1771

Question 41.
If there is an integral root between -3 and +3, by inspection and then by using synthetic division solve x³ +15x² – 72x + 76 = 0
Answer:
Given x³ + 15x² -72x + 76 = 0
x = 2 is the multiplier.

∴ Quotient is x2 + 1 7x – 38 = 0
Solving this we get (x + 19)(x – 2) = 0
x = -19, 2
∴ the roots of the given equation are 2, 2, -19

Question 42.
If ₹ 9,000 amounts to Rs. 10,418.625 in 3 years, And the compound interest rate present. .
Answer:
Given P = 9000, A= 10,418.625 .
n = 3 yrs,, r = ?
r = A.L\(\left\{\frac{\log A-\log P}{n}\right\}\) -1 = r = A.L\(\left\{\frac{\log 10418.625-\log 9000}{3}\right\}\) – 1 = A.L[0.021067] – 1 = 1.050 – 1 = 0.05
∴ r = 0.05 x 100 = 5%

Question 43.
Preritha wants to buy a house after 5 years when it is expected to cost Rs. 50 lakhs. How much should she save annually if her savings earn a compound interest of 12% p.a?
Answer:
Given: F = 50,00,000; n = 5, i = 0.12
F = \(\frac{A\left[(1+i)^{n}-1\right]}{i}\)
50,00,000 = A(6.3528)
A = \(\frac{50,00,000}{6.3528}\)
∴ A = ₹ 787054.5

Question 44.
Find the equations of the medians of the triangle formed by the points (-1, 3), (-3, 5), and (7, -9).
Answer:
Let D as the midpoint of BC = \(\left(\frac{-3+7}{2}, \frac{5-9}{2}\right)\)
Equation of the median AD is
\(\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
\(\frac{y-3}{x+1}=\frac{-2-3}{2+1}=\frac{-5}{3}\)
3y – 9 = -5x – 5
5x + 3y – 4 = 0

Let E is the mid-point of AC = \(\left(\frac{-1+7}{2}, \frac{3-9}{2}\right)\) = (3, 3)

The equation of the median BE is
\(\frac{y-5}{x+3}=\frac{-3+5}{3+3}=\frac{-8}{6}=\frac{-4}{3}\)
3y – 15 = -4x – 12
4x + 3y – 3 = 0

Let F is the midpoint of AB = \(\left(\frac{-1-3}{2}, \frac{3+5}{2}\right)\) = (-2, 1)
The equation of the median CF is
\(\frac{y+9}{x-7}=\frac{-9+4}{7+2}=\frac{-13}{9}\)
9y + 81 = -13x + 91
13x + 9y – 10 = 0

Question 45.
Evaluate: \(\frac{213.781 \times 7.434}{6.321}\) using logarithmic tables.
Answer:
Let x = \(\frac{213.781 \times 7.434}{6.321}\)
logx = log\(\left[\frac{213.781 \times 7.434}{6.321}\right]\)
logx = log213.781 + log7.434 – log6.321
= 2.3298 + 0.87 12 – 0.8008
= 2.4002
x = A.L (2.4002) = 251.3

Question 46.
Find the foot of the perpendicular drawn from the point (-2, -1) on the line 3x + 2y – 5 = 0.
Answer:
Let P(x,y) = (-2, -1), Q (h, k) be the foot of the perpendicular on the line 3x + 2y – 5 = 0
\(\frac{\mathrm{h}-x_{1}}{\mathrm{a}}=\frac{\mathrm{k}-\mathrm{y}_{1}}{\mathrm{~b}}=\frac{-\left(\mathrm{ax}_{1}+\mathrm{by}_{1}+\mathrm{c}\right)}{\mathrm{a}^{2}+\mathrm{b}^{2}}\)

Here a = 3, b = 2, c = 5 & (x₁, y₁)(-2, -1) .
Substituting all these values in the above expression we get
\(\frac{h-(-2)}{3}=\frac{k-(-1)}{2}=-\left(\frac{3(-2)+2(-1)-5}{3^{2}+2^{2}}\right)\)
\(\frac{h+2}{3}=\frac{k+1}{2}\) = -(-1) = 1
\(\frac{h+2}{3}\) = 1 if \(\frac{\mathrm{k}+1}{2}\) = 1
h = 1, k = 1
The foot of the perpendicular is Q (1, 1).

Question 47.
If x = arsin AcosB, y = brsin AsinB and z = crcosA , prove that \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}\) = r²
Answer:
Given x = arsin AcosB ; y = brsinASinB; z = crcosA
L.H.S. = \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}\)
= \(\frac{a^{2} r^{2} \sin ^{2} A \cos ^{2} B}{a^{2}}+\frac{b^{2} r^{2} \sin ^{2} A \sin ^{2} B}{b^{2}}+\frac{c^{2} r^{2} \cos ^{2} A}{c^{2}}\)
= r² sin² A (cos² B + sin² B) + r² cos2 A
= r² sin² A.1 + r² cos² A = r² (sin² A + cos² A) = r².1 = r² = R.H.S.

Question 48.
Find the equation of the locos of a point that moves such that its distance from the point (-4, 0) is 4 times its distance from (0, -2).
Answer:
Let P(x,y) be any point on the locus
A = (-4, 0) B = (0, -2)
Given PA = 4PB PA² = 16PB²
(x + 4)²+(y-0)² = 16[(x-0)²+(y + 2)²]
x² + 16 + 8x + y² = 16[x² + y² + 4 + 4y]
16x²+ 16y²+ 64 + 64y = x² + y² + 8x +16
15x² + 15y² – 8x + 64y + 48 = 0 is the required equation of the locus. ‘

PART-E

V. Answer any ONE question : (6 × 5 = 30)

Question 49.
(a) Find the equation of a line passing through (-2, 6) and the sum of intercepts on the coordinate axis is 5.
Answer:
Let x-intercept = a, y-intercept = b
Given: a + b = 5
The equation of the line is \(\) = 1
x(5 – a) + y(a) = a(5 – a).
But this passes through the point (-2, 6)
-2(5 – a) + 6a = 5a – a²
-10 + 2a + 6a = 5a – a²
a² – 3a – 10 = 0
(a – 5)(a + 2) = 0
a = 5 or a = -2
If a = 5 then b = 5 – 5 = 0,
If a = -2, b = 5 – (-2) = 7
the required equations of the lines are
\(\frac{x}{-2}+\frac{y}{7}\) = 1
7x – 2y = -14
or 7x – 2y + 14 = 0

(b) Evaluate : \(\frac{\sin 150^{\circ}-5 \cos 300^{\circ}+7 \tan 225^{\circ}}{\tan 135+3 \sin 210^{\circ}}\)
Answer:
\(\frac{\sin 150-5 \cos 300+7 \tan 225}{\tan 135+3 \sin 210}=\frac{\sin (180-30)-5 \cos (360-60)+\tan (180+45)}{\tan (180-45)+3 \sin (180+30)}\)
= \(\frac{\sin 30-5 \cos 60+7 \tan 45}{-\tan 45-3 \sin 30}=\frac{\frac{1}{2}-5 \cdot \frac{1}{2}+7.1}{-1-\frac{3}{2}}=\frac{\frac{1-5+14}{2}}{\frac{-2-3}{2}}=\frac{10}{-5}\) = -2

(c) If log6 = 0.7782, find the number of zeroes between the decimal point and the first significant figure in (0.6)30.
Answer:
Let x = (0.6)30
logx = 30log0.6
= 30 × 1.7782 = 30 (-1 + 0.7782)
= -30 + 23.346
= -7̄.346
= 7̄.346
Since the characteristic is 7̄. Hence the required number of zeros is 6.

Question 50.
(a) Three numbers whose sum is 12 are in A.P. If 1, 4, 11 are added to them the resulting numbers are in G.P. Find the numbers.
Answer:
Let the 3 numbers be a – d, a, a + d
By data a-d + a + a + d = 12
3a = 12 ⇒ a = 4
Also by data 4-d + 1, 4 + 4, 4 + d + 11 are in G.P.
5 – d, 8, 15 + d are in G.P.
∴ \(\frac{8}{5-d}=\frac{15+d}{8}\)
64 = (5 – d)(15 + d)
64 = 75 + 5d – 15d – d².
d² + 10d – 11 = 0
(d + 11 )(d – 1) = 0
d = 1 and d = -11
When a = 4, d = 1 the required numbers are 3, 4, 5
When a = 4, d = -11 then required numbers are 15, 4, -7.

(b) A watch manufacturer produced 100 watches for a total cost of Rs. 20,000 and when the production is increased to 200 watches the total cost increases to Rs. 30,000. Assuming that the costs and outputs are linearly related, find the cost equation and find the cost of manufacturing 150 watches.
Answer:
Total cost = TVC + TFC
C(x) = ax + b
When x = 100 watches
C(x) = 100a + b
20,000 = 100a + b ……………(1)
Similarly when x = 200 watches
30,000 = 200a + b. ……………..(2)
Solving equations 1 and 2 we get
100a = 10,000
10,000
∴ a = \(\frac{10,000}{100}\) = 100
20,000 = 100 × 100 + b
∴ b = 20,000 – 10,000 = 10,000
∴ when x = 150, T.C 100 × 150 + 10,000
TC = ₹ 25,000

(c) Form the cubic equation whose roots are -1, 4, 6.
Answer:
The cubic equation with the roots -1, 4 & 6 is given by
(x + 1)(x – 4)(x – 6) = 0

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2016 (North), Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2016 (North)

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any TEN questions. (10 × 1 = 10)

Question 1.
Write the imaginary part of the complex number \(\frac{-3+4 i}{5}\)
Answer:
Imaginary part of \(\frac{-3}{5}\) + i\(\frac{4}{5}\) is \(\frac{4}{5}\).

Question 2.
If set A has 16 subsets, find the number of elements in A.
Answer:
A has 4 elements.

Question 3.
If a function f(x) is defined by f(x) = 3x – 1, find f(\(\frac{1}{3}\)) .
Answer:

Question 4.
Find the 8th term of the sequence \(\frac{1}{2}\), \(\frac{1}{4}\), \(\frac{1}{6}\)………..
Answer:
8th term = T₈ = \(\frac{1}{16}\)

Question 5.
Find the value of log01 100
Answer:
Let log0_0.1 100 = x
100 = (0.1)x
102 = 10 – x ⇒ x = -2
log_0.1 100 = -2

Question 6.
Simplify \(\left(\frac{64}{81}\right)^{-1 / 4}\)
Answer:
\(\left(\frac{64}{81}\right)^{-1 / 4}=\left(\frac{81}{64}\right)^{1 / 4}=\frac{\left(3^{4}\right)^{1 / 4}}{\left(2^{6}\right)^{1 / 4}}=\frac{3}{2^{3 / 2}}=\frac{3}{2 \sqrt{2}}\)

Question 7.
Solve for x : x + a(x + b) = ax + b.
Answer:

x = b – ab
x = b(1 – a)

Question 8.
What is the present value of an income of Rs. 10,000 a year to be received forever at 10% p.a.?
Answer:
P_∞ = \(\frac{a}{r}=\frac{10,000}{0.1}\) = 1,00000

Question 9.
Convert 315° into radians.
Answer:

Question 10.
Find the slope of a line perpendicular to the line 3x – 4y + 5 = 0.
Answer:
Slope of the given line = \(\frac{3}{4}\)
Slope of the perpendicular line = –\(\frac{4}{3}\)

Question 11.
Find the coordinates of the centroid of the triangle formed by the points (1, 1), (-2, 1), and (-5, -5).
Answer:
Centroid = G(x,y) = \(\left[\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right]=\left[\frac{1-2-5}{3}, \frac{1+1-5}{3}\right]=\left[\frac{-6}{3}, \frac{-3}{3}\right]\) = [-2, -1]

Question 12.
Mr. Prem bought a cycle at Rs. 5000 and sold the same at a loss of 20% to Mr. Bheem. Find the cost price of Bheem.
Answer:
Cost price = 5000 – 20% of 5000 = 5000 – 1000 = 4000 Rs.

PART-B

II. Answer any ten questions: (10 × 2 = 20)

Question 13.
Find the sum of positive divisors of 198.
Answer:
198 = 2¹ × 3² × 11¹
T(198) = (1 + 1)(2 + 1)(1 + 1) = 2 × 3 × 2 = 12.

Question 14.
Find greatest number which can divide 60,72 and 84 without leaving any remainder.
Answer:
60 = 2² × 3¹ × 5¹
72 = 2³ × 3²
84 = 2² × 3¹ × 7¹
∴ greatest number is nothing but HCF = 22 × 31 = 12.

Question 15.
If A = {x / x ∈ N, x < 4}, B = {x – 25 = 0, x ∈ N} find B x A.
Answer:
A = (1, 2, 3} B = {5}
B × A = {(5, 1) (5, 2) (5, 3)

Question 16.
Prove that \(\frac{1}{1+\log _{b} a}+\frac{1}{1+\log _{a}}\) = 2
Answer:
L.H.S = \(\frac{1}{\log _{b}^{b}+\log _{b}^{a}}+\frac{1}{\log _{a}^{a}+\log _{a}^{b}}=\frac{1}{\log _{b}^{b a}}+\frac{1}{\log _{a}^{a b}}\) = logab b + laga ab = logab ab = 1 = R.H.S

Question 17.
Insert 3 geometric means between \(\frac{1}{4}\) and \(\frac{1}{64}\)
Answer:
Insert 3 Gms G₁, G₂, G₃. between \(\frac{1}{4}\) & \(\frac{1}{64}\)
\(\frac{1}{4}\), G₁, G₂, G₃, \(\frac{1}{64}\) are in G.P
a = \(\frac{1}{4}\), ar⁴ = \(\frac{1}{64}\) ⇒ \(\frac{1}{4}\)r⁴ = \(\frac{1}{64}\) ⇒ r⁴ = \(\frac{4}{64}\) ⇒ r⁴ = \(\frac{1}{16}\) = \(\left(\frac{1}{2}\right)^{4}\) ⇒ r = \(\frac{1}{2}\)

Question 18.
If sum of 2 numbers is 107 and their difference is 17, find the numbers.
Answer:
Let the two number be x and y
Given x + y = 107 …. (1)
x – y = 17 …. (2) solving we get
2x = 124 ⇒ x = 62, y = 107 – 62,y = 45
∴ the 2 numbers are x = 62 and y = 45.

Question 19.
Solve : 3x – 2 ≤ 2x + 1, x ∈ R and represent on the number line.
Answer:
3x – 2 ≤ 2x + 1, x ∈ R .

3x – 2x ≤ 2 +1
x ≤ 3

Question 20.
In how many years will a sum of money double itself at 10% simple interest p.a.?
Answer:
Let principle = 100 Rs. ∴ Amount = 200 Rs.
P = 100, F = 200, r = 10% = 0.1 T = ?
SI = F – P= 100 Rs.
w.k.t S.I = \(\frac{\text { PTR }}{100}\)

∴ In 10 years sum of money double itself.

Question 21.
The average age of 10 students is 6 years. The sum of the ages of 9 of them is 52 years. Find the age of the 10th Student.
Answer:
Average age of 10 students = 6yrs
∴ Total age of 10 students = 10 × 6 = 60yrs
Total age of 9 students = 52 yrs
∴ Age of the 10th students = 60 – 52 = 8yrs.

Question 22.
The difference between the cost price and the selling price is Rs. 225. If the profit percentage is 15% find the selling price.
Answer:
Let the cost price be x
As there is profit SP > CP
SPx + 225
SP = \(\frac{100+\text { Profit } \%}{100}\) × CP
x + 225 = \(\left(\frac{100+15}{100}\right)\)x
100x + 22500 = 100x + 15x
15x = 22500
x = = 1500
SP = x + 225 = 1500+225 = 1725

Question 23.
Prove that cotθ + tanθ = secθ cosecθ.
Answer:
LHS = \(\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}=\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}=\frac{1}{\sin \theta \cdot \cos \theta}\) = cosecθ sec θ = RHS

Question 24.
IfA, B, C are angles of a triangle, then prove that cos\(\left(\frac{\mathbf{A}+\mathbf{B}}{\mathbf{2}}\right)\) = sinC/2.
Answer:
W.K.T A + B + C = 180°
\(\frac{A+B+C}{2}\) = 90 ⇒ \(\frac{A+B}{2}\) = 90 – \(\frac{\mathrm{C}}{2}\)
Taking sin on both sides
We get sin\(\left(\frac{A+B}{2}\right)\) = sin \(\left(90-\frac{\mathrm{C}}{2}\right)\)
sin\(\left(\frac{A+B}{2}\right)\) = cos\(\frac{\mathrm{C}}{2}\)

Question 25.
Derive two point form of equation of straight line in the form y – y₁ = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)(x – x₁)
Answer:
Let A(x₁, y₁), B(x₂, y₂) be the two given points, let p(x, y) be any point on the line. Here A, B, and P tie on the same line
∴ The slope of AP = Slope of AB

i.e., \(\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
(y – y₁) = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)(x – x₁)
This is called two point formula.

PART-C

III. Answer any ten questions. (10 × 3 = 30)

Question 26.
Prove that \(\sqrt{5}\) is irrational.
Answer:
If possible let \(\sqrt{5}\) is a rational number.
∴ .\(\sqrt{5}\) = p, q∈z, q ≠ 0
we shall also assume that p and q do not have any common factor
∴ 5 = \(\frac{\mathrm{p}^{2}}{\mathrm{q}^{2}}\)
∴ p² is a multiple of 5
∴ 5 divides p²
∴ 5 divides p
∴ p = 5k where k∈z, k≠O
p² = 25k²
5q² = 25k² ∵ p² = 5q²
q² = 5k²
∴ q² is a multiple of 5
∴ 5 divides q²
∴ 5 divides q
∴ 5 divides both p and q
This is a contradiction
∴ Our assumption is wrong
∴ \(\sqrt{5}\) is irrational.

Question 27.
Let f {(1, 1), (2, 3), (0, -1)) be a function from Z to Z defined by f(x) = ax + b a, b ∈ Z determine a and b.
Answer:
Given f(x) = ax + b
When x = 1, f(x) = 1 ⇒ a + b = 1
When x = 0, f(x) = -1 ⇒ 0 + b = -1 ⇒ b = -1
if b = 1 then a – 1 = 1 ⇒ a = 2
∴ a = 2, b = -1.

Question 28.
If A = {0, 1,2,3), B = {2, 3, 4), C = {4,5,6) then prove that A∩(B∪C)=(A∩B)∪(A∩C)
Answer:
A∩(B∩C) = {0, 1, 2, 3} n {2, 3, 4, 5, 6} {2, 3} ………………(1)
(A ∩ B) ∪ (A ∩ C) = {2, 3} ∪ {Q} = {2, 3) …………………..(2)
From 1 and 2 we have
A∩(B∪C) = (A∩B)∪(A∩C)

Question 29.
If x² + y² = 12xy. Show that 2 log (x – y) = log 10 + log (xy)
Answer:
Given x² + y² = 12xy
x² + y² – 2xy = 10xy
(x – y)² = 10xy
Taking logm both sides
log(x – y)² = log(10xy)
2 log (x – y) = log 10 + log xy

Question 30.
Find three numbers in A.P, whose sum is 24 and product is 440.
Answer:
Let the 3 number in AP is {a – d) a, a + d
Given Sum = 24

3a = 24 ⇒ a = 8
Also given product = 440
(a – d) (a) (a + d) = 440
a² – d² =55 ⇒ d² =64 – 65 = 9 ⇒d = ±3
∴ the 3 number are 5, 8,11 of a = 8, d = 3 the 3 numbers are 11,8, 5 if a = 8, d = -3

Question 31.
If α and β are the roots of the equation 3x² – 6x + 4 = 0 find the value of
(i) α³ + β³
(ii) α/β² + β/α²
Answer:
Given 3x² – 6x + 4 = 0
Here α + β= \(\frac{-b}{a}=\frac{6}{3}\) = 2
αβ = \(\frac{c}{a}=\frac{4}{3}\)

Question 32.
Solve the system of inequalities 2x + y ≥ 8, and x + y ≥ 10 graphically.
Answer:
Consider 2x + y = 8
Put x = 0, y = 8 (0, 8)
Put y = 0, x = 4 (4, 0)

x + y = 10
put x = 0, y = 10 (0, 10)
put y =0, x = 10 (10, 0)

Plot the above point on the graph.

Question 33.
Mr X bought a Walkman for Rs. 1800. If it depreciates at the rate of 15% per year, how much is it worth after 3 years?
Answer:
C = Rs 1800, r = 15% = 0.15, n = 3yrs, B = ?
B = C(1 – r)n
B = 1800 (1 – 0.15)3 = (1800) × (0.85)3 = 1105.425
worth after 3 years = 1105.425

Question 34.
A batsman finds that, by scoring a century in the 11th innings of his test matches, he has bettered his average of preview test innings by 5 runs. What is the average after the 11 innings?
Answer:
Let the average of 10 inninngs = x
∴ Total runs after 10 inninngs = lOx
Average after 11 inninngs x +5
∴ Total runs after 11 inninngs = 11 (x + 5)
11 (x + 5) – 10x = 100
11x + 55 – 10x = 100
x + 55 = 100
x = 100 – 55 = 45
Average after the 11th inninngs = x + 5 = 45 + 5 = 50 runs

Question 35.
By how many percent should the use of coffee be increased if the price of coffee is decreased by 10% so that the expenditure remains unchanged.
Answer:
Let the price of coffee =100
Let the quantity used be 100
∴ Total expenditure = 100 x 100 = 10,000
∴ New price = 100 – 10 = 90
Let the new quantity used = x
the total expenditure = 90x
For unchanged expenditure 90x = 10000

∴ the increase in consanption of coffee =111.11 – 100 = 11.11

Question 36.
Find x, if xsin\(\frac{\pi}{4}\) xtan \(\frac{\pi}{3}\) = \(\frac{\sin \frac{\pi}{6} \times \cot \frac{\pi}{6}}{3 \cos \frac{\pi}{6} \times \cos s e \frac{\pi}{4}}\)
Answer:

Question 37.
Find the equation of the locus of the point which moves such that its distance from the line 3x – 4y + 1 = 0 is equal to its distance from (1, -1).
Answer:
A = (1,-1)
Let p(x, y) be any point on the locus
Given: PA = distance of P from the line 3x- 4y + 1 = 0
\(\sqrt{(x-1)^{2}+(y+1)^{2}}=\left|\frac{3 \dot{x}-4 y+1}{\sqrt{3^{2}+(-4)^{2}}}\right|\)
\(\sqrt{(x-1)^{2}+(y+1)^{2}}=\frac{3 x-4 y+1}{\sqrt{25}}\) S.B.S
(x – 1)² + (y – 1)² = \(\frac{(3 x-4 y+1)^{2}}{25}\)
25[(x – 1)² + (y + 1)²] = (3x – 4y + 1)²
25[x² + y² – 2x + 2y + 2] = 9x² + 16y² +1 – 34xy – 8y + 6x
⇒ 25x² + 25y² – 50x + 50y + 50 = 9x² + 16y² + 1 – 24xy – 8y + 6x
16x² + 9y² + 24xy – 42y – 44x + 49 – 0 is the required equation.

Question 38.
Find the future value of an annuity of Rs. 5000 at 12% p.a. for 6 years.
Answer:
A = Rs. 5,000 r = 0.12 n = 6
F = \(\frac{\mathrm{A}\left[(1+\mathrm{r})^{\mathrm{n}}-1\right]}{\mathrm{r}}=\frac{5000\left[(1+0.12)^{6}-1\right]}{0.12}=\frac{4869.11}{0.12}\) = 40,575.9

PART-D
IV. Answer any six questions : (6 × 5 = 30)

Question 39.
In a class of 50 students, 15 do not participate in any game, 25 play cricket, and 20 play football. Find the number of students who play.
(i) both cricket and football.
(ii) Only football.
Represent the results through the Venn diagram.
Answer:
Given n(u) = 50

n(C ∪ F) = 50 – 15 = 35, r(C) = 25, n(F) = 20
n(C ∩ F) = n(C) + r(F) – r(C ∪ F) = 25 + 20 – 35 = 10
From venn diagram we get only
Number of students who play only foot ball = 10.

Question 40.
Evaluate \(\frac{123.4 \times 54.8}{0.562 \times 7346}\) using logarithmic tables.
Answer:
Let x = \(\frac{123.4 \times 54.8}{0.562 \times 7346}\)
Then log x = log\(\left[\frac{123.4 \times 54.8}{0.562 \times 7540}\right]\)
= log123.4 + log54.8- log0.562 – log7346
x = Anti log (value)

Question 41.
The first term of G.P exceeds the second term by \(\frac{1}{2}\) and the sum to infinity is 2. Find the GP.
Answer:
a, ar, ar² ……………. GP
a – ar = 1/2
S_∞ = 2 …….(2)

a² = 1 ⇒ a = 1
\(\frac{\mathrm{a}}{1-\mathrm{r}}\) = 2 \(\frac{1}{1-r}\) = 2
1 – r = 1/2
1 – 1/2 = r
r = 1/2

Question 42.
Two years ago, a father’s age was three times the square of his son’s age. After 3 years, his age will be four times his son’s age. Find the present age of father and son.
Answer:
Let the age of the father be x yrs and age of the son be y yrs.
Given: x-2 = 3(y-2)² ⇒ x-2 = 3(y² + 4 – 4y)
Also given x + 3 = 4(y + 3)
x + 3 = 4y + 12

3y² – 16y + 5 = 0
3y² – 15y – y + 5 = 0
3y(y – 5) – 1(y – 5) = 0
(3y – 1)(y – 5) = 0
y = 51

Sons age is 5 yrs and fathers age is
x = 4y + 9 = 20 + 9
x = 29yrs

Question 43.
A certain amount invested at 4% p.a. compounded semi-annually amounts to Rs. 78,030 at the end of one year. Find the sum.
Answer:
F = 78030, i = 0.04, n= 1
F = P\(\left(1+\frac{i}{2}\right)^{n \times 2}\)
78030 = P\(\left(1+\frac{0.04}{2}\right)^{1 \times 2}\)
P = \(\frac{78030}{(1.02)^{2}}\) = 75,000
sum = 75,000.

Question 44.
If a person wants to have Rs. 80,000 after 5 years, how much should he deposit every year if the bank offers 12% p.a. interest compounded quarterly.
Answer:
F = 80,000, n = 5yrs, r = 12% = 0.12, i = \(\left(1+\frac{0.12}{4}\right)^{4}\) – 1 = 0.125
F = \(\frac{a\left[(1+i)^{n}-1\right]}{i}\)
a = \(\frac{\mathrm{F} \times i}{(1+i)^{n}-1}=\frac{80,000 \times 10.125}{(1.125)^{5}-1}=\frac{80,000}{6.416}\) = ₹12,468.8

Question 45.
In a dance competition, 70% of the participants were girls, 35% of the boys, and 65% of the girls got qualified for the next round. If 49 girls were eliminated find the number of boys who were eliminated.
Answer:
Let the tota! number of participants = x
Total number of girls = \(\frac{70}{100}\)x = 0.7x ⇒ Total number of boys = \(\frac{30}{100}\) × x = 0.3x
If 65% of girls got qualified then 35% of girls got eliminated
∴ The total girls eliminated = \(\frac{35}{100}\) × 0.7x = 0.245x
Hence 0.245. x = 49
x = \(\frac{49}{0.245}\)
Hence total number of boys eleminated = \(\frac{35}{1000}\) × 0.3 × 200 = 2

Question 46.
Prove that \(\frac{\operatorname{cosec}\left(180^{\circ}+0\right) \sin \left(360^{\circ}-0\right) \tan \left(36^{\circ}+\theta\right)}{\sin \left(90^{\circ}+\theta\right) \cos \left(180^{\circ}-\theta\right) \tan (-\theta)}\) = sec2 θ
Answer:

\(\frac{1}{\cos ^{2} \theta}\) = sec2θ = RHS

Question 47.
Show that the points (4, 4), (3, 5), and (-1, -1) are the vertices of a right-angled triangle. Also, find its area.
Answer:
Let A= (4, 4) B = (3, 5) and C = (-1, -1)
AB = \(\sqrt{(4-3)^{2}+(4-5)^{2}}=\sqrt{1^{2}+(-1)^{2}}\) ⇒ AB² = 2
BC = \(\sqrt{(4-3)^{2}+(4-5)^{2}}=\sqrt{1^{2}+(-1)^{2}}\) ⇒ AB² = 2
CA = \(\sqrt{(3+1)^{2}+(5+1)^{2}}=\sqrt{16+36}=\sqrt{50}\) ⇒ CA² = 50

From the above AB2 + CA2 = BC2

Question 48.
Find the equation of line passing through (3, 4) and having intercepts on the axes whose sum is 14.
Answer:
Req. Eqn is \(\frac{x}{a}+\frac{y}{b}\) = 1
Give a + 6=14 ⇒ b = 14 – a
eqn becomes \(\frac{x}{a}+\frac{y}{14-a}\)
(14 – a)x + ay = a(14 – a)
By data this must pass the (3, 4)
(14 – a) 3 + 4a = a(14 – a)
42 – 3a + 4a = 14a – a²
a² + a – 14a + 42 = 0
a² -13a + 42 = 0
(a – 7)(a – 6) = 0 ⇒ a = 7
If a = 7 then b = 7 If a = 6 then b = 8.
∴ the required equations are
\(\frac{x}{7}+\frac{y}{7}\) = 1 & \(\frac{x}{6}+\frac{y}{8}\) = 1

PART-E

V. Answer any one question : (1 × 10 = 10)

Question 49.
(a) Find the equation of a line passing through the point of intersection of the lines x – 2y + 4 = 0 and 4x – 3y + 1 = 0 inclined at angle of 135° with positive x-axis. Also write the x intercept of the line. (4M)
Answer:
Point of inter section of x – 2y + 4 = 0 and 4x – 3y + 1 = 0
Multiply equation one by 4 we get

y = \(\frac{15}{-5}\) = 3
x – 2y + 4 = 0
x = 2y – 4
x = 6 – 4 = 2
∴ point of intersection is (2,3)
Slope = m = tan (135°) = – 1
The required equation of the line with slope = -1 and the point (2,3) is given by
y – y₁ = m(x – x₁).
y- 3 = -1(x – 2)
∴ x + y – 5 = 0

(b) If sin θ = -8/17, π< θ < \(\frac{3 \pi}{2}\) find the value of \(\frac{\tan \theta-\cos \theta}{\sec \theta+{cosec} \theta}\)
Answer:
If sinθ = \(\frac{-8}{17}=\frac{O p p}{1+y p}\) Here adj = \(\sqrt{17^{2}-8^{2}}=\sqrt{225}\) = 15
θ lies in Quadrant

(c) Two numbers are in the ratio 7:5 and their difference is 17. Find the numbers. 2
Answer:
Let the two numbers be 7x & 5x
Given 7x – 5x = 17
2x = 17
x = 8.5
∴ the two number are 7 × 8.5 & 5 × 8.5 = 59.5 and 92.5.

Question 50.
(a) A confectioner makes and sells biscuits. 11e sells one pack of biscuits at 80. His cost of manufacturing is 40 per pack as a variable cost and 3000 as a fixed cost.
Find (i) his profit function.
(ii) break-even point. also
(iii) if he limits his Production o 100 packets, can be.make profit? 4
Answer:
T.R (x) = Price x Quantity = 80x
T.C C(x) = ax + b = 40x + 3000
Profit function = P(x) = R(x) – C(x) = 80x – 40x – 3000 = 40x – 3000
If x = 100 then P(100) = 40 × 100 – 3000 = ₹1 ,000

(b) Find the sum of n terms the series 5 + 35 + 555 + ………….
Answer:
Let S = 5 + 55 + 555 +…………… to n terms
\(\frac{\mathrm{S}}{5}\) = 1+ 11 + 111+………..to n terms
\(\frac{9 \mathrm{~S}}{5}\) = 9 + 99 + 999 +…………… to n terms
=(10 – 1) + (100 – 1) + (1000 – 1) +……….. to n terms
= (10 + 100 + 1000 +………… to n terms) – (1 + 1 + 1………. to n ternis)
= \(\frac{10\left(10^{n}-1\right)}{9}\) – 1 × n
S = \(\frac{5}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]\)

(c) If 3^x = 5^y = 15^z, Show that z(x + y) = xy.
Answer:
Let 3^x = 5^y = 15^z = k(say)
3^x = k ⇒ 3 = k^1/x, 5^y = k ⇒ 5 = k^1/y, 15^z = k⇒ 15 = k^1/z
Now 3 x 5 = 15
k^1/x. k^1/y = k^1/z ⇒ K^1/x+1/y = K^1/z
⇒ \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\)
⇒ \(\frac{\mathrm{y}+\mathrm{x}}{\mathrm{xy}}=\frac{1}{\mathrm{z}}\)
⇒ z (y + x) = xy ⇒ z (x + y) = xy

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2016 (South), Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2016 (South)

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any TEN questions. (10 × 1 = 10)

Question 1.
Give the canonical representation of 385.
Answer:
385 = 5′ × 7¹ × 11¹

Question 2.
If R = {(1, 2), (2, 3), (3, 3} a relation defined on the set A = {1, 2, 3} find R^-1.
Answer:
R^-1 = {(2, 1), (3,2), (3, 3)}

Question 3.
If A = (3, 5, 7}, B = {7, 9, 11} find B-A.
Answer:
B – A = {9, 11}

Question 4.
Simplify \(\sqrt[4]{x^{-4 / 3}}\)
Answer:
(x^-4/3)^1/4 = x^-1/3 = x^1/1/3 = \(\frac{1}{\sqrt[3]{x}}\)

Question 5.
Find the value of x if log\(\sqrt{2}\) x = 4
Answer:
(\(\sqrt{2}\))4 = x ⇒ x = 22 ⇒ x = 4

Question 6.
Find the 8th term of the sequence 1, \(\frac{1}{2}\), \(\frac{1}{3}\)
Answer:
T₈ = \(\frac{1}{8}\)

Question 7.
Form a quadratic equation with roots ‘O’ and ‘-1′
Answer:
(x – 0)(x + 1) = 0 ⇒ x2 + x= 0

Question 8.
What is the simple interest on Rs. 1000 for half a year at 10% p.a.?
Answer:

Question 9.
The sum of the ages of 10 girls in a class is 120 years. If one more girl of age 1 year is added to the group, find the average of 11 girls.
Answer:
X̄₁₁ = \(\frac{120+1}{11}=\frac{121}{11}\) = 11yrs

Question 10.
Convert 15% to the ratio form.
Answer:
3:20

Question 11.
Write the value of cosec² 10° – cot²10°.
Answer:
cosec²10 – cot²10 = 1

Question 12.
Find the distance from the origin to the line 3x – 4y + 5 = 0.
Answer:
\(\frac{5}{\sqrt{3^{2}+4^{2}}}=\frac{5}{\sqrt{25}}=\frac{5}{5}\) = 1

PART-B

II. Answer any TEN questions: (10 × 2 = 20)

Question 13.
Find the total number of positive divisors of 180.
Answer:
180 = 2² × 3² × 5
T(180) = (2 + 1) (2 + 1) (1 + 1) = 3 × 3 × 2 = 18

Question 14.
If f(x) = x², g(x) = x – 1, find go f(x) and gof(a).
Answer:
gof (x) = gf(x) = g(x²) = x² – 1
gof(a) = a² – 1

Question 15.
Simplify: \(\frac{2^{7 a-2 b} \cdot 8^{2 a-b}}{16^{a+b}}\)
Answer:
\(\frac{2^{7 a-2 b} \cdot\left(2^{3}\right)^{2 a-b}}{\left(2^{4}\right)^{a+b}}=\frac{2^{7 a-2 b} \cdot 2^{6 a-3 b}}{2^{4 a+4 b}}\) = 2^{7a-2b+a-3b-4a-4b} = 2^9a-9b

Question 16.
Prove that log \(\sqrt{a / b}\) + log \(\sqrt{b / c}+\) + log \(\sqrt{c / a}+\) = 0
Answer:
LHS = \(\frac{1}{2}\)log\(\frac{a}{b}\) + \(\frac{1}{2}\)log\(\frac{b}{c}\) + \(\frac{1}{2}\)log\(\frac{c}{a}\)

= \(\frac{1}{2}\) × 0 = 0 = R.H.S

Question 17.
Find the sum to infinity of GP. 3, -1, \(\frac{1}{3}\), –\(\frac{1}{9}\) ………. if it exists.
Answer:
S_∞ = \(\frac{a}{1-r}\) Given a = 3, r = –\(\frac{1}{3}\)
S_∞ = \(\frac{3}{1+\frac{1}{3}}=\frac{9}{4}\)

Question 18.
Find two numbers whose sum is 64 and the difference is 16.
Answer:
Let 2 numbers be x and y
∴ Given x + y = 64 & x – y = 16
Solving the above equations we get
x = 40 & y = 24
∴ The two numbers are 24 and 40

Question 19.
Solve the unequalities 5x – 3 > 3x + 1, x∈R and represent on the number line.
Answer:
5x – 3 > 3x + 1 & x ∈ R

5x – 3x > 1 + 3
2x > 4 ⇒ x > 2

Question 20.
A certain sum of money amounts to Rs.24,200 in 2 years
Answer:
Given F = 24,200, n = 2yrs r = 10%, P = ?
F = P(i + r)ⁿ
P = \(\frac{\mathrm{F}}{(1+r)^{n}}=\frac{24,200}{(1+0.1)^{2}}=\frac{24,200}{(1.1)^{2}}\) = ₹20,000
∴ The Sum = ₹20,000

Question 21.
Ram and Shyam went up a hill at a speed of 20kmph. Both of them came tumbling down the same distance at a speed of 30 km ph. Find the average speed for the round trip.
Answer:
Average speed = \(\frac{2 d}{\frac{d}{20}+\frac{d}{30}}=\frac{2 d}{\frac{5 d}{60}}=\frac{120}{5}\) = 24 kmph

Question 22.
While taking measurement a tailor writes 34 instead of 24. What is the percentage error?
Answer:
For 24 → 10 Errors
∴ 100 → ?
\(\frac{10 \times 100}{24}\) = 41.66
Hence percentage error = 41.66

Question 23.
If tan A = 5/12 acute find the value of \(\frac{\sin A+\cos A}{\sin A-\cos A}\)
Answer:
Given tan A = \(\frac{5}{12}=\frac{\mathrm{opp}}{\mathrm{adj}}\)
∴ HYP = \(\sqrt{12^{2}+5^{2}}=\sqrt{169}\) = 13
∴ sinA = \(\frac{5}{13}\), & cos A = \(\frac{12}{13}\)

Question 24.
Show that the points A(4, -2), B(2, -4), and C(7, 1) are collinear using the slope method.
Answer:
A(4, -2) B(2, -4) C(7, 1)
Slope AB = \(\frac{y_{2}-y_{1}}{y_{2}-x_{1}}=\frac{-4-(-2)}{2-4}=\frac{-2}{-2}\) = 1
Slope of BC = \(\frac{1-(-4)}{7-2}=\frac{5}{5}\) = 1
Slopes are equal ∴ the points are collirear

Question 25.
Find the equation of the line through (1, 2) and making an angle of 60° with the positive direction of x-axis.
Answer:
Given = (1, 2) and θ = 60°
m = tan θ = tan 60° = \(\sqrt{3}\)
Required equation is y – y₁ = m(x – x₁)
y – 2 = \(\sqrt{3}\)(x – 1)
y – 2 = \(\sqrt{3}\)x – \(\sqrt{3}\)
⇒ \(\sqrt{3}\)x – y + 2 – \(\sqrt{3}\) = 0

PART-C

II. Answer any TEN questions: (10 × 3 = 30)

Question 26.
Find the greatest number which divides 98, 120, and 153 leaving the same remainder.
Answer:
We have to find the H.C.F of 120 – 98, 153 – 120, 153-98
Required is H.C.F of 22, 33, 55
∴ H.C.F is 11
Hence the greatest number is 11.

Question 27.
Show that the relation “Congruent to” on the set of all triangles is an equivalence relation.
Answer:
Every triangle is congruent to itself
∴ the relation is reflexive
∆T₁ is congruent to ∆T₂ ⇒ ∆T₂ is congruent to ∆T₁ ⇒ R is symmetric
If triangle T, is congruent to T₂, ∆T₂ is congruent to ∆T₃ then ∆T₁ is congruent to ∆T₃
∴ the relative is transitive Hence the relation “congruent to” is an equivalence relation.

Question 28.
If A = {1, 3, 5}, B = {5, 7}, C = {7} verify that A ∪ (B ∩ C) – (A ∪ B) ∩ (A ∪ C)
Answer:
A ∪ (B ∩ C) = {1, 3, 5} ∪ {7} = {1, 3, 5, 7}
(A ∩ B) ∩ (A ∪ C) = {1, 3, 5, 7} ∩ {1, 3, 5, 7} = {1, 3, 5, 7}

Question 29.
If p^x = q^y = r^z = s^w and pq = rs then prove that \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}+\frac{1}{w}\)
Answer:
Let p^x = q^y = r^z = s^w = k
⇒ p = k^1/w, q = k^1/y, r = k^1/z, s = k^1/w
Also given pq = rs
⇒ k^1/xk^1/y= k^1/zk^1/w
⇒ k^1/x+1/y = k^1/z+1/w
⇒ \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}+\frac{1}{w}\)

Question 30.
Find the nature of roots of the equation 3x² + 2x + 1 = 0 without solving. If α and β are the roots of 3x² + 2x + 1 = 0 find the value of \(\).
Answer:
Δ = b² – 4ac = 4 – 12 = -8 < 0 ∴ the roots are imaginary
α + β = \(\frac{-b}{a}=\frac{-2}{3}\) & α.β = \(\frac{c}{a}=\frac{1}{3}\)

Question 31.
The sum of four numbers which are in A.P. is 28 and 10 times the least numbers is 4 times the greatest. Find the numbers.
Answer:
Let the 4 numbers are a – 3d, a-d, a+d, a+3d
Given sum = 28

4a – 28 ⇒ a = 1
Also given 10(a – 3d) – 4(a + 3d)
10a – 30d = 4a + 12d
6a = 42 d
42 = 42d ⇒ d = 1
∴ the 4 numbers are 4, 6,8,10

Question 32.
Solve the linear inequalities x + 3y ≤ 3 2x + y, 2, x,y ≥ 0 graphically.
Answer:
Consider x + 3y = 3 .
If x = 0, then y = 1 (0,1)
If y = 0 then x = 3 (3, 0)

2x + y – 2
If x = 0, y = 2 (0, 2)
If y = 0 then x = 1, (1, 0)
Plot the two lines on the graph there

Question 33.
XYZ suppliers buy a machine for Rs. 20,000 the rate of depreciation is 10%. Find the depreciated value of the machine after 3 years. Also, find the amount of depreciation.
Answer:
B = C (1 – r)ⁿ
Given C = 20,000, r = 0.1, n = 3 B = ?
B = 20,000 (1~0.1)³ = ₹ 14,580
the amount of depreciation = 14,580 ₹

Question 34.
The average age of A and B is 18 years, the average age of B and C is 17 years and that of C and A is 20 years. Find A’s age.
Answer:
Given a + b = 36 – 0 → (1)
b + c = 34 (2) → (2)
c + a = 40 → (3)
Adding we get 2{a + 6 + c) = 110 ⇒ a + b + c = 55 → (4)
Equ 4 – Equ 2 gives a = 55 – 34 = 21
Where A’s age = a = 21 yrs

Question 35.
A certain amount invested at 4% p.a. compounded semi-annually amount to Rs. 78030 at the end of one year. Find the sum.
Answer:
F = 78,030, r = 4% = 0.04, n = 1 yr
F = P\(\left(1+\frac{r}{2}\right)^{n \times 2}\)
78,030 = P\(\left(1+\frac{0.02}{2}\right)^{2}\)
P = \(\frac{78.030}{(1.01)^{2}}\) = ₹75,000

Question 36.
Prove that \(\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}=\frac{1+\cos \theta}{\sin \theta}\)
Answer:
L.H.S = \(\sqrt{\frac{\sec \theta+1}{\sec \theta-1}} \times \frac{\sqrt{\sec \theta+1}}{\sqrt{\sec \theta+1}}\)
\(\frac{(\sqrt{\sec \theta+1})^{2}}{\sqrt{\sec ^{2} \theta-1}}=\frac{\sec \theta+1}{\tan \theta}=\frac{\frac{1}{\cos \theta}+1}{\frac{\sin \theta}{\cos \theta}}=\frac{1+\cos \theta}{\sin \theta}\) = R.H.S

Question 37.
Derive slope-intercept form of line y = mx + c. Also, write the equation of passing through origin with slope m.
Answer:
Let the line passess thr y-axis at A(0, C) and x-axis at B, and m = tanθ is slope of the line by using one point formula we get y – y₁ = m(x – x₁)
y – c = m(x – 0)
⇒ y – c – mx
⇒ y = mx + c is the required equation of the line.
Also equation passing thr (0, 0) is y = mx

Question 38.
Find the equation of the line which passes through the point (-4, 5) and whose intercepts are equal in magnitude but opposite in sign.
Answer:
Intercept form is \(\frac{x}{a}+\frac{y}{b}\) = 1
Given b = -a
∴ \(\frac{x}{a}+\frac{y}{-a}\) – 1
of x – y = a
This equn passess through (-4, 5)
-4 – 5 = a ⇒ a = – 9
∴ The required equation is x – y = 9

PART-D

IV. Answer any SIX questions: (6 × 5 = 30)

Question 39.
In a college, 2/5th of the students play basketball, and 3/4th play volleyball. If 50 students play none of these two games and 125 play both find the numbers of students who play.
(a) at least one of the two games.
(b) exactly one
Represent the results using the Venn diagram.
Answer:
Let total students = x,
Let A = basket ball and B = volley ball
∴ n(A) = \(\frac{2}{5}\)x
n(B) = \(\frac{3}{4}\)x

Given n (AUB)’ = 50 n(AnB) = 125
n(AUB) = x – x(AUB)’ = x – 50

(a) n(AuB) = n(A) + n(B) – n(A n B)
x – 50 = 0.4x + 0.75x – 125
∴ 75 = 1.15x – 1x = 0.15x
x = \(\frac{75}{0.15}\) = 500
n( A u B) = x – 50 = 500 – 50 = 450

(b) n(only A) + n(only B)
n(A) = \(\frac{2}{5}\) × 500 = 200
= 200 – 125 + 375 – 125

= 325

Question 40.
Evaluate \(\frac{1.234 \times .0 .8921}{43.43 \times 0.0092}\) using logarithm table.
Answer:
Let x = \(\frac{1.234 \times .0 .8921}{43.43 \times 0.0092}\)
Taking logm both sides we get
logx = log 1.234 + log 0.8921 – log43.43 – log 0.0092
= 0.931 + 1.9504 – 1.6378 – 3.9638
logx = 0.4401
x = A.L(0.440l)
x = 2.755.

Question 41.
Ms. Sneha buys a second-hand bike for Rs. 18,000 she pays Rs. 12,000 cash and agrees to pay the balance in 12 annual installments of Rs. 500 each plus 10% interest on the unpaid amount. How much will the bike cost her?
Answer:
Balance amount 18,000 – 12,000 = 6,000 Rs.
No. of instalment = 12, Each unit amt = Rs. 500
1st instalment = 500 + 6000 × 0.1 = 1,100
2nd instalment = 500 + 5000 × 0.l = 1050
3rd instalment = 500 + 5000 × o. 1 = 1000
Cost of instalments = 1,100 + 1050 + 1000 +……………. 12 instalment
The terms are in A.P with d = -50, a = 1,100
∴ S_m = \(\frac{12}{2}\)[1,100 + (12-1) × -50] = 6(1,100 – 550) = 6 × 550 = ₹3300
∴ Total cost= 12,000 +3,300 = ₹15,300

Question 42.
Obtain a root of the equation x³ – 2x² -2x + 3- 0 by inspection and then solve using synthetic division method.
Answer:
f(x) = x³ – 2x² – 2x + 3
f(1) = 0 ∴ 1 is a root
∴ by sythetic division we have

∴ Reduced equation is x² – x – 3 = 0
∴ Roots are x = \(\frac{1 \pm \sqrt{13}}{2}\)
Hence the 3 roots are 1, \(\frac{1+\sqrt{13}}{2}\), \(\frac{1-\sqrt{13}}{2}\)

Question 43.
A person spent 30% of his wealth and thereafter ₹ 20,000 and a further 10% of the remainder. If 29,250 is still remaining, what was his total wealth amount?
Answer:
Let total wealth be Rs. x

0.63x = 18,000
x = ₹ 75,000

Question 44.
A father wants to send his child for higher studies after 15 years. He expects the cost of higher studies to be Rs. 1,00,000. How much should he share annually to have one lakh after 15 years if the interest rate ¡s 12% C.I. per annum?
Answer:
Given data F = 1,00000, a = ?, n = 15, r = 0.12
F = \(\frac{a\left[(1+i)^{n}-1\right]}{i}\)
1,00,000 = a\(\frac{\left[(1+0.12)^{15}-1\right]}{0.12}\)
a = \(\frac{1,00000}{(1.12)^{75}-1}\) = ₹ 2682.47

Question 45.
Find x if \(\frac{x \sin ^{2} 300^{\circ} \cdot \sec 240^{\circ}}{\cos ^{2} 225^{\circ} \cdot {cosec} 240^{\circ}}\) = cot 135° tan 315°
Answer:
sin 300 = sin (360 – 60) = – sin 60 = \(\frac{\sqrt{3}}{2}\)
sec 240 = sec(180 + 60)= – sec60°= – 2
cos 225 = cos(180 + 45°) – cos45°= – \(\frac{1}{\sqrt{2}}\)
cosec 240 = cosec (180 + 60°) = – cosec 60 = – \(\frac{2}{\sqrt{3}}\)
cot 315° = cot (360 – 45°) = – cot 45° = – 1
tan 300 = tan (360 – 60°) = – tan 600 = – \(\sqrt{3}\)

∴ The given equation becomes:
\(\frac{x \cdot \frac{3}{4} \cdot 4}{\frac{1}{2}, \frac{4}{3}}\) = 1.3
⇒ x = \(\frac{6}{9}=\frac{2}{3}\)

Question 46.
In what ratio is the line segment joining the points (4, 5) and (1, 2) is divided by the x-axis. Also, find the coordinates of the point of division.
Answer:
Point on x-axis is y = 0
Let P(x, y) divide AB in
the ratio k: 1 ratio
∴ P(x, y) = \(\left(\frac{4 k+1}{k+1}, \frac{5 k+2}{k+1}\right)\)
(x, 0) = \(\left(\frac{4 k+1}{k+1}, \frac{5 k+2}{k+1}\right)\)
⇒ \(\frac{5 k+2}{k+1}\) = 0 ⇒ k = \(\frac{-2}{5}\)
ratio is 2: 5 externaily
∴ P = (-1, 0)

Question 47.
Find the equation for the locus of a point that moves such that the sum of its distances from (0, 3) and (0, -3) is 8 units.
Answer:
Let P(x, y)be a moving point
Given |PA| + |PB| = 8
|PA| = \(\sqrt{x^{2}+(y-3)^{2}}\) & |PB| = \(\sqrt{x^{2}+(y+3)^{2}}\)
|PA| = 8 – |PB|
S.B.S We get
x² + (y – 3)² = \(\left(8-\sqrt{x^{4}+(y+3)^{2}}\right)^{2}\)
Simplifying we get the required equation of the locus 16x² + 7y² = 112.

Question 48.
Show that the points (2, -2), (8, 4), (5, 7), and (-1, 1) are taken murder the vertices of a rectangle.
Answer:
AB = \(\)
Similarly BC = \(\sqrt{18}\) , CD = \(\sqrt{72}\) , DA = \(\sqrt{18}\) , AC = BD = \(\sqrt{90}\)
Opposite sides are equal and diagonals are equal

PART-E

V. Answer any ONE question : (6 × 5 = 30)

Question 49.
(a) If x = a r sin A cos B, y = br sin A sin B, z = cr cos A, prove that \(\)
Answer:

= r²[sin² A(cos²B + sin²B) + cos²A] = r² (sin² A + cos² A)
= r² = RHS.

(b) Prove that the lines x + y + 4 = 0, 2x = 3y + 7 and 3x + y = -6 are concurrent. Also find the point of concerrency.
Answer:
Solving equations 1 and 2 we get
x + y + 4 = 0 …(1)
2x – 3y – 7 = 0 …(2)
x = -1 & y = -3
Put x = -1 and y = -3 in eqn (3) We get
3x + y = -6
3(-1) + -3 = -6 – 6 = – 6
Hence the 3 lines are concurrent and the point of concurrency (-1, -3)

(c) Find the present value of a perpetuity of Rs. 3000 to be received forever at 4 p.a.
Answer:
Given A = 3000, r = 0.04, P∞ = ?
P∞ = \(\frac{A}{r}=\frac{3000}{0.04}\) = 75,000
∴ Present value of perpetuity = 75,000

Question 50.
(a) Find the sura of all numbers between 50 and 200 which are divisible by 11.
Answer:
Numbers are 55, 66, 77, ……………….. 198
which are in A.P
Here a = 55, T_n = 198, d= 11
198 = 55 + (n – 1)11
n = 14
∴ S_n = \(\frac{14}{2}\)(55 + 198) = 1771

(b) A company sells ‘x’ tins of talcum powder per day at Rs. 10/tin. The cost of manufacturing a tin is Rs. 6 and distributor charge per tin is Rs. 1. Besides, the daily overhead cost is Rs. 600. Determine
(i) Cost function
(ii) Revenue function
(iii) Profit function
(iv) Break even level of production.
Answer:
(i) C(x) =V.C + E.C
= 6x + x + 600
∴ C(x) = 7x + 600

(ii) R(x) = Price x Quantity
R(x) = 10x

(iii) P(x) = R(x) – C(x)
= 10x – 7x – 600
P(x) = 3x – 600

(iv) At BEP:P(x) = 0 or R(x) = C(x)
3x – 600 = 0
x = 200 units

∴ Break even level of production 200 units

(c) Form the LCM of 2/3, 4/5, 6/7
Answer:
LCM of fractions = \(\frac{\text { LCM of NR }}{\text { HCF of DNR }}=\frac{\text { LCM of } 2,4,6}{\text { HCF of } 3,5,7}=\frac{12}{1}\) = 12

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2017 (North), Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2017 (North)

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any TEN questions. (10 × 1 = 10)

Question 1.
Give the canonical representation of 825.
Answer:
825 = 3¹ × 5² × 11.

Question 2.
Convert the set A = {a, e, i, o, u} from roster form into rule form.
Answer:
A = {x/x is a vowel in English alphabet}

Question 3.
If A = {a, b, c}; B = {c, d}. Find B × A.
Answer:
B × A = {c, d) x {a, b, c}
= c, a) (c, b) (c, c) (d, a) (d, b) (d, c)}.

Question 4.
Simplify: \(\frac{a^{m+n} \cdot a^{2 m-n}}{a^{m-n}}\)
Answer:

Question 5.
Express Y = 27 in the .Lograrithmic Form.
Answer:
log₃ 27 = 3

Question 6.
Find the 8th term of an A.P -2, -4, -6 ………..
Answer:
T = a + (n – 1)d
T₈ = -2 + (8 – 1)(-2)
T₈ = -2 – 14 = -16

Question 7.
Solve: \(\frac{x+2}{x-2}=\frac{5}{2}\)
Answer:
2(x + 2) = 5(x – 1)
2x + 4 = 5x – 5
4 + 5 = 5x – 2x ⇒ 3x = 9 ⇒x = 3.

Question 8.
Define perpetuity.
Answer:
An annuity that is payable forever is called a perpetuity.

Question 9.
Convert 42% into a decimal.
Answer:
42% = \(\frac{42}{100}\) = 0.42.

Question 10.
The average score of 35 girls is 80 and the average score of 25 boys is 68. Find the average score of both boys and girls together.
Answer:
X̄₁₂ = \(\frac{35 \times 80+25 \times 68}{35+25}=\frac{2800+1700}{60}\) = 75

Question 11.
Convent 75° into radians.
Answer:

Question 12.
If the distance between the points (3, -2) and (-1, a) is 5 units. Find the values of
Answer:
\(\sqrt{(3+1)^{2}+(-2-a)^{2}}\) = 5
42 + 4 + a² + 4a = 5
a² + 4a + 15 = 0
Using formula method we get the values of a.

Part – B

II. Answer any TEN questions. (10 × 2 = 20)

Question 13.
Find the L.C.M of 12, 21 and 24.
Answer:

∴ L.C.M = 2³ × 3 × 7 = 24 × 7 = 168

Question 14.
Find the H.C.F of 18 and 24.
Answer:
18 = 2 × 3², 24 = 2³ × 3
H.C.F = 2 × 3 = 6

Question 15.
If A = {a, b,c, d}\ B = {d, e, f, g, h, i}. Find A – B & B – A.
Ans.wer:
A – B = {a, b, c}, B – A = {e, f g, h, i}

Question 16.
Find the value of x so that the slope of the line joining the points (2, 5) and (x 13) is 20.
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
\(\frac{13-5}{x-2}\) = 20 ⇒ \(\frac{8}{x-2}=\frac{20}{1}\)
⇒ 20x – 40 = 8 ⇒ 20x = 48
⇒ x = \(\frac{48}{20}=\frac{12}{5}\) = 2.4

Question 17.
Find the common difference of an A.P. whose first term is 6 and 12th term is 72.
Answer:
Given a = 6 and T₁₂ = a + 11d = 72
6 + 11d = 72,
11d = 72 – 6 = 66 ⇒ d = 6.

Question 18.
The sum of two consecutive numbers is 23. Find them.
Answer:
Let the 2 consecutive numbers be x and x + 1
Given sum = 23
x + x + 1 = 23
2x = 22 ⇒ x = 11
∴ The 2 consecutive numbers are 11 & 12.

Question 19.
Solve 2x + 6 < 0; x ∈ Z inequalities in one variable and represent the solution on the number line.
Answer:
Answer:
Given 2x + 6 < 0
∴ 2x ≤ -6 ⇒ x ≤ -3

Solution set is [-∞, -3].

Question 20.
Find the interest on Rs. 1,500 at 4% pa. for 145 days.
Answer:
S.I.= 1500 × \(\frac{145}{365} \times \frac{4}{100}=\frac{870000}{36500}\) =23.836.

Question 21.
If x = a secθ; y = btanθ prove that \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}\) = 1
Answer:

⇒ sec² θ – tan² θ = 1 = R.H.S

Question 22.
The weight of 6 men is 90 Kg, 70.5, 56 Kg, 45.5Kg, 85 Kg, and 78 Kg. Find average weight.
Answer:
X̄ = \(\frac{90+70.5+56+45.5+85+78}{6}=\frac{425}{6}\) = 70.83

Question 23.
Karthik received a scholarship of ₹ 5,000 in 2011 and ₹ 8,000 in 2012. Find the percentage increase.
Answer:
Diff in the amount = 8000-5000 = 3000₹

Question 24.
Prove that tan² A(1 – sin²A) = sin² A
Answer:

Question 25.
Find the area of the triangle whose vertices are A(3, 4); B(2, -1), and C(4, -6).
Answer:
Area of ∆ABC = \(\frac{1}{2}\)[x₁(y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac{1}{2}\)[3 (-1 + 6) + 2 (-6 – 4)+4 (4 – (-1)].
A = \(\frac{1}{2}\)[3(5) + 2(-10) + 4(5)]

Part – C

III. Answer any TEN questions. (10 × 3 = 30)

Question 26.
If the product of two numbers is 216 and their L.C.M is 36. Find their H.C.F.
Answer:
Given L.C.M of 36 and Product = 216, H.C.F = ?
H.C.F = \(\frac{a \times b}{L . C . M}=\frac{216}{36}\) = 6

Question 27.
If A= {1, 2}; B = {2, 3}; C = {3, 4}. Find (A × B) ∩ (A × C).
Answer:
A × B = {1,2} × {2,3}{(1,2) (1, 3) (2,2) (2, 3)}
A × C = {1, 2} × {3, 4} – {(1,3) (1, 4) (2,3) (2,4)}
(A × B) ∩ (A × C) = {(1, 3) (2, 3)}.

Question 28.
If A = {2, 3, 4} write ail the proper subsets of ‘A’.
Answer:
Proper subsets of A
X = P(A) = {{Q} (2) (3) (4) (2, 3) (3, 4) (4, 2)}.

Question 29.
Prove that (x^b-c)^a . (x^c-a)^b . (x^a-b)^c = 1.
Answer:

Question 30.
If the second term of the G.P is 6 and 5th term is 162. Then find the GP.
Answer:
T₂ = ar = 6 ….(1), T₅ = ar⁴ = 162 ….(2), G.P. = ?

ar = 6 ⇒ 3a = 6 ⇒ a = 2
Terms of G.P are a, ar ar², ar³
2, 2 × 3, 2 × 3², 2 × 3³………….
2, 6, 18…………..

Question 31.
The sum of 6 times a number and 5 times the number is 55. Which is that number?
Answer:
Let the number be x
Given 6x + 5x = 55
11x = 55 ⇒ x = 5
∴ The number is 5.

Question 32.
Solve graphically:
3x – 7 > 2(x – 6) and 6 – x > 11 – 2x; x∈R
Answer:
3x – 7 > 2x – 12
3x – 2x > -12 + 7
x > -5

6 – x > 11 – 2x
-x + 2x >11 – 6
x > 5

Question 33.
Find the present value of an annuity due of 8000 for 5 years at 5% p.a.
Answer:
P = \(\frac{a\left[(1+i)^{n}-1\right]}{i(1+i)^{n}}\)(1 + i)
P = \(\frac{8000\left[(1+0.05)^{5}-1\right]}{0.05(1+0.05)^{5}}\) × (1 + 0.05)
= \(\frac{8000\left[(1.05)^{5}-1\right](1.05)}{0.05 \times(1.05)^{5}}=\frac{8000[(1.2763-1)](1.05)}{0.05 \times 1.2763}\) = ₹ 36369.50

Question 34.
Calculate the arithmetic average mark from the following data.

Answer:
X̄_w = \(\frac{45 \times 11+75 \times 10+60 \times 15+55 \times 12+93 \times 2}{11+10+15+12+2}\)
= \(\frac{495+750+900+660+186}{50}=\frac{2991}{50}\) = 59.82 ≈ 60

Question 35.
In an election, the winning candidate got 4,800 votes which are 80% of the total votes. Calculate the total number of votes.
Answer:

∴ Total number of votes = 6000

Question 36.
If tan θ = \(\frac{a}{b}\) show that \(\frac{a \sin \theta-b \cos \theta}{a \sin \theta+b \cos \theta}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\)
Answer:

Question 37.
A point ‘P’ moves such that PA² = 3PB². If A = (5, 0) and B = (-5, 0). Find the equation of the Locus of ‘P’
Answer:
Let P(x, y) be the point on the locus
Given PA² = 3PB² and A = (5, 0), B = (-5, 0)
(x – 5)² + (y – 0)² = 3[(x + 5)² + (y – 0)²]
x² + 25- 10x + y² = 3[x² + 25 + 10x + y²]
3x² + 75 + 30x + 3y² = x² + y² – 10x + 25
2x² + 2y² + 40x + 50 = 0 is the required equation.

Question 38.
Find the point of intersection of the lines 3x = 2y – 5 = 0 and 4x – y – 3 = 0.
Answer:
3x + 2y – 5 = 0
4x – y – 3 = 0 → Multiplying by 2

11x – 11 = 0 ⇒ x = \(\frac{11}{11}\) = 1
y = 4x – 3 = 4 – 3 = 1.
∴ Point of intersection is (1,1).

Part – D

IV. Answer any SIX questions. (6 × 5 = 30)

Question 39.
Out of 50 people, 20 people drink tea, 10 take both tea and coffee. How many take at least one of the two drinks.
Answer:
Given n(T) = 20, n(T∪C) = 50, n(T∩C) = 10, n(C) = ?

n(T∪C) = n(T) + n(C) – n(T∩C)
50 = 20 + n(C)-10
n(C) = 40
∴ No. of people taking atleast one of the two drinks = 10 + 10 + 30 = 50.

Question 40.
If log\(\left(\frac{a-b}{4}\right)\) = log\(\sqrt{a}\) + log\(\sqrt{b}\) show that (a + b)2 = 20ab.
Answer:
Given log\(\left(\frac{a-b}{4}\right)\) = log\(\sqrt{a}\) + log\(\sqrt{b}\)
log\(\left(\frac{a-b}{4}\right)\) = log\(\sqrt{ab}\)
\(\frac{a-b}{4}=\sqrt{a b}\) S.B.S
(a – b)² = 16ab
a² + b² – 2ab = 16ab (Add 4ab both sides)
a² + b² + 2ab = 20ab,
(a + b)² = 20ab.

Question 41.
Find the sum of n terms of GP. 5 + 55 + 555 +……………
Answer:
Let S = 5 + 55 + 555 +…………… to n terms
\(\frac{\mathrm{S}}{5}\) = 1+ 11 + 111+………..to n terms
\(\frac{9 \mathrm{~S}}{5}\) = 9 + 99 + 999 +…………… to n terms
=(10 – 1) + (100 – 1) + (1000 – 1) +……….. to n terms
= (10 + 100 + 1000 +………… to n terms) – (1 + 1 + 1………. to n ternis)
= \(\frac{10\left(10^{n}-1\right)}{9}\) – 1 × n
S = \(\frac{5}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]\)

Question 42.
Find the integral root between -3 and 3 by inspection and then using synthetic division, solve the equation x³ – 2x² – 5x + 6.
Answer:
Let f(x) = x³ – 2x² – 5x + 6,
f(1)= 1 – 2 – 5 + 1 = 0
∴ x = +1 is a root of the given equation. Let us remove this root by synthetic division.

∴ The resulting equation is x² – x – 6 = 0 is the quotient and remainder is 0.
x² – x – 6 = 0
(x – 3) (x + 2) = 0 ⇒ x = -2 or 3
Thus x = 1, -2, 3 are the roots of the given equation.

Question 43.
At what time a sum of ₹1,200 will earn ₹573 as compound interest at the rate of 5% p.a. if the interest is added annually?
Answer:
Given P=1200, C.I = 573, A = 1200 + 573 = 1773
i = \(\frac{5}{100}\) = 0.05, n = ?
A = P(1 + i)ⁿ
1773 = 1200(1 +0.05)ⁿ
\(\frac{1773}{1200}\)1.4775 = (1.05)ⁿ
(1.05)n = (1.05)8 ⇒ n = 8.

Question 44.
Preritha wants to buy a house after 5years when it is expected to cost 50 lakhs. How much should she save annually if her savings earn a compound interest of 12 percent?
Answer:
Given F = 50,00,000, n = 5, i = 0.12, a = ?
F = \(\frac{a\left[(1+i)^{n}-1\right.}{i}\) ⇒ 50,00,000 = \(\frac{a\left[(1+0.12)^{5}-1\right]}{0.12}\)
50,00,000 = 6.3528a ⇒ a = \(\frac{50,00,000}{6.3528}\) = ₹ 787054.5

Question 45.
If θ = \(\frac{5}{2}\) and θ is acute then prove that \(\frac{3 \cos \theta+2 \sin \theta}{3 \cos \theta-4 \sin \theta}=\frac{19}{7}\)
Answer:
Given cotθ = \(\frac{5}{2}=\frac{a d j}{o p p}\)
∴ Hyp = \(\sqrt{5^{2}+2^{2}}=\sqrt{29}\)
cosθ = \(\frac{5}{\sqrt{29}}\), sinθ = \(\frac{2}{\sqrt{29}}\)
L.H.S = \(\frac{3 \cos \theta+2 \sin \theta}{3 \cos \theta-4 \sin \theta}\)

Question 46.
The price of a pair of trousers was decreased by 22% to 390. Whast was the original price of the trousers?
Answer:
Let the price of trouser = x
It was decreased by 22% to ₹ 390
∴ x – \(\frac{22 x}{100}\) = 390
\(\frac{100 x-22 x}{100}\) = 390 ⇒ 78x = 390 × 100
x = \(\frac{39,000}{78}\) =500
∴ Original price of the trouser = ₹500.

Question 47.
Find the coordinates of the vertices of the triangle. Given the midpoints of the sides as (4, -1); (7, 9) & (4, 11).
Answer:
Let A, B, C be the vertices of the A,e, and D, E, and F be the mid-points of the sides BC, CA, and AB respectively.
D = Midpoint of BC
(4, -1) = \(\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)\)

⇒ x₂ + x₃ = 8, y₂ + y₃ = 2 …………….(1)
E = Mid point of CA
(7, 9) = \(\left(\frac{x_{3}+x_{1}}{2}, \frac{y_{3}+y_{1}}{2}\right)\)

⇒ x₃ + x₁ = 14, y₃ + y₁ = 18 ……………..(2)
F = Mid point of AB
(4,11) = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)
⇒ x₁ + x₂ = 8, y₁ + y₂ = 22 ………………..(3)

Solving equations 1,2 and 3 we get
x₁ = 7, x₂ = 1, x₃ = 7
y₁ =21, y₂ = 1, y₃ = -3
Thus A = (7, 21), B = (1, 1) and C = (7, -3).

Question 48.
Find ‘a’ so that the lines x – 6y + a = 0; 2x + 3y + 4 = 0 and x + 4y + 1 = 0 concurrent.
Answer:
Given 2x + 3y + 4 = 0 ………….. (1)
x + 4y + 1 = 0 …………. (2) × 2
x – 6y + a = 0 …………. (3)
Solving 1 and 2 we get

= \(\frac{-8}{5}\) – 1 = \(\frac{-8-5}{5}=\frac{-13}{5}\)

Put x = \(\frac{-13}{5}\) and y = \(\frac{2}{5}\) in equation 3
We get x – 6y + a = 0
a = 6y – x
a = 6.\(\frac{2}{5}-\left(\frac{-13}{5}\right)\)
= \(\frac{12}{5}+\frac{13}{5}=\frac{25}{5}\) = 5
∴ a = 5

Part – E

V. Answer any ONE question. (1 × 10 = 10)

Question 49.
(a) If U = {1, 2, 3, 4, 5, 6, 7, 8, 9} A = {1, 2, 3, 4, 5} B = {3, 4, 5, 6, 7}
Show that (A ∩ B)’ =A’∪B’
Answer:
A ∩ B = {3,4, 5}
(A ∩ B)’ = {1,2, 6, 7, 8,9} ………….(1)
A’∪B’ = {6,7, 8,9} ∪ {1,2, 8, 9}
= {1,2, 6,7, 8, 9} ………………(2)
From 1 and 2 we get A’∪B’ = (A ∩ B)’.

(b) Insert G. Means [Geometric Mean] between \(\frac{1}{4}\) and \(\frac{1}{64}\).
Answer:
G.M = \(\sqrt{\frac{1}{4} \times \frac{1}{64}}=\sqrt{\frac{1}{4^{4}}}=\frac{1}{4^{2}}=\frac{1}{16}\)

(c) If a^x = b; & b^y = c; c^z = a. Show that xyz = 1.
Answer:
Consider a^x = b’
(c^z)^x = b
c^zx = b
(b^y)^zx = b’ ⇒ & bxyz = b’⇒ xyz = 1.

Question 50.
(a) Find the equation of the locus of the point which moves such that its distance from 3x – 4y + 1 = 0 is equal to its distance from (1, -1).
Answer:
Let A = (1, -1) and P(x, y) be any point on the locus, then PA = distance from the line 3x – 4y +1 = 0
\(\sqrt{(x-1)^{2}+(y+1)^{2}}=\left|\frac{3 x-4 y+1}{\sqrt{9^{2}+16^{2}}}\right|\) S.B.S

(x – 1)² + (y + 1)² = \(\frac{(3 x-4 y+1)^{2}}{25}\)
25(x² – 2x + 1 + y² + 2y + 1) = 9x² + 16y² + 1 – 24xy – 8y + 6x
16x² + 24xy + 9y – 56x + 58y + 49 =0 is the required equation of the locus.

(b) A manufacturer produced and sells balloons at 8 per unit. His fixed cost is ₹6,500 and the variable cost per balloon is ₹3.50. Calculate
(i) Revenue Function (ii) Cost Function
(iii) Profi Function (iv) Break even point.
Answer:
i.e. (i) R(x) = 8x

(ii) C(x) = 3.5x + 6500

(iii) P(x) = 4.5x – 6500
i.e.,P(x) = 6500 = 0

(iv) At BEP P(A) = 0
4.5x – 6500 = 0
4.5x = 6500
x = 144.4 units

(c) What is the present value of an income of 3000 to be received forever if the interest rate if 14% p.a.
Answer:
P_∞ = \(\) = ₹21428.5

Students can Download 1st PUC Basic Maths Previous Year Question Paper March 2017 (South), Karnataka 1st PUC Basic Maths Model Question Papers with Answers helps you to revise the complete syllabus.

Karnataka 1st PUC Basic Maths Previous Year Question Paper March 2017 (South)

Time: 3.15 Hours
Max. Marks: 100

Instructions:

1. The questions paper consists of five parts A, B, C, D, and E.
2. Part – A carries 10 marks, Part – B carries 20 marks, Part – C curries 30 marks, Part – D carries 30, and Part – E carries 10 marks.
3. Write the question numbers properly as indicated in the questions paper

PART-A

I. Answer any TEN questions. (10 × 1 = 10)

Question 1.
Write the imaginary part of 3 – Imaginary part of 3 – i is -1.
Answer:
The imaginary part of 3 – i is -1

Question 2.
If A has 4 elements, how many elements, how many elements will P(A) have.
Answer:
P(A) will have 16 Elements.

Question 3.
If R-1 = {(2, 4) (1, 2) (3, 1) (3, 2)} find R.
Answer:
R= {(4, 2) (2,1) (1,3) (2, 3)}.

Question 4.
Simplify: (2²)² – 2^(3°)
Answer:
= 1 – 2′ = 1 – 2 = -1

Question 5.
Solve for x if log₇ x = 2.
Answer:
x = 7² = 49.

Question 6.
Find the sum to infinity of the GP 3,1, \(\frac{1}{3}\), ……………
Answer:
S_n = \(\frac{a}{1-r}=\frac{3}{1-\frac{1}{3}}=\frac{9}{3-1}=\frac{9}{2}\)

Question 7.
Form the quadratic equation whose roots are 2, 3.
Answer:
(x – 2)(x – 3) = 0 ⇒ x² – 5x + 6 = 0.

Question 8.
What is the simple interest on 650 for 14 weeks at 6% p.a?
Answer:
S.I = 650 × \(\frac{14}{52} \times \frac{6}{100}\) = 10.5

Question 9.
The average score of 35 girls is 80 and the average score of 25 boys ¡s 68. Find the average score of both boys and girls together.
Answer:

Question 10.
By selling a book at’ 250 the profit made is ₹ 50. What is the cost price of the book?
Answer:
C.P = S.P – Profit = 250 – 50 = 200₹.

Question 11.
Convent 315° to radians.
Answer:

Question 12.
Find the slope of the line 2x + 5y – 11 = 0.
Answer:
Slope = –\(\frac{2}{5}\)

Part – B

II. Answer any TEN questions. (10 × 2 = 20)

Question 13.
Find the number of positive divisors of 360.
Answer:
360 = 2³ × 3² × 5¹
n = P₁^α₁. P₂^α₂. P₃^α₃
T(n) = T(360) = (1 + α₁) (1 + α₂) (1 + α₃) = (3 + 1) (2 + 1) (1 + 1) = 24.
∴ If has 24 positive divisors.

Question 14.
Find the HCF of two numbers if their LCM is 1260 and product is 52920.
Answer:
H.C.F = \(\frac{a \times b}{\mathrm{LCM}}=\frac{52920}{1260}=\frac{5292}{126}\) = 42

Question 15.
If f(x) = x, g(x) = x³ + 1 find (a) fog (2) (b) gof (1)
Answer:
fog (2) = f(g (2)) = f( 2³ + 1) = f( 9) = 9
gof(1) – g (f(1)) = g (1) = 1³ + 1 = 1 + 1 = 2.

Question 16.
Find the domain and range of the relation R = {(x, y): y = x², x is a positive prime number less than 10}
Answer:
R= {(2,4)'(3, 9) (5,25) (7, 49)}
Domain – {2, 31 5, 7}, Range = {4,9,25, 40}.

Question 17.
If a = 3^x b = 3^y c = 3z and ab – c². P.T. x + y = 2z.
Answer:
Given ab = c²
3^x.3^y = (3^2z) ⇒ 3^x+y = 3^2z
x + y = 2z.

Question 18.
Insert 3 Arithmetic means between -2 and -10.
Answer:
Let A₁, A₂, A₃ are 3 A.m between -2 and -10
∴ -2, A₁, A₂, A₃………. -10 are in A.P .
a a + d a + 2d a + 3d a + Ad are in A.P
∴ a = – 2 and a + 4d = – 10 ⇒ 4d = -10 + 2 = -8
d = – 2 .
Hence A₁ = -2 – 2 = -4,
A₂ = a + 2d = – 2 – 4 = – 6,
A₃ = a + 3d = -2 – 6 = -8
Hence 3 Ams are -4, -6, and -8.

Question 19.
If a and b are the roots of the equation 2x² + 5x + 5 = 0 the find the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\)
Answer:
Given 2x² + 5x + 5 = 0
α + β = \(\frac{-b}{a}=\frac{-5}{2}\)
αβ = \(\frac{c}{a}=\frac{5}{2}\)
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{-5 / 2}{5 / 2}\) = -1

Question 20.
Solve 7x + 3 < 5x + 9, x∈R and represent on number line.
Answer:
7x + 3 < 5x + 9

7x – 5x <9 – 3
2x < 6
x < 3
∴ Solution set is (- ∞, 3)

Question 21.
Find the effective rate of interest when a sum lent at 12% is computed half yearly.
Answer:
r = \(\left(1+\frac{i}{q}\right)^{q}\) – 1 = \(\left(1+\frac{0.12}{2}\right)^{2}\) – 1
= (1 + 0.06)2 – 1 =(1.06)2 – 1 = 12.36.

Question 22.
The average age of 12 boys is 8 years. Another boy of 21 years joins the group. Find the average age of the new group.
Answer:
Ave age of 12 boys = 8 yrs.
∴ Total age of boys = 8 x 12 = 96 yrs.
When a boy of 21 yrs joins the group
∴ Total age of the 13 boys = 96 + 21 = 117 yrs.
∴ Average age of the new group = \(\frac{117}{13}\) = 9 yrs.

Question 23.
Find the value of sin (480°) + tan 135°.
Answer:
sin (480°) + tan (135°) = sin (360 + 120) + tan (180 – 45)
= sin 120-tan 45
= sin (180 – 60) – 1 = sin 60 – 1 = \(\frac{\sqrt{3}}{2}\) – 1

Question 24.
The CEO the triangle ABC is the point (2, 3). The coordinates of A are (5, 6) and B(-l, 4). Find the coordinates of C.
Answer:
G(x, y) = \(\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\), C(x₃, y₃) = ?
(2, 3) = \(\left(\frac{5-1+x_{3}}{3}, \frac{6+4+y_{3}}{3}\right)\)
(2, 3) = \(\left(\frac{4+x_{3}}{3}, \frac{10+y_{3}}{3}\right)\) ⇒ 6 = 4 + x₃,
10 = y₃ = 9
⇒ x₃ – 6 – 4, y₃ = 9 – 10
⇒ x₃ = 2, y₃ = -1
Hence co-ordinates of C(x₃, y₃) = C(2, – 1)

Question 25.
Find equation of line passing through (0, -4) and making an angle of 30° with the x-axis.
Answer:
Required equation is y – y₁ = m(x – x₁)
i.e., (y – (-4) = tan 30° (x = 0)
y + 4 = \(\)(x)
x = \(\sqrt{3}\)y + 4\(\sqrt{3}\) = 0

Part-C

III. Answer any TEN questions. (10 × 3 = 30)

Question 26.
Find the number which when divided by 36, 40 and 48 leaves the same remainder 5.
Answer:
Let us find out the L.C.M of 36,40 and 48
36 = 2² × 3², 40 = 2³ × 5¹, 48 = 2⁴ × 3
∴ LCM = 2⁴ × 3² × 5¹
= 16 × 9 × 5 = 720
Since 5 has to be the remainder. We have to add 5 to 720 i.e., 720 + 5 = 725

Question 27.
If A = {1, 2, 3} and R = {(1, 1) (1, 2) (2,1) (2, 2) (3, 3)} Prove that R is an equivalence relation on A.
Answer:
R = { (1, 1) (1, 2) (2, 1) (2, 2) (3, 3)}
R is reflexive ∵ (1, 1) (2, 2) (3, 3) ∈ R
R is Symmetric ∵ (1, 2) ∈ R => (2, 1) ∈ R
R is transitive ∵ (1, 2) (2, 1) ∈ R => (1, 1) ∈ R

Question 28.
If a^1/3 + b^1/3 = c^1/3 Prove that (a + b + c)³ = 27 abc.
Answer:
Given a^1/3 + b^1/3 = -c^1/3
Cubing both sides we get
(a^1/3 + b^1/3)³ = -c
a + b + 3a^1/3b^1/3(a^1/3 + b^1/3)³ = -c
a + b + c = -3a^1/3, b^1/3, c^1/3.
Cubing both sides we get
{a + b + c)³ = 27 abc.

Question 29.
Solve: log₂ x + log₄ x = 3.
Answer:
\(\frac{\log x}{\log 2}+\frac{\log x}{2 \log 2}\) = 3
\(\frac{2 \log x+\log x}{2 \log 2}\) = 3
⇒ 3 log x = 6 log 2 ⇒ log x³ = log 2⁶ ⇒ x³ =(2²)³ ⇒ x = 4

Question 30.
The sum of 3 numbers in A.P. is 15 and their product is 105. Find the three numbers.
Answer:
Let the 3 numbers be a – d, a, a + d
Given

3a = 15 ⇒ a = 5
Also (a – d) (a) (a + d) = 105
5(25 – d²) = 105
25 – d² = 21 ⇒ 25 – 21 = d² ⇒ d² = 4 ⇒ d = 2
The 3 numbers are 5 – 2, 5, 5 + 2 = 3, 5, 7.

Question 31.
The age of a father is 5 times that of a son. 3 years ago the age of the father was 8 times that of his son. Find their present ages.
Answer:
Let the present age of the son be x yrs.
Let the present age of father be 5x yrs.
3yr ago, the age of son is x – 3 and that of father is 5x – 3
Given 5x – 3 = 8 (x – 3)
5x – 3 = 8x – 24
24 – 3 = 8x – 5x = 3x
x = 21 ⇒ x = 7
∴ The present age of son = 7 yrs and father is 35 yrs.

Question 32.
Solve graphically:
x + 3y ≥ 3; 2x + y ≤ 2; x ≥ 0, y ≥ 0.
Answer:
Let x + 3y = 3,
Put x = 0, y= 1, (0,1)
Put y = 0 x = 0, x = 3, (3,0)

2x + y = 2,
Put x = 0, y = 2, (0,2)
Put y = 0, x = 1, (1, 0)
Plot the two lines in the graph we get

Question 33.
Find the future value of an annuity of 5000 at 12% p.a. for 6 years.
Answer:
F = \(\frac{a\left[(1+i)^{n}-1\right]}{i}\) = 500\(\frac{\left[(1+0.12)^{6}-1\right]}{0.12}\) = 500\(\frac{\left[(1.12)^{6}-1\right]}{0.12}\) = 40575.9 Rs
Future value = 4.575.9

Question 34.
Ritu’s salary was increased by 10% and then again by 5%. If the present salary is ₹ 9,240. What was Ritu’s previous salary?
Answer:
Let his salary be x

\(\frac{220 x+11 x}{200}\) = 9240 ⇒ 231x = 9240 × 200
x = \(\frac{9240 \times 200}{231}\) = 8000

Question 35.
Prove that \(\frac{1}{1+\cos A}+\frac{1}{1-\cos A}\) = 2cosec²A
Answer:
\(\frac{1}{1+\cos A}+\frac{1}{1-\cos A}\)

Question 36.
Find area of the quadrilateral whose vertices are A(1, 2) B(6, 2) C(5, 3) and D(3, 4) in order.
Answer:
Area of the Quadrilateral ABCD
= Area of triangle ∆ ABC + Area of ∆ACD
= \(\frac{5}{2}\) + 3 = \(\frac{11}{2}\) sq units
But Area of ∆ABC = \(\frac{1}{2}\)Σx₁ (y₂ – y₃)

= | Area of ∆ABC| + |Area of ∆ACD| = \(\left|\frac{5}{2}\right|\) + |3| = \(\frac{5}{2}\) + 3 = \(\frac{5+6}{2}=\frac{11}{2}\) sq.units

Question 37.
Find the equation of the locus of the point which moves such that it ¡s equidistant from (4, 2) and x-axis.
Answer:
Let P(x, y) be the point on the locus and A = (4,2)
∴ PA = Y = (PA)² = y²
⇒ (x – y)² + (y – 2)² = y²

⇒ x² – 8x – 4y + 20 = 0 is the required equation.

Question 38.
Find k so that the distance from (2, 3) to the line 8x + 15y + k = 0. may be equal to 4 units.
Answer:
\(\left|\frac{8(2)+15(3)+k}{\sqrt{8^{2}+15^{2}}}\right|\) = 4
\(\left|\frac{16+45+k}{\sqrt{64+225}}\right|\) = 4
16 + k = 4\(\sqrt{289}\) = 4 × 17 = 68
k = 4(17) – 61 = 68 – 61 = 7. ∴ k = 1.

Part-D

IV. Answer any SIX questions. (6 × 5 = 30)

Question 39.
In a survey of 100 persons, it was found that 28 read magazine A, 30 read magazine B, 42 read magazine C, 8 need magazines A and B, 10 read magazines A and C, 5 read magazines B and C while 3 read all the three magazines.
Find: (a) How many read none of the three magazines
(b) How many read-only magazines
(c) How many read exactly one magazine only.
Answer:
Given U = 100, n(A) = 28, n(B) = 30, n(C) = 42
n(A∩B) = 8, n(A∩C) = 10, n(B∩C) = 5, n(A∩B∩C) = 3

(i) Number of people who read none of the magazines
= 100 – (13 + 5 + 20 + 7 + 3 + 2 + 30)
= 100 – 80 = 20

(ii) No. of people who read only magazines C = 30

(iii) No. of people who read exactly one magazines only = 13 + 20 + 30 = 63

Question 40.
Using log table find the value of \(\frac{21.54 \times 72.6}{6.5}\).
Answer:
Let log = log\(\left(\frac{21.54 \times 72.6}{6.5}\right)\)
logx = log 21.54 + log 72.6 – log 6.5
= 1.3332+ 1.8609-0.8129
x = A.L (2.3812) = 240.58.

Question 41.
Find the sum of all even integers from 40 to 160.
Answer:
S_n = 40, 42, 44……………160
a = 40, d = 2, T_n = 160
T_n = a + (n – 1)d
160 = 40 + (n – 1)2
120 = 2n – 2
122 = 2n ⇒ n = 61
Consider Sn = \(\frac{n}{2}\)(a + l) = \(\frac{61}{2}\)(40+160)= \(\frac{61 \times 200}{2}\) =6100
∴ Sum of all even integers = 6100.

Question 42.
Solve x³ – 10x² + 29x – 20 = 0. Using synthetic division given that it has an integral root between -3 and 3.
Answer:
f(x) = x³ – 10x² + 29x – 20
f(1)= 1 – 10 + 29 – 20 = 0
x = 1 is a root of the given equation. Let us remove this root by synthetic division.

∴ The resulting equation is x² – 9x + 20 = 0 is the quotient and remainder is 0.
x² – 9x + 20 = 0 (x – 5) (x – 4) = 0
x = 4 or 5
Thus x = 1, 4, 5 are the roots of the given equation.

Question 43.
The difference between simple interest and compound interest on a certain sum of money invested for 3 years at 6% p.a. is f 110.16. Find the sum.
Answer:
Let the sum = x
S.I = \(\frac{x \times 3 \times 6}{100}\) = 0.18x
C.I. = A – P where A = P(1 + i)ⁿ
= x(1 +0.06)³ = 1.191016x
C.I.= A – P = 1.191016x – 1x = 0.191016x
Given C.I. – S.I = 110.16
0.191016x – 0.18x = 110.16
0.011016x = 110.16
x = \(\frac{110.16}{0.011016}\) = 10,000
∴ The sum= 10,000.

Question 44.
Find x if x³ sin45°cos 60° = \(\frac{\tan ^{2} 60^{\circ} \cdot \cos e c 30^{\circ}}{\sec 45^{\circ} \cdot \cot ^{2} 30^{\circ}}\)
Answer:
x³. \(\frac{1}{\sqrt{2}} \cdot\left(\frac{1}{2}\right)^{2}=\frac{(\sqrt{3})^{2} \cdot 2}{\sqrt{2} \cdot\left(\frac{1}{\sqrt{3}}\right)^{2}}\)
x³.\(\frac{1}{\sqrt{2}} \cdot \frac{1}{4}=\frac{3.2}{\sqrt{2} \cdot \frac{1}{3}}\)

x³. \(\frac{1}{4 \sqrt{2}}=\frac{18}{\sqrt{2}}\) ⇒ x3 = \(\frac{72 \sqrt{2}}{2}\) ⇒ x = 3\(\sqrt{72}\)

Question 45.
Raj wants to invest a lump-sum amount in the bank so that he can get an annual income of Rs.15,000 every year for the next 10 years. If the bank offers 16% p.a. compound interest, what is the amount he should invest today?
Answer:
Given, n = 10yrs, a = 15,000, i = 0.16, P = ?
P = \(\frac{a\left[(1+i)^{n}-1\right]}{i(1+i)^{n}}\)
= \(\frac{15,000\left[(1+0.16)^{10}-1\right]}{0.16(1+0.16)^{10}}=\frac{15,000\left[(1.16)^{10}-1\right]}{0.16 \times(1.16)^{10}}\) = 72498.2
∴ He should invert today 7249.8

Question 46.
A person spent 30% of his wealth and thereafter Rs. 20,000 and further 10% of the remainder. If Rs. 29,250 is still remaining, what was his total wealth?
Answer:
Let his wealth be x Rs

Again he spent 20,000 \(\frac{7 x}{10}\) – 20,000 \(\frac{7 x-2,00,000}{10}\)
Again he spent 10% = \(\left(\frac{7 x-2,00,000}{10}\right)-\frac{7 x-2,00,000}{10} \times \frac{1}{10}\)x
\(\left(\frac{7 x-2 ; 00,000}{10}\right)\left(1-\frac{x}{10}\right)=\frac{(7 x-2,00,000)(10-x)}{100}\)
\(\frac{(7 x-2,00,000)(10-x)}{100}\) = 29,250 =x 75,000
∴ Total wealth = 75,000.

Question 47.
Prove that A(-3, 6) B(-2,11) C(3,12) and D(2, 7) are the vertices of a rhombus. Also And its area.
Answer:
A(-3, 6) B(-2, 11) C(3, 12) and D(2, 7)
AB = \(\sqrt{(-2+3)^{2}+(11-6)^{2}}=\sqrt{1^{2}+5^{2}}=\sqrt{1+25}=\sqrt{26}\)
BC = \(\sqrt{(3+2)^{2}+(12-11)^{2}}=\sqrt{5^{2}+1^{2}}=\sqrt{25+1}=\sqrt{26}\)
CD = \(\sqrt{(2-3)^{2}+(7-12)^{2}}=\sqrt{(-1)^{2}+(-5)^{2}}=\sqrt{1+25}=\sqrt{26}\)
DA = \(\sqrt{(2+3)^{2}+(7-6)^{2}}=\sqrt{5^{2}+1^{2}}=\sqrt{25+1}=\sqrt{26}\)
d₁ = AC\(\sqrt{(3+3)^{2}+(12-6)^{2}}=\sqrt{(6)^{2}+(6)^{2}}=\sqrt{36+36}=\sqrt{72}\) = 6\(\sqrt{2}\)
d₂ = BD\(\sqrt{(3+3)^{2}+(12-6)^{2}}=\sqrt{(6)^{2}+(6)^{2}}=\sqrt{36+36}=\sqrt{72}\) = 4\(\sqrt{2}\)
All sides are equal = \(\sqrt{26}\) ∴ ABCD forms Rhombus.
Area = \(\frac{1}{2}\) d₁d₂ = \(\frac{1}{2}\).6\(\sqrt{2}\).4\(\sqrt{2}\) = 24 sq. units

Question 48.
Find the equation of a straight line passing through the point (2, 2) such that the sum of its intercepts on the axes is 9.
Answer:
Intercept equation is \(\frac{x}{a}+\frac{y}{b}\) = 1
Also given a + b = 9
b = 9 – a
∴ Equation becomes \(\frac{x}{a}+\frac{y}{9-a}\) = 1. But this equation passes through (2,2)
\(\frac{2}{a}+\frac{2}{9-a}\) = 1

18 = 9a – a²
a² – 9a + 18 = 0
(a-6) (a-3) = 0
a = 6 or 3,
b = 3 or 6
The required equations are \(\frac{x}{6}+\frac{y}{3}\) = 1 or \(\frac{x}{3}+\frac{y}{6}\) = 1

Part-E

V. Answer any ONE question. (1 × 10 = 10)

(a) If A = {3, 5, 7, 8} B = {4, 5, 7, 9} C = {1, 2, 3, 4, 5}
Find (i) (A – B) × (B – C)
(ii) (A ∩ B) × (B ∩ C)
Answer:
(A – B)= {3, 8}, B – C = {7, 9}
(i) (A – B) × (B – C) = {(3, 7) (3, 9) (8, 7) (8, 9)}
(ii) (A ∩ B) × (B ∩ C) = {5, 7} × {4, 5}
= {5, 4) (5, 5) (7, 4) (7, 5)}.

(b) A manufactures sells his product at Rs. 8.35 per unit, he is able to sell his entire production. His fixed cost is Rs. 2116 and his variable cost per unit is ₹ 7.20 Find
(i) Level of production at which he can make a profit of ₹ 4, 600
(ii) Level of output at which he will incur a loss of ₹1150
(iii) Break-even level of production.
Answer:
C(x)= 7.20x + 2116,
R(x)= 8.35x
P(x) = R(x) – C(x)

(i) P(x) = 4,600,
4600 = 8.35x – [7.20x + 2116] = 1.15x – 2116
1.15x = 4600 + 2116 = 6716 = 5840 .

(ii) C(x) – R(x) = 1150,
7.20x + 2116 – 8.35x = 1150
115x = 966,
x = 840 units

(iii) C(x) = R(x)
720x + 2116 = 8.35x,
2116 = 8.35x – 7.20x
2116 = 1.1 5x
x = 1840 units BEP

(c) Using log find the number of digits in 350.
Answer:
Consider log3²⁰ =20 log 3 = 20 × 0.477 1 = 9.542
Since the characteristic is 9, there are 10 digits in 3²⁰.

Question 50.
(a) If the lines 2x – y = 5,
Kx-y = 6 and 4x – y = 1 are concurrent find k.
Answer:
Solving equations 2x – y = 5, & 4x – y = 7 we get

∴ y = 2x – 5 = 2 – 5 = -3

Since lines are concurrent put x = 1 and y = -3 in kx – y = 6
Kx – y = 6 ⇒ K(1) – (-3) = 6
⇒ K + 3 = 6
⇒ K = 6 – 3
⇒ K = 3

(b) Find sum of 7 + 77 + 777 + ……… n terms.
Answer:
Let S_n = 7 + 77 + 777 +…………….n terms
= 7 [1 + 11 + 111 + –—–— to n terms] = \(\frac{7}{9}\)[9 + 99 + 999 + …………. to n terms]
= \(\frac{7}{9}\)[(10 – 1) + (100 – 1) + (1000 – 1)+……………to n terms]
= \(\frac{7}{9}\)[(10 + 10² + 10³ +…………to n terms) – (1 + 1 + 1 …………to n terms]
= \(\frac{7}{9}\left[\frac{10\left(10^{n}-1\right)}{10-1}-n\right]\)[… 10 + 10² + 10³……….. to n terms is in GP Where a = 10, r = 10]
S_n = \(\frac{7}{9}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right]\)

(c) A scholarship of Rs. 2000 every year has to be instituted. How much should be invested today ¡f the rate of interest, is 8% p.a. .
Answer:
P = \(\frac{a}{i}=\frac{2000}{0.08}\) = 25,000