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

Karnataka State Syllabus SSLC Science Model Question Paper 1 with Answers in Kannada Medium

Students can Download Karnataka SSLC Science Model Question Paper 4 with Answers in Kannada, Karnataka SSLC Science Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus SSLC Science Model Question Paper 4 with Answers in Kannada Medium

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

Karnataka State Syllabus SSLC Science Model Question Paper 1 With Answers

Time: 3 Hours
Max Marks: 80

I. Four alternatives are provided for each of the following questions or incomplete statements. Choose the most appropriate alternative and write with its alphabet. ( 8 × 1 = 8 )

Question 1.
Which of the following groups contain only biodegradable items
A) Grass, flowers, and leather
B) Grass, wood and plastic
C) Fruit, peels, cake and lime juice
D) Cake, wood and glass
Answer:
C) Fruit, peels, cake and lime juice

Question 2.
The embryo gets nutrition from the mother’s blood with the help of a special tissue called
A) Zygote
B) Utreus only
C) Placenta
D) None of these
Answer:
C) Placenta

Question 3.
One of the main aim of conservation is to try and preserve the ______ we have inherited.
A) Ecosystem
B) Biodiversity
C) Environment
D) Universal
Answer:
B) Biodiversity

Question 4.
Which of the following terms does not represent electrical power in a circuit?
A) I²R
B) IR²
C) VI
D) V²/R
Answer:
B) IR²

Question 5.
By which reaction metal is obtained from metal oxide
A) Liquification
B) Reduction
C) Calcination
D) Roasting
Answer:
B) Reduction

Question 6.
Conversion of electrical signal to chemical signal occurs at:
A) Dendrite of a neuron
B) Axon of a neuron
C) Nerve ending of a neuron
D) Cell body of a neuron
Answer:
C) Nerve ending of a neuron

Question 7.
Which of the following reaction will never occur
A) Cu + H₂SO₄ → CuSO₄ + H₂
B) Mg + H₂SO₄ → MgSO₄ + H₂
C) 2Al + 6HCl → 2AlCl₃ + 3H₂
D) Fe + 2HCl → FeCl + H₂
Answer:
A) Cu + H₂SO₄ → CuSO₄ + H₂

Question 8.
Sodium carbonate is a basic salt because it is a salt of
A) Strong acid and strong base
B) Weak acid and weak base
C) Strong acid and weak base
D) Weak acid and strong base
Answer:
D) Weak acid and strong base

II. Answer the following questions ( 8 × 1 = 8 )

Question 9.
What are autosomes and sex chromosomes?
Answer:
Human cell contain 23 pairs of chromosomes out of 23 pairs, 22 pairs are called autosomes, rest of 1 pair, which determine the sex of child is called sex chromosomes.

Question 10.
What does the high level of total coliform count in river Ganga indicate?
Answer:
The high level of total coliform count in river Ganga indicates that the water is contaminated by disease causing microorganisms (mainly sewage).

Question 11.
Why are we not able to see the things clearly when we come out of a dark room?
Answer:
When we are in dark, pupil size is bigger, as we come out of dark room, its size needs to become smaller. For that time interval the person is unable to see.

Question 12.
What is an electric current? Write its SI unit.
Answer:
The rate of flow of charges through a conductor is called an electric current. Its SI unit is ampere (A).

Question 13.
In a bakery, baking powder was not added while preparing cake. The cake obtained was hard and small in size. What is the reason for this.
Answer:
Baking powder is used for baking cakes. It contains sodium hydrogen carbonate, which breaks down when heated to form CO₂ gas. The CO₂ help to make the cakes light and fluffy.

Question 14.
Why do we consider tungsten as suitable material for making the filament of a bulb?
The filament of an electric bulb is usually made of tungsten because of its higher resistivity value (5.56 × 10 Ω m) and its high melting point (3422°C).

Question 15.
Where do plants get each of the raw materials required for photosynthesis?
Answer:

• Carbon dioxide : Air, through stomata.
• Water : Soil through roots.
• Sunlight; From the sun, present in green parts of the body absorbed by chlorophyll molecules plant body.

Question 16.
What is combustion?
Answer:
The complete oxidation of a carbon compound leading to the formation of CO₂ and H₂O is called combustion.

III. Answer the following questions. ( 8 × 2 = 16 )

Question 17.
“Two areas of study namely evolution and classification are interlinked” Justify the statement?
Answer:

• Different forms of organisms or life have evolved during the course of evolution. Classification deals with grouping of there organisms into groups and subgroups based on their similarities and differences.
• The characteristics commonly seen in any two species will be closely related and they will be closely related and they will have a more recent or common ancestor.
• Thus, classification helps tracing the evolutionary relationships between the two organisms. Hence it can be said that classification and evolution are interlinked.

Question 18.
Draw the diagram of a simple electric motor. Label the following parts.
i) Split rings ii) Brushes
Answer:

Question 19.
What is the advantages of disposable paper cups over disposable plastic cups?
Answer:
Disposable paper cups are biodegradable as they are degraded by saprophytic bacteria and fungi. The disposable plastic cups remain in the environment for longer period of time causing pollution as they are non-biodegradable.

OR

What does energy flow diagram indicate?
Answer:
Energy flow diagram indicates the following:

• Energy flow is undirectional.
• As much energy is wasted only a limited number of trophic levels (3 or 4) are found in the food chain.
• Biological magnification of harmful non – biodegradable substances occur in food chain.

Question 20.
Draw the diagram of the apparatus used in electrolytic refining of copper and label the electrode where pure copper is deposited.
Answer:

Question 21.
Why are decomposition reaction called the opposite of combination reactions? Write equations for these reactions.
Answer:
In decomposition reaction, a single compound splits to give rise to two or more simple substances. In combination reaction, two or more simpler substances combine together to form a single compound. Hence they are opposite to each other.
Example of decomposition reactions :

Example of combination reaction

OR

Give an example of redox reaction, naming the substances which are oxidised and reduced.
Answer:

Question 22.
Draw the structure of a neuron and label and following parts.
Answer:

Question 23.
What is short circuit? State one condition that leads to it. Name the device in the household that acts as a safety measure for it.
Answer:
When live wire and neutral wire come in direct contact, a heavy current in the circuit is called short circuiting.
Factor : Insulation of wire is damaged.
Safety device : Electric fuse.

Question 24.
Draw the ray diagram showing the position of the object and image, to get the virtual erect image whose size is enlarged by using convex lens.
Answer:

IV. Answer the following questions ( 9 × 3 = 27 )

Question 25.
a) Define Resistance of conductor.
Answer:
The property of the conductor to oppose the flow of charges through it.

b) Calculate the equivalent resistance between A and B from the following combination of resistors.
Answer:

Here 1 Ω and 2 Ω are connected in parallel. So equivalent resistance

Two resistor of \(\frac{2}{3}\) Ω are in series is parallel to \(\frac{2}{3}\) Ω resistor.
Equivalent resistance between A and B

Question 26.
What is Geothermal energy? How can it be harnessed to produce electrical energy.
Answer:
The energy obtained from the earth’s crust is called Geothermal energy.
When underground water comes in contact with hot spots, steam is generated. The steam is taken out through pipe to a turbine to produce electricity.

OR

List three factors responsible for the wind. State three limitations in harnessing wind energy.
Answer:
Three factors responsible for wind:

1. Uneven heating of earth’s surface.
2. Rotation of the earth.
3. Local condition – change in pressure, temperature.

Three limitations:

1. Can be used where wind blows for a greater part of the year.
2. Wind speed should be higher than 15 km / hr.
3. Establishment is expensive and requires large area.

Question 27.
Define and show in a diagram, the following terms relating to concave mirrors.
i) Aperture
ii) Radius of curvature
List any four uses of concave mirror.
Answer:
i) The diameter of the reflecting surface of the mirror is called aperature.

ii) The radius of the sphere of which reflecting surface of the spherical mirror forms a part is called the radius of curvature of the spherical mirror.

Uses of concave mirror are :

• Shaving mirror.
• Reflectors in automobile head light and torches.
• By dentists to see back side of tooth.
• Reflector of solar furnaces.

Question 28.
What happens when a solution of baking soda is heated? Write chemical equation for the same. Name the product which is responsible for making the bread or cake spongy and fluffy.
Answer:
When baking soda is heated it decomposes to produce sodium carbonate, water and carbondioxide gas.

CO₂ gas produce during the reaction makes the cake or bread spongy and fluffy.

OR

i) Write the name given to bases that are highly soluble in water. Give an example.
ii) How is tooth decay related to pH.
iii) Why does bee – sting cause pain and irratiation?
Rubbing of baking soda on the sting area gives relief. How?
Answer:
i) Alkali, NaOH
ii) Tooth decay starts when the pH of the mouth is lower than 5.5. It can be prevented by using toothpaste which are generally basic.
iii) Bee sting has acid that cause pain and irritation.
Baking soda being alkaline, Neutralises acid and given relief.

Question 29.
How does a solenoid behave like a magnet? Can you determine the north and south poles of a current carrying solenoid with the help of a bar magnet?
Answer:
When electric current flows through a solenoid, magnetic field is set up around the solenoid. The pattern of the magnetic field is same as that of the magnetic field of a bar magnet. One end of the solenoid behaves as north pole and the other end of the solenoid behaves as south pole.

To determine the north and south poles of a current carrying solenoid with the help of a bar magnet, suspend it with a strong thread. Now bring the north pole of a bar magnet towards one end of the solenoid. If the solenoid attracts towards the magnet then the face of the solenoid is south pole. If the bar magnet moves away from the solenoid, then the face of the solenoid is the north pole.

Question 30.
a) What is meant by least distance of distinct vision?
Answer:
It is the minimum distance up to which eye can see clearly and is called the least distance of distinct vision.

b) What happens to the image distance in the eye when we increase the distance of an object from the eye.
Answer:
The size of the eye can change, so the image distance is fixed when we increase the distance if the eye does not change, due to power of accommodation of the eye, focal length of the eye lens is changed, which compensates the increase in object distance. Hence image distance remains fixed and image is formed on the retina of the eye.

OR

a) Why are we not able to see the things clearly when we come out of a dark room?
Answer:
When we are in dark, pupil size is bigger, as we come out of dark room, its size needs to become smaller. For that time interval person is unable to see.

b) Why does the sun appear reddish early in the morning.
Answer:
During sunrise, the light rays coming from the sun have to travel a greater distance in the earth’s atmosphere before reaching our eyes.

In this journey, the shorter wave length of lights are scattered out and only longer wavelength are able to reach our eyes. Since blue colour had a shorter wavelength and red colour has longer wave length, the red colour is able to reach our eyes after the atmospheric scattering of light. Therefore the sun appear reddish early in the morning.

Question 31.
a) Define speciation?
Answer:
Speciation is arising of a new species from a sub-population of a species which is geographically or reproductively isolated over a long period of time from the other population of the same species.

b) Mention the sex chromosomes present in human male and human female with the help of a flow chart determine genetically in human beings the sex of the offspring if the sperm carrying x chromosomes fertilizers the egg.
Answer:

Question 32.
a) Define 1 dioptre of power of a lens.
Answer:
The SI unit of power of lens is dioptre which is denoted by the letter D. 1 Dioptre is defined as the power of a lens of focal length 1 meter.

b) The refractive index of diamond is 2.42. What is the meaning of this statement.
Answer:
The refractive index of diamond is 2.42 this means that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in air. In other words the speed of light in diamond is 2.42 times less than the normal speed of light in vacuum.

c) In which type of lens linear magnification is always less than one?
Answer:
Concave lens always has linear magnification less than one, because it always gives diminished images.

Question 33.
a) What will be the formula and electron dot structure of cyclopentane.
Answer:
The formula for cyclopentane is C₅H₁₀. Its electron dot structure is given below.

b) How would you distinguish experimentally between an alcohol and carboxylic acid.
Answer:
We can distinguish between an alcohol and a carboxylic acid on the basis of their reaction with carbonate and hydrogen carbonates. Acids reacts with carbonate and hydrogen carbonate to evolve CO₂ gas that turns lime water milky.
Metal carbonate / metal hydrogen carbonate + carboxylic acid.
↓
Salt + water + Carbondioxide.
Alcohols on the other hand, do not react with carbonates and hydrogen carbonates.

OR

What are addition reaction? What are catalysts? Illustrate with an example.
Answer:
Reactions which involve addition of two reactants to form a single product are called addition reactions. Catalysts are the substances which can change, usually increase the speed of a chemical reaction without being used up in that reaction.

For example, vegetable oils having long unsaturated carbon chains are converted into vegetable ghee by heating them in presence of Nickel, platinum or palladium metals used as catalysts.

V. Answer the following questions. ( 4 × 4 = 16 )

Question 34.
a) Define Mendeleev’s periodic law.
Answer:
Mendeleev’s periodic law states that the properties of elements are a periodic function of their atomic masses, b) The elements of the third period of the periodic table are given below :

a) Which atom is bigger, Na or Mg? Why?
Answer:
Sodium is bigger than magnesium at it has lesser nuclear charge so there is less force of attraction between nucleus and valence electrons and less effective nuclear charge. It is, therefore bigger in size.

b) Identify the most i) Metallic and ii) Non-metallic element in period 3.
Answer:
i) Sodium is the most metallic as it can lose electrons easily due to its larger atomic size.
ii) Chlorine is the most non – metallic element because it can gain electrons easily due to its smallest atomic size.

OR

a) What are groups and period in the periodic table?
b) Two element M and N belong to group I and II respectively and are in the same period of the periodic table. How do the following properties of M and N vary?
1) Size of their atoms
2) This metallic characters
3) Their valencies informing oxides.
4) Molecular, formulae of their chlorides.
Answer:
a) The vertical columns in the periodic table are called group. The horizontal rows in the periodic table are called periods.
b) 1. M and N belong to the same period but group I and li. Therefore N will be smaller than M as a atomic size is decreases from left to right.
2. M is more metallic than N. Metalic character goes on decreasing from left to right as tendency to lose electrons.
3. Their valencies are 1 and 2 respectively in forming oxides. Valency goes on increasing first and then decreases.
4. MCl, NCl₂ are molecules formulae of their chlorides.

Question 35.
Draw the diagram showing the structure of human alimentary canal and label the following parts.
a) The part which stores bile juice
b) The longest part of the alimentary
Answer:

Question 36.
a) Oil and fat containing food items are flushed with nitrogen. Why?
Answer:
To prevent the oxidation of oil and fat present in food items, nitrogen gas is flushed in food packets products formed due to oxidation of oil and fat have unpleasant smell and taste due to rancidity. Flushing food items containing fat and oil with nitrogen prevents rancidity.

b) Write the balanced chemical equations for the following reactions.
Answer:
i) Calcium hydroxide + Carboridioxide → Calcium carbonate + water
Answer:
Ca(OH)_2(aq) + CO_2(g) → CaCO_3(s) + H₂O_(l)

ii) Zinc + Silver nitrate → Zinc nitrate + silver
Answer:
Zn_(s) + 2AgNO₃ → Zn(NO₃)_2(aq) + 2Ag_(s)

iii) Aluminium + Copper chloride → Aluminium chloride + Copper
Answer:
2Al_(s) + 3CuCl_2(aq) → 2AlCl_3(aq) + 3Cu

Question 37.
a) List the physical properties of metals.
Answer:
Physical properties of metals are :

• Metals are usually hard.
• They are Sonorous.
• They are lustrus
• Metals exhibit malleability and ductility
• They exhibit high tensil strength and have high densities.

b) Differentiate between roasting and calcination. Explain the two with the help of suitable chemical equations.
Answer:
Roasting : It is a process in which sulphide ore is heated in the presence of oxygen to convert into oxide.
2Zns + 3O₂ → 2ZnO + 2SO₂

Calcination : It is a process in which carbonate ore is heated in the absence of air to form oxides.

By reduction process Zn can be extracted from its ore.
Reduction : 2ZnO + C → ZnO + CO₂

VI. Answer the following question. ( 1 × 5 = 5 )

Question 38.
a) What is fertilisation? Distinguish between external fertilisation and internal fertilisation. What is the site of fertilisation in human beings.
Answer:
Fertilisation is defined as the fusion of a male gamete (sperm) with a female gamete (an ovum or egg) to form a zygote during sexual reproduction external fertilisation.

• The fusion of male gamete (sperm) and female gamate (ovum) occur outside the body.
• Both individuals discharge their gamates outside the body.
• Development occurs outside the body.
  Ex: Frog.

Internal fertilisation :

• The fusion of gametes occurs inside the body.
• Only the male discharge sperm into female gentital tract.
• Development occurs inside the body.
  Ex : Humans, cattle etc.

The site of fertilisation in human beings is in the fallopian tube of female reproductive system.

b) How does growing embryo get nutrition from the mother’s blood?
Answer:
The embryo gets nutrition from the mother’s blood with the help of a special tissue called placenta. This is a disc which is embedded in the uterine wall and transfer glucose and oxygen from the mother to the embryo.

Students can Download Karnataka SSLC Maths Model Question Paper 3 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus SSLC Maths Model Question Paper 3 with Answers

Time: 3 Hours
Max Marks: 80

I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )

Question 1.
In the following numbers, irrational number is
a) 0.232332333……
b) 0.23233
c) 0.232323
d) 0.2323
Answer:
a) 0.232332333……

Question 2.
10 sec²A – 10tan²A =
a) sec²A
b) 10
c) 1
d) 0
Answer:
b) 10
Solution:
10 sec²A – 10tan²A
10(sec²A – tan²A) = 10(1)= 10

Question 3.
The length of the tangent drawn to a circle of radius 3cm from 5cm away from the centre is
a) 4cm
b) 5cm
c) 3cm
d) 2cm 4
Answer:
a) 4cm
Solution:
d² = r² + t²
t² = d² – r²
5² – 3²
= 25 – 9
∴ t = 16 = 4cm

Question 4.
A solid piece of copper of dimension 24 × 49 × 33 cms is moulded and recast into a sphere. The radius of the
sphere formed is ________
a) 49 cm
b) 24 cm
c) 21 cm
d)33 cm
Answer:
e) 21 cm
l × b × h = \(\frac{4}{3} \pi r^{3}\)

Question 5.
The degree of the polynomial in the graph given below is

a) 4
b) 3
c) 1
d) 2
Answer:
a) 4
Solution:
4 since it is intersecting the x – axis at 4 points.

Question 6.
The sum of the n terms of an AP is 2n2 + 5n and its common difference is 6, then its first term is
a) 0
b) 5
c) 2
d) 7
Answer:
d) 7
Solution:
S_n = 2n² + 5n
S₁ = 2(1)² + 5(1)
= 2 + 5 = 7

Question 7.
In ∆ PQR, PR = 12cm, QR = 6√3 cm, PQ = 6cm. The angle Q is
a) 45°
b) 90°
c) 30°
d) 120°
Answer:
b) 90°
Solution:
PR² = PQ² + QR²
12²= (6√3 )² + 6²
144 = 108 – 36
144=144
∴ ∠Q = 90°

Question 8.
A cube numbered 1 to 6 is thrown once, the probability of getting a number divisible by 3 is
a) 2/3
b) 0
c) 1/3
d) 1
Answer:
Event A = {3,6}
n(s) = 6, n(A) =2
P(A) = \(\frac{n(A)}{n(S)}=\frac{2}{6}=\frac{1}{3}\)

II. Answer the following Questions : (1 x 8 = 8)

Question 9.
Given g(x) = 2x + 1, q(x) = (x³ + 3x² – + 1), r(x) = 0, Find p(x)
Answer:
p(x) = g(x) q(x) + r(x) 1 650 1170
= (2x + 1) (x³ + 3x² – x + 1) + 0
p(x) = 2x (x³ + 3x² -x + 1) + 1(x³ + 3x² – x + 1)
= 2x⁴ + 6x³ – 2x²+ 2x + x³ + 3x² – x + 1
= 2x⁴ + 7x³ + x² + x + 1

Question 10.
If the sum of first n odd natural number is 1225, find the value of n.
Answer:
Sum of “n” odd natural number = n² = 1225
n = √1225
n = 35

Question 11.
In the fig. ∠AOD is divided into 2 parts which are in A.P. the smallest angle ∠AOB = 20° . Find the common difference between each angle.
Answer:
∠AOB +∠BOC + ∠COD = 180°
a + a + d + a + 2d= 180°
3a + 3d= 180°
3(20) + 3d = 180°
60 + 3d = 180°
3d = 180 – 60
d = 120/3 = 40°

Question 12.
In ∆ ABC, if DE || BC, then \(\frac{\mathbf{A B}}{\mathbf{A D}}=\frac{\mathbf{A C}}{\mathbf{A} \mathbf{E}}=\frac{\mathbf{B C}}{\mathbf{D E}}\) , state the theorem to justify this.
Answer:
In a triangle, if a line is parallel to one of the sides then the sides of given triangle are propotional to sides of the intercepted triangle.

Question 13.
Find the largest number which divides 650 and 1170.
Answer:
H.C.F of (650, 1170) = 130

Question 14.
If sin θ = 7/25, cos θ = 24/25 find the value of sin² θ + cos² θ
Answer:

Question 15.
Find the value of cos60° cos 30° sin60° sin 30°
Answer:
cos 60° cos30 – sin 60 sin 30

OR

This is in the form of
cos A cos B – sin A sin B = cos (A + B)
cos 60 cos 30 – sin 60 sin 30
= cos (60 + 30)
= cos (90)
– =0

Question 16.
The T.S.A of a solid hemisphere of radius 21 mm.
Answer:
T.S.A of hemisphere = 3 πr²
= 3 × \(\frac{22}{7}\) × 21 × 21 = 4158 mm²

III. Answer the following : ( 2 x 8 = 16 )

Question 17.
Prove that if x and y are odd positive integers, then x² + y² is even but not divisible by 4.
Answer:
Let x = 2m + 1 and y = 2n + 1 for some integers m and n.
x² + y² = (2m + 1)² + (2n + l)2
x² + y² = 4m² + 4m + 1 + 4n² + 1 + 4n
= 4m² + 4n² + 4m + 4n + 2.
x² + y² = 4(m² + n²) + 4 (m + n) + 2
x² + y² = 4 {(m² + n²) + (m + n)} +2
x² + y² = 4q + 2, where q = (m² + n² ) + (m+n)
⇒ x² + y² is even and leaves the remainder 2 when divided by 4.
⇒ x² + y² is even but not divisible by 4.

Question 18.
Solve : 100x + 200y = 700
200x + 100y = 800
Answer:
Consider 100x + 200y = 700 ….(1)
200x + 100y = 800 ….(2)
Add (1) & (2) 300x + 300y= 1500
Divide by 300, x + y = 5 …… (3)
Subtract (1) and (2)
-100x + 100y= -100 Divide by 100.
-x + y = -1 …(4)
Solve (3) and (4)

y = 2
Substitute the value of y in equation (3)
x + y = 5
x + 2 = 5
x = 5 – 2 = 3
x = 3

Question 19.
Find the roots of the quadratic
equation 3x² – 2√6x + 2 = 0 by formula method.
Answer:

Question 20.
Find the value of x in which the points (1, -1) (x, 1) and (4, 5) are collinear.
Answer:
A(1, -1) B(x, 1)C(4,5)
Area of the triangle = 0, when points are collinear.
0 = x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃ (y₁ – y₂)
0 = (1) (1 – 5) + x(5 + 1) + 4(-1 – 1)
0 = -4 + 6x – 8
6x – 12 = 0
x = 12/6
x = 2

Question 21.
ABC is a right angled triangle, having [B = 90°. If BD = DC, Show that AC² = 4AD² – 3AB²
Answer:

AC² = AB² + BC² (Pythagoras)
AC² = AB² + 2(BD)²
AC² = AB² + 4BD²
In right angled ∆ ABD,
AD² = AB² + BD²
BD² = AD² – AB²
Consider AC² = AB² + 4BD²
= AB² + 4(AD² – AB²)
= AB² + 4AD² – 4AB²
= – 3AB² + 4AD²
= 4 AD² – 3AB²
AC² = 4 AD² – 3AB²

OR

Prove that area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.
Answer:

Data : ABCD is a square.
Equilateral triangles ∆ BCE and ∆ ACF have been described on side BC and diagonal AC respectively.
T.P.T. : Area of ( ∆ BCE) = 1/2 Area of ∆ ACF
Proof : Since ∆ BCE and ∆ ACF are equilateral.
∴ They are equiangular.
∆ BCE ~ ∆ ACF

Question 22.
A box contains 90 dices which are numbered from 1 to 90. If one dise is drawn at random from the box, find the probability that it bears
(i) two digit number
(ii) a perfect square number.
Answer:
The numbers in the dise form the sample space
S = {1,2, 3,4,…….. 90}
One dise can be drawn out of 90 in 90 ways. .
n(s) = 90

i) There are 90 -9 = 81, two digit numbers, out of which one dise can be drawn in 81 ways.
∴ n(A) = 81
P(A) = \(\frac{n(A)}{n(S)}=\frac{81}{90}=\frac{9}{10}\)

ii) The perfect square numbers are B= {1,4,9, 16,25,36,49,64,81}
∴ n(B) = 9
P(B) = \(\frac{n(B)}{n(S)}=\frac{9}{90}=\frac{1}{10}\)

Question 23.
Draw a pair of tangents to a circle of radius 5cm which are inclined to each
other at an angle of 60°.
Answer:

Question 24.
Prove that \(\frac{\tan \theta+\sin \theta}{\tan \theta-\sin \theta}=\frac{\sec \theta+1}{\sec \theta-1}\)
Answer:

OR

Prove that (cosecθ – cotθ)² = \(\frac{1-\cos \theta}{1+\cos \theta}\)
Answer:
Consider (cosecθ – cotθ)²

IV. Answer the following : ( 3 × 9 = 27 )

Question 25.
The sum of the ages of A and B is 85 years. 5 years ago, the age of A was twice that of B. Find the present ages.
Answer:
Let the present age of A be x years.
Let the present age of B be y years.
Sum of their ages = 85 x + y = 85 …….. (1)
Five years back, age of A was (x – 5) and that of B was (y – 5).
The age A is twice that of B.
∴ x – 5 = 2(y – 5) ……(2)
x – 5 = 2y – 10
x – 2y = -10 + 5
x – 2y = -5 ……(2)
Solve (1) and (2)

Substitute y in (1)
x + y=85
x + 30 = 85
x = 85 – 30 = 55
x = 55
∴ Present age of A is 55 years and B = 30 years.

OR

A piece of work can be done by 2 men and 7 boys in 4 days. The same piece of work can be done by 4 men and 4 boys in 3 days. How long it would take to do the same work by one man or one boy?
Answer:
Let x and y be the number of days in which one man can complete the work.
∴ In 1 day a man can do 1/x th work and
In 1 day a man can do 1/y th work.
2 men and 7 boys can complete the work in 4 days.
∴ They can complete in 1/4 th of the work in one day.
∴ \(\frac{2}{x}+\frac{7}{y}=\frac{1}{4}\) …… (1)
4 men and 4 boys together complete the work in 3days, and they can complete it in 1/3 rd of work.

4(2y + 7x) = xy
8y + 28x = xy …….(3)
3(4y + 4x) = xy
12y + 12x = xy ……..(4)
Solve (3) and (4)
(8y + 2x = xy) x 12
(l2y+ 12x = xy) x 18

4 y = 240
y = 240/4
y = 60

3(4y + 4x) xy
12y+12x = xy ………..(4)
4y = 240
y = 240/4 = 60
Substitute y in equation (3)
8y + 28x = xy
8(60) + 28x = xy
480 + 28x = x(60)
480 + 28x = 60x
480 = 60x – 28x
480 = 32x
x = \(\frac{480}{32}=\frac{120}{8}=\frac{30}{2}\) = 15
x = 15
Thus, one man can complete the work in 15 days and one boy can do the work in 60 days.

Question 26.
Find the zeroes of the polynomial p(y) = y3 – 5y2 – 2y + 24 if it is given that Sum of the two zeros is Answer:
Let α, β,γ be the zeros of the given polynomial.
p(y) = y³ – 5y² – 2y + 24
α + β = 7
Sum of the zeros

α + β + γ = 5
7 + r = 5
r = 5 – 7 = -2
r = -2
Sum of product of zeroes taken two at a

αβ + βr + rα = -2/1 = – 2
αβ + r(β + α) = -2
αβ + (7) (-2) = -2
αβ-14 = -2
αβ = -2+ 14
αβ = 12
(α – β )² = (α + β)²– 4αβ
= (7)² – 4(12)
(α – β)² = 49 – 48
α – β = ±1
Solve for α and β. When α – β = +1
α + β = 7
α – β = 1
2α =8
α = 8/2 = 4
α = 4
Substitute the value of a in α + β =7
4+p =7
β =7-4 = 3
β = 3
Solve for α and β when α – β = -1

α = 6/2 = 3
α = 3
α + β = 7
3 + β=7
β = 7 – 3
a + β = 7
β = 4
When α – β = 1,
the values are α = 4, β = 3, r = -2.
When α – β = -1,
The values are α = 3, β = 4, r = -2.

Question 27.
The diagonal of a rectangular field is 60 metres more than the shorter side, if the longer side is 30 metres more than the shorter side, find the sides of the field.
Answer:

Let the shorter D side be x mtr.
Longer side is 30m more than the shorter side. A Longer side = (x + 30)m.
Diagonal is 60m more than the shorter side.
Diagonal = (x + 60) m
AC² = AB² + BC² (Pythagoras)
(x + 60)² = (x + 30)² + x²
x²+(60)² + 2x (60) = x² + (30)² + 2(x)(30)+ x²
x² + 3600 + 120x = x² + 900 + 60x + x²
-x² + 60x + 2700 = 0
x² – 60x – 2700 = 0
x² – 90x + 30x – 2700 = 0
x(x – 90) + 30(x – 90) = 0
(x – 90) (x + 30) = 0
x – 90 = 0 or x + 30 = 0
x = 90 or x = – 30
∴ Shorter side = 90m = BC = x
Longer side = x + 30 = 90 + 30 = 120m
Diagonal = x + 60 = 90 + 60 = 150m

OR

Sum of the areas of two squares is 468m2. If the difference of their perimeters is 24m, find the sides of two squares.
Answer:
Let the sides of the two squares be x&y. Sum of the areas of two squares = 468
x² + y² = 468 ……(1)
Difference of their perimeters = 24m.
4x – 4y = 24 x – y = 6
x = 6 + y …….(2)
Substitute the value of x in (1)
x² + y² = 468
(6 + y)² + y² = 468
36 + y²+ 12y + y² = 468
2y² + 12y + 36 = 468
Divide by 2
y² +6y+ 18 = 234 y2 + 6y = 234 – 18
y² + 6y – 216 = 0
y²+18y – 12y -216 = 0
y(y + 18)-12(y + 18) = 0
(y + 18) (y – 12) = 0
y+18 = 0, y -12 = 0
y = -18,y = 12
∴ x = 6+y = 6 + 12 = 18m
The sides of two squares are 18m and 12m.

Question 28.
Show that the points x(2, -2) y(-2, 1) and z(5, 2) are the vertices of a right angled triangle XYZ and also calculate its area.
Answer:


⇒ XYZ is an isosceles right angled triangle.
Area of a right angled triangle = \(\frac{1}{2}\) x base x height
\(\frac{1}{2}\) × 5 × 5 = \(\frac{25}{2}\) = 12.5 cm

OR

Find the values of k for which the points A(k + 1, 2k) B(3k, 2k+3) and C(5k – 1, 5k) are collinear.
Answer:
A(k + 1, 2k) = (x₁, y₁)
B(3k, 2k + 3) = (x_2, y₂)
C(5k-1, 5k) = (x₃, y₃)
Area of the triangle = 0, for the points to be collinear.
Area =[x₁(y₂ – y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)]
0 = \(\frac{1}{2}\){(k + 1) (2k+3 – 5k)+3k(5k- 2k) + (5k-1) [2k-2k-3]}
0= \(\frac{1}{2}\)]{(k+ 1) (-3k+ 3) + 3k (3k) + (5k- I)(-3)i
0 =\(\frac{1}{2}\){-3k²-3k + 3k + 3 + 9k2 – 15k+3}
0= \(\frac{1}{2}\){6k²– 15k + 6}
0 = 6k² – 15k + 6
Divide by 3
2k² – 5k + 2= 0
2k² – 4k – 1k + 2 = 0
2k(k – 2) -1(k – 2) = 0
(k – 2) (2k -1) = 0
k – 2 = 0, or 2k- 1=0
k = 2 or k = 1/2

Question 29.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Answer:

Data: ‘O’ is the centre of the circle PA and PB are the two tangents drawn from an external point P. OA and OB are radii of the circle.
To prove that: ∠APB+ ∠AOB = 180°
Proof: In AAPO and ABPO
∠OAP = ∠OBP = 90°
(∵ Angle between the radius and tangent at the point of contact is 90°)
OP = OP (∵ Common side)
OA = OB (∵ Radii of the same circle) According to RHS postulate AAPO ABPO
∠OAP = ∠OBP = 90°
∠OAP + ∠OBP = 90° + 90° = 180°
⇒ Opposite angles of OAPB quadrilateral are supplementary
∴ OAPB is a cylic quadrilateral
⇒ ∠APB + ∠AOB = 180°

Question 30.
Find the area of the shaded region where a circular arc of Radius 6 cm has been drawn with the vertex ‘O’ of an equilateral Triangle OAB of side 12cm as centre.
Answer:
Area of the shaded Region = Area of circle area of sector OCDE + Area of equilateral
∆ le OAB

OR

From each corner of a square of side 4cm a quadrant of a circle of radius 1cm is cut and also a circle of diameter 2cm is cut as shown in Fig, Find the area of the remaining portion of the square.
Answer:


Area of shaded region = Area of square – (Area of circle + Area of 4 quadrants)
The following table gives production yield per hectare of wheat of 100 farms of a

Question 31.
The following table gives production yield per hectare of wheat of 100 farms of a
village.

Change the distribution to a more than type distribution and draw its ogive.
Answer:

Question 32.
If the median of the distribution given below is 28.5. Find the valucs of x and y.

Answer:
Meadian = 28.5, lies in the C.I , 20 – 30
The median class = 20 – 30
l = 20, f = 20, c.f = 5 + h, h = 10, n = 60

Question 33.
Construct a triangle ABC with side BC = 7cm, ∠B = 45°, ∠C = 105°. Then construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.
Answer:

V. Answer the following ( 4 × 4 = 16 )

Question 34.
Solve the pair of equations graphically.
x+y=3 and 3x-2y=4
Answer:
x + y = 3
y = 3- x

3x – 2y = 4
-2y = 4 – 3x .

Question 35.
If the sum of first 8 terms of an Arithmetic progression is 136 and that of first 15 terms is 465, then find the sum of first 25 terms.
Answer:
Given S₈ = 136, S₁₅ = 465, S₂₅ = ?
Use the formula S_n = \(\frac{n}{2}\) [2a + (n -l)d]
s₈ = \(\frac{8}{2}\)[2a + (s-l)d]
136 = 4 (2a + 7d)
∴ 2a + 7d = 136/4
2a + 7d= 34…….(1)
S₁₅ = \(\frac{15}{2}\)[2a + (15-l)d]
465 = \(\frac{15}{2}\)[2a + 14d]
465= \(\frac{15}{2}\) x 2(a + 7d)
∴ a + 7d = 465/15
a + 7d = 31 …….(2)
From (1) and (2)

Consider a + 7d = 31
3 + 7d = 31
7d = 31 – 3
d = 28/7
d = 4
∴S₂₅ = \(\frac{25}{2}\)[2a + (25-1)d]
S₂₅ = \(\frac{25}{2}\)[2a + (25-1)d]
S₂₅ = \(\frac{25}{2}\)[2a + (24)d]
S₂₅ = \(\frac{25}{2}\)[2(3) + 24(4)d]
S₂₅ = \(\frac{25}{2}\) × 102
S₂₅ = 1275
‘5 ‘ 9

OR

The sum of the 5th and 9th terms of an A.P is 40 and the sum of the 8th and 14th term is 64. Find the sum of the first 20 terms.
Answer:
Given T₅+ T₉ = 40 and T₈ + T₁₄ = 64
a + 4d + a + 8d=40
2a + 12d = 40….(1)
a + 7d + a + 13d = 64
2a + 20d = 64…. (2)

Consider 2a + 12d = 40
2a + 12(3) = 40
2a + 36 = 40
2a = 40 – 36
a = 4/2
a = 2
S_n = \(\frac{n}{2}\)[2a + (n-1)d]
S₂₀ = \(\frac{20}{2}\)[2(2) + (20-1)3]
= 104(4 + 57)
= 10 × 61
= 610

Question 36.
Two poles of equal heights are standing opposite to each other on either side of the road, which is 80m wide. From a point between them on to the road, the angles of elevation of the top of the poles re 60 and 30’, respectively. Find the height of the poles and distances of the point from the poles.
Answer:

In ∆ ABE.
tan60 = \(\frac{\mathrm{AB}}{\mathrm{BE}}\)
√3 = x/BE
x = √3BE ….. (1)
tan30 = DC/EC
\(\frac{1}{\sqrt{3}}=\frac{x}{E C}\)
EC = x√3
EC = BE × √3 .√3 [∵ x = √3.BE]
FC = 3BE
∴ EC=3BF
We know that
BE + EC = 80 rn
BE 3BE 80m
4BE = 80
BE = 80/4 = 20
⇒ EC = 3BE 3(20)= 60m
∴ x = BE.
x = 20√3
⇒ AB = CD = 20√3 m
∴ The distance from the point to the pole arc 20m towards left and 60m towards the right
poles
Height of the poles 20√3m

Question 37.
Prove areas of similar triangles.
Answer:
Areas of similar triangles are proportional
to the squares on the corresponding sides.

In ∆ABM and ∆DEN
∠AMB = ∠DNE = 90 [construction]
∠B =∠E (Data)
∠BAM = ∠EDN (Remainingang1e)
∆ABM ∼ ∆ DEN
∆ABM & ∆DEN are equiangular

VI. Answer the following : ( 5 × 1 = 5 )

Question 38.
A circus tent is made of canvas and is in the form of a right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are 126m and 5m respectively. The total height of the tent is 21m. Find the total cost of the canvas used to make the tent when the cost per m2 of the canvas is ?15.
Answer:

Total canvas used=CS A of cylindrical part -1- CS A of conical part
=2πrh + πrl
= 2 × \(\frac{22}{7}\) × 63 × 5 + \(\frac{22}{7}\) x 63 x 65
= 1980 m² + 12870 m²
= 14850 m²
Total canvas used = 14850 m²
* Cost of canvas at the rate of ₹16 per m²
= 14850 × 15= ₹ 2,22.750.
Cost of canvas = ₹ 2,22,750

Students can Download Karnataka SSLC Maths Model Question Paper 2 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus SSLC Maths Model Question Paper 2 with Answers

Time: 3 Hours
Max Marks: 80

I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )

Question 1.
For some integer n every odd integer is of the form
a) 2n + 1
b) n + 1
c) 2n
d) n
Answer:
2n + 1

Question 2.
The value of sin215° + sin225° + sin265° + sin275° is
a) 0
b) 1
c) 2
d) 3
Answer:
2

Question 3.
If chord AB subtends an angle 50° at the centre of a circle then the angle between the tangents at A and B is
a) 40°
b) 100°
c) 130°
d) 120°
Answer:
130°

Question 4.
The formula used to find the volume of a sphere
a) \(\frac{4}{3} \pi r^{3}\)
b) \(\frac{2}{3} \pi r^{3}\)
c) \(\frac{1}{3} \pi r^{3}\)
d) πr³
Answer:
a) \(\frac{4}{3} \pi r^{3}\)

Question 5.
α + β are the zeroes of the polynomial x² – 6x + 4, then the value of \(\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha \beta}\) is
a) 7
b) 8
c) -7
d) -8
Answer:
a) 7
Solution:

Question 6.
If 29th term of an A.P is twice its 19th term, then the 9th term is
a) -1
b) 0
c) 1
d) 2
Answer:
b) 0
Solution:
T₂₉ = 2T₁₉
a + 28d= 2(a + 18d)
a + 28d = 2a + 36d
0 = 2a + 36d-a-28d
0 = a + 8d
T₉ = 0

Question 7.
In ∆ ABC, AB = 6√3 cm, AC = 12 cm, BC = 6cm, The angle B is
a) 45°
b) 90°
c) 60°
d) 30°
Solution:
AC² = AB² + BC²
12²= (6√3 )² + 6²
144= 108 + 36 144=144
∴ ∠B = 90°

Question 8.
If the probability of an event is P(A) then the probability of its complimentary event will be
a) 1+P(A)
b) 1 – P(A)
c) P(A) – 1
d) 1 / P(A)
Solution:
We know that for any two complimentary events A and Ā
P(A) + P(Ā) = 1 → P(Ā) = 1 – P(A)

II. Answer the following questions : ( 1 × 8 = 8 )

Question 9.
If α and β are the zeroes of the quadratic polynomial 2 – 3x – x² then what is the value of α + β + αβ ?
Answer:
α + β = -b/a , αβ = c/a
= \(\frac{-(-3)}{-1}\) αβ = \(\frac{2}{-1}\) = – 2
= – 3
α + β + αβ = -3 -2 = -5

Question 10.
What are the roots of the quadratic equation x² + (√3 + 1)x + √3 = 0?
Answer:
² + (√3 + 1)x + √3 = 0 = 0
x² + x + √3x + √3 = 0
x(x + 1)+ √3 (x + 1) = 0
(x + 1)(x + √3) = 0
x + 1 = 0, x + √3 = 0
x = -1, x = -√3

Question 11.
If the nth terms of the two AP 9, 7, 5, ….. and 24, 21, 18, ……… are same. Find n.
Answer:
9, 7, 5, …….
a = 9, d = 7-9 = -2 T = a + (n – 1)d
T_n = 9 + (n – 1) (-2)
T_n = 9 + (-2n + 2) ‘
T_n = 9 – 2n + 2
T_n = 11 – 2n
24,21,18,
a = 24, d = 21 – 24 = -3
T_n = a + (n – 1) d T=24 + (n – 1)(-3)
T_n = 24 – 3n + 3
T_n = 27 – 3n
Given : nth term of both A. P. is same.
∴ 11 – 2n = 27 – 3n
-2n +3n = 27 -11
n = 16

Question 12.
State A – A criterian theorem.
Answer:
“If two triangles are equiangular, then the corresponding sides are proportional.”

Question 13.
Find the H.C.F of 455 and 42 with the help of Euclid’s division algorithm.
Answer:
Step (1): a = 455 b = 42
a = bq + r
455 = 42 × 1 + 35

Step (2): a = 42, b = 35
a = bq + r
42 = 35 × 1 + 7

Step (3): a = 35, b = 7
a = bq + r
35 = 7 × 5 + 0
∴ H.C.F. (455, 42) = 7

Question 14.
Find θ if sin (θ + 56) = cosθ, where θ and ( θ + 56) are less than 90°.
Answer:
sin (θ + 56) = cos θ
sin (θ + 56) = sin (90 – θ)
θ + 56 = 90 – θ
θ + θ = 90 – 56
2θ = 34
θ = 34/2
θ = 17

Question 15.
If x = a sin θ and y = b tan θ, then find the value of \(\frac{a^{2}}{x^{2}}-\frac{b^{2}}{y^{2}}\)
Answer:
Consider

Question 16.
Calculate the height of a right circular cone where C.S.A. and base radius are 12320 cm2 and 56 cms. respectively.
Answer:

C.S.A of a cone = π rl
12320 = \(\frac{22}{7}\) x 56 x 1
1 = \(\frac{12320 \times 7}{22 \times 56}\)
1 = 70 cm
But 1² = r² + h²
70² = 56² + h²
h2² = 70² – 56²
h² = (70 + 56) (70 – 56)
h² =(126) (14)

III. Answer the following ( 2 × 8 = 16 )

Question 17.
Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5 where q is an integer.
Answer:
Let nbe any positive, integer using division algorithm,
a = bq + r
Taking a = n, b = 6
n = 6q + r
If any number is divided by 6, then the possible remainders are 0, 1, 2, 3, 4 or 5.
∴If n is odd, then r = 1, 3, 5
⇒ 6q + 1, 6q + 3, 6q + 5 are the positive odd integers.

Question 18.
Solve : 2x + 3y = 9
3x + 4y = 5
Answer:

Substitute y = 17 in 2x + 3y = 9
2x + 3(17) = 9
2x + 51 =9
2x = 9 – 51
x = \(\frac{-42}{2}\)
= -21

Question 19.
Solve : (x – 2)² + 1 = 2x – 3
Answer:
x² + 4 – 4x + 1 = 2x – 3
x² – 4x + 5 – 2x + 3 = 0
x² – 6x + 8 = 0 x2 – 4x – 2x + 8 = 0
x(x – 4) -2(x – 4) = 0
(x – 4) (x – 2) = 0
x – 4 = 0, x -2 = 0
x = 4 or x = 2

Question 20.
Show that the points (-2, 1) (2, -2) and (5, 2) are the vertices of a right angled triangle.
Answer:

∴ ∆ ABC is a right angled triangle at B.

Question 21.
The equilateral triangles are drawn on the sides of a right triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of the triangles on the other two sides.
Answer:


∴ Area of ∆ XAB + Area of ∆ YBC = Area of ∆ ZAC .

OR

In the given figure, PA, QB and RC are each perpendicular to AC. Prove \(\frac{1}{x}+\frac{1}{z}=\frac{1}{y}\)
Answer:

Question 22.
Two dice are thrown simultaneously. Find the probability that the sum of the numbers on the faces is neither divisible by 4 nor by 5.
Answer:
S = {(1, 1) (1,2) (1, 3) (1,4) (1, 5) (1, 6) (2, 1) (2, 2) (2,3) (2,4) (2, 5) (2,6) (3.1) (3, 2) (3, 3) (3, 4) (3, 5) (3,6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4,6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5,6) (6.1) (6,2) (6, 3) (6,4) (6, 5) (6,6)}
n(S) = 36.
An event of getting sum of the numbers neither divisible by 4 nor by 5.
A={ 1,1) (1,2) (1,5) (1,6) (2,1) (2, 4)
(2, 5) (3, 3) (3, 4) (3, 6) (4, 2) (4,3) (4, 5) (5,1) (5, 2) (5, 4) (5, 6) (6,1) (6, 3) (6, 5)} n(A) = 20
n(A) = 20
P(A) = \(\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{20}{36}\)

Question 23.
Draw a circle of radius 3 cm. Take a point P outside the circle without using the centre of the circle, draw to tangents to the circle from an external point P.
Answer:

Question 24.
Prove that (cosecθ-cotθ)² = \(\frac{1-\cos \theta}{1+\cos \theta}\)
Answer:
Consider LHS
(cosecθ-cotθ)² = cosec² θ +cot² θ – 2cosecθ.cotθ

OR

If sinθ + cos θ = √2 sin (90 -θ) determine cot θ
Answer:
sin θ + cos θ = √2 sin (90 – θ)
sin θ + cos θ = √2 cos θ
sin θ = ypi cos θ – cos θ
sin θ = cos θ (√2 -1)
sinθ /cosθ = √2 -1

IV. Answer the following : ( 3 × 9 = 27 )

Question 25.
Asha is 5 times as old as her daughter Usha, 5 years later Asha will be 3 times as old as her daughter Usha. Find the present ages of Asha and Usha.
Answer:
Let the age of Asha be x years
Let the age of Usha be y years
According to question x=5y
x – 5y = 0 …….(1)
Five years later,
(x + 5) = 3(y + 5)
x + 5 = 3y + 15
x – 3y = 15 – 5
x – 3y = 15 – 5
x – 3y = 10 (2)
Substitute x = 5y in equation (2)
5y – 3y = 10
2y = 10
y = 10/2
y = 5
∴ x = 5y
= 5 × 5
x = 25
∴ The present age of Asha is 25 years.
The present age of Usha is 5 years.

OR

The sum of 2 digits of a 2 digit number is 12 the number obtained by interchanging the digits exceeds by the given number by 18. Find the number.
Answer:
Let the two digits be x and y
The 2 digit number will be 10x + y
The sum of digits = x + y – 12 ……..(i)
The number obtained by interchanging the digits =10y + x
= 10x + y + 18
10y + x – 10x – y = 18
9y – 9x = 18
Divide by 9
y – x = 2
y = x + 2 ….(ii)
Consider equation (1)
x + y = 12
Substitute y = x + 2
x + x + 2 = 12 .
2x =12 – 2
2x= 10
x= 5
y = x + 2
y = 5 +2
y = 7
∴ The number is : 10x +y
= 10(5)+7 = 50+ 7
= 57

Question 26.
Find the other two zeroes of the polynomial y⁴ + y³ – 9y² – 3y + 18 if the
zeroes are √3 and -√3
Answer:
y = √3 and y = -√3
(y + √3) = o (y-V3) = o
(y + √3 )(y – √3 ) = o
y(y – √3 ) + √3 (y – √3 ) = 0
y² – √3y + √3 y – 3 = o
y² – 3 = 0

y² + y – 6 = 0
y² + 3y – 2y – 6 = 0
y(y + 3)-2(y + 3) = 0
(y + 3) = 0 & (y – 2) = 0
y=-3 & y = 2
The four zeros of polynomial are √3 , -√3 , -3 & 2

Question 27.
Solve for x.
\(\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}\) (Where a ≠ 0, b ≠ 0 x ≠ 0, x ≠ -a,-b
Answer:

(-a – b)ab = (a + b)x (a + b + x)
-a²b – ab² = (a + b) (ax + bx + x²)
-a²b – ab² = a²x + abx + ax² + abx + b²x + x²b
-a²b – ab²= a²x + 2abx + ax² + b²x + x²b
-ab(a+b) = ax(a+x) + xb(b+x)+2abx
=x[a(a+x)+b(b+x)+2ab]
=x[a² + ax+b² +bx + 2ab]
= x[(a+b)² + x(a+b)]
-ab(a + b) = x(a+b)[a+b+x]
-ab = x(a + b + x)
-ab = ax + bx + x²
x2 + x (a + b) + ab = 0
x² + ax + bx + ab = 0
x(x+a) + b(x + a) = 0
(x + a) (x +b) = 0
=>x + a = 0, x + b = 0
=> x = -a, x = -b

OR

The diagonal of a rectangular field is 60m more than the shorter side. If the larger side is 30m more than the shorter side, find the sides of the field.
Answer:
Let the shorter side of the rectangular field be x m.
Longer side is x + 30 and the diagonal is x+60

In ∆ ABC,
AC² = AB² + BC²
(x + 60)² = ( x + 30)²+ x²
x² + 3600 + 120x = x² + 900 + 60x + x²
x² + 900 + 60x – 3600 – 120x = 0
x² – 60x – 2700 = 0
x² – 90x + 30x – 2700 = 0
x(x – 90) + 30(x – 90) = 0
(x – 90) (x + 30) = 0
x – 90 = 0, x + 30 = 0
x = 90, x = – 30.
∴ The shorter side = x = 90m
The longer side = x + 30 = 90+30 = 120m.

Question 28.
If the points (7, -2) (5, 1) and (3, 5) are collinear. Find the value of k.
Answer:
Since the give points are collinear, the area of the triangle formed by them must be O i.e.
= \(\frac { 1 }{ 2 }\) [x₁(y₂-y₃) + x₂(y₃-y₂) + x₃(y₁-y₂)]
x₁ = 7, x₂ = 5, x₃ = 3
y₁ = – 2, y₂ = 1, y₃ = k
= \(\frac { 1 }{ 2 }\) [7(1 – k) + 5 (k + 2) + 3(-2-1)] = 0
= \(\frac { 1 }{ 2 }\) [7 – 7k + 5k + 10 – 9]-0
= \(\frac { 1 }{ 2 }\) [7 – 2k + 1] = 0
= \(\frac { 1 }{ 2 }\) [8 – 2k] = 0
8-2k = 0 or 2k=8
k = 8/2 = 4
k = 4

OR

Find the area of Rhombus if its vertices are (3, 0) (4, 5) (-1, 4) and (-2, -1) taken in order.
Answer:

Area of rhombus = \(\frac { 1 }{ 2 }\) d₁ × d₂
Let A = (3, 0) B = (4, 5)
C = (-1, 4) D = (-2, -1)
Diagonal

Question 29.
Prove the tangents drawn from an external point to a circle are equal.
Answer:

Data: A is the centre of the circle B is an external point.
BP and BQ are the tangents.
AP, AQ and AB are joined.
To prove that: a) BP = BQ
b) ∠PAB-∠QAB.
c) ∠PBA = ∠QBA
Proof: In ∆ APB and ∆ AQB
AP = AQ [Radii of same circle]
∠ APB – ∠AQB = 90°
AB = AB [common side]
∴  ∆ APB ≅ ∆ AQB [RHS]
a) PB = QB
b) ∠PAB = ∠QAB
c) ∠PBA = ∠QBA

Question 30.
Find the area of the shaded region in the figure, where ABCD is a square of side 14 cm
Answer:

Area of square ABCD
= 14 × 14=196 cm²

Hence, area of the shaded region
= Area of square – Area of four circles
= 196 – 154
= 42cm²

OR

Find the area of the shaded regions. Given PQRS a square of sides 14cm. Soln: S R
Answer:

Question 31.
The following table gives the weight of 120 articles :

Change the distribution to a “more than type” distribution and draw its ogive.
Answer:

Question 32.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Weight (in kg)

Answer:

Question 33.
Construct a right triangle whose hypotenuse and one side measures 10cm and 8cm respectively. Then construct another triangle whose sides are 4/5 times the corresponding sides of the triangle .
Answer:

V. Answer the following ( 4 × 4 = 16 )

Question 34.
Solve the pair of equations graphically.
x + y = 8 and x – y = -2
Answe:
x + y = 8
y = 8 – x

x – y = -2
x = -2 + y

x = 6/2 = 3
x + y = 8
3 + y = 8
y = 8 – 3
y = 5

Question 35.
Devide 20 into four parts which are in arithmetic progression and such that the product of first and fourth is to the product of second and third in the ratio 2:3.
Answer:
Let the four parts in A.P be :
a – 3d, a – d, a +d, a + 3d
Their sum = (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20
⇒ 4a = 20 or a = 5
Also, it is given that

or 75 -27d² = 50 – 2d²
25d² = 25 .
⇒ d² = 1 or d = +√1 = ±1
When a = 5, d = 1, the four parts are 2, 4, 6,8
When a = 5, d = -1; the four parts are 8, 6, 4,2
Hence the four parts are (2, 4, 6, 8) or (8, . 6,4,2)

OR

The angles of a quadrilateral are in AP such that the greatest is double the least calculate all the angles of the quadrilateral.
Answer:
Let the quadrilateral be ABCD
The angles of the quadrilateral are in AP.
∴ The four angles are A, B, C and D which are a -3d, a – d, a + d, a + 3d respectively.
⇒ The least angle is a – 3d and the greatest angle is a + 3d.
It is given that
a + 3d = 2(a – 3d)
a + 3d = 2a – 6d
2a – a – 6d – 3d = 0
a – 9d ……(1)
9d = a
But ∠A + ∠B+∠C + ∠D = 360°
a – 3d + a – d + a + d + a +.3d = 360°
4a = 360 / 4 = 90°
But 9d = a
d = \(\frac{a}{9}=\frac{90}{9}\) = 10°
∴ ∠A=a- 3d = 90 – 3(10) = 90 -30 = 60°
∠B = a – d = 90 – 10 = 80°
∠C = a + d = 90 + 10= 100
∠D = a + 3d = 9 + 3(10) = 90 + 30 = 120

Question 36.
A person on the lighhouse of height 100m above the sea level observes that the angle of depression of a ship sailing towards the light house changes from 30° to 45°. Calculate the distance travelled by the ship during the period of observation. (Take √3 ≈ 1.73)

Answer:
P is the top of the light house PQ. We are given that its height PQ = 100m PX is horizontal line through P
∠APX = 30° and ∠BPX =45°
Then, ∠PAQ =30° and ∠PBQ = 45°
Suppose, AB = x metre and BQ = y metre
Karnataka SSLC Maths Model Question Paper 2 with Answers – 40
⇒ x + y =100√3 = 100√3 = 100× 1.732
⇒ x +y = 173.2
⇒ x + 100 = 173.2
⇒ x = 73.2
Therefore, the distance travelled = 73.2m

Question 37.
Prove that “the ratio of areas of two similar triangles is equal to the square of the ratio of their altitudes.
Answer:

Data : ∆ABC ∼ ∆DEF
∠A = ∠D
∠B = ∠E
∠C = ∠F
\(\frac{A B}{D E}=\frac{A C}{D F}=\frac{B C}{E F}\)
T.P.T : \(\frac{\text { Area of } \Delta \mathrm{ABC}}{\text { Area of } \Delta \mathrm{DEF}}=\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}=\frac{\mathrm{AC}^{2}}{\mathrm{DF}^{2}}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}\)
Construction:Draw AM ⊥ BC and AN ⊥ EFim

VI. Answer the fallowing : ( 5 × 1 = 5 )

Question 38.
The radii of the circular ends of the frustrum of height – 6cm are 14 cm and 6cm respectively. Find the lateral surface area and total surface area of frustrum.
Answer:
R = 14cm, r = 6cm and h =6cm
Now, let 1 be the slant height of the frustrum;

Now, lateral surface area of the frustrum
= π(R + r)l
= π (14 + 6) 10
= 628.57 cm²
And total surface area of the frustrum = π[(R+r)l + R²+ r²]
= π [(14 + 6)10 + (14)² + (6)²]
= π(200 + 196 + 36)
= π(432)
= 1357.71 cm².