Category: Class 7

KSEEB Solutions for Class 7 English Poem Chapter 7 Dear Grandma and Grandpa is available Online for students. Revise all the concepts of Chapter 7 Dear Grandma and Grandpa Questions and Answers easily taking help from the KSEEB Solutions. Download the KSEEB Solutions for Class 7 English PDF free of cost and get good scores in the board exams. English Chapter 7 Dear Grandma and Grandpa Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 7 English Karnataka State Board Solutions help you to revise complete Syllabus.

Karnataka State Board Class 7 English Poem Chapter 7 Dear Grandma and Grandpa

Enhance your subject knowledge through the KSEEB Solutions for Class 7 English Dear Grandma and Grandpa Questions and Answers lay a stronger foundation of your basics. Verify your answers with the Karnataka State Board Class 7 English Chapter 7 Dear Grandma and Grandpa with Answers provided and know where you went wrong.

Dear Grandma and Grandpa Questions and Answers, Summary, Notes

Before You Read:

Speaking practice

Look at the following picture and speak as many sentences as you can to describe what they are doing.
Recall Activity: Write as many words as you can.
He/she is the most
Self-assessment.

C1. Answer the following questions in one-two sentences each:

Question 1.
Who is the speaker in the poem?
Answer:
The grandchildren are the speakers in the poem.

Question 2.
Who does ‘you’ refer to in the poem?
Answer:
‘You’ refer to the grandparents in the poem

Question 3.
How do grandparents make their grandchildren feel they are special?
Answer:
The grandparents are very much affectionate to their grandchildren. They talk kindly, treat with love, and made more fun. At that moment they were also behaving like children. Totally the grandchildren always feel their grandparents are special.

Question 4.
How is the company of grandchildren helpful to grandparents?
Answer:
Grandparents feel lonely in their old age. Nobody accompanied them. But grandchildren are depending on them, they are very eager to give love and take love. Both of them enjoyed and in this way, they are helpful and give new hopes in their life.

Question 5.
According to the poet, what is the best thing grandparents can do for their grandchildren?
Answer:
Showing love and kindness and giving. hugs are the best thing that grandparents can do.

Question 6.
What is the secret that the grandchildren share with their grandparents?
Answer:
The grandparents give the best hugs and it is better than anyone. The grandchildren love their grandparents with their full heart and they believe their hearts belong to the grandparents.

C2. Discuss with your partner and answer the following question.

Question 1.
How do you express your love for your grandparents?
Answer:
Self-assessment.

C3. Observe that the word ‘two’ inline-five rhymes with the word you inline-six. Can you find the other rhyming words in the poem?
Answer:
The other rhyming words in the poem are ones-smiles, best-rest, true-you.

Dear Grandma and Grandpa Summary in English

This is the poem of an unknown author. It is a very short poem. The main theme of this poem is the relationship between grandparents its guild children. towards their grandchildren. They see the things and world through the eyes of the child. The grandchildren need love and affection, their company, hugs, etc., The grandchildren get all these from their grandparents. In the company of children, they can also feel young and happy. For all the kind deeds the grandchildren gave their heart to them. Their relationship was really unforgettable.

Dear Grandma and Grandpa Summary In Kannada

The above furnished information regarding KSEEB Solutions for Class 7 English Chapter 7 Dear Grandma and Grandpa Questions and Answers is true as far as our knowledge is concerned. If you have any doubts feel free to reach us via the comment section and we will reach you at the soonest possible.

KSEEB Solutions for Class 7 English Prose Chapter 7 A Tribute to Netaji is available Online for students. Revise all the concepts of Chapter 7 A Tribute to Netaji Questions and Answers easily taking help from the KSEEB Solutions. Download the KSEEB Solutions for Class 7 English PDF free of cost and get good scores in the board exams. English Chapter 7 A Tribute to Netaji Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 7 English Karnataka State Board Solutions help you to revise complete Syllabus.

Karnataka State Board Class 7 English Prose Chapter 7 A Tribute to Netaji

Enhance your subject knowledge through the KSEEB Solutions for Class 7 English A Tribute to Netaji Questions and Answers lay a stronger foundation of your basics. Verify your answers with the Karnataka State Board Class 7 English Chapter 7 A Tribute to Netaji with Answers provided and know where you went wrong.

A Tribute to Netaji Questions and Answers, Summary, Notes

I. Take turns with your partner to ask and answer the questions given below.

Explain to him/her why you think your answer is correct. Quote from the text if necessary. Write down your answers.

Question 1.
How does Subha’s uniqueness come out in his struggle for freedom? How does he differ from other national leaders? Give two examples.
Answer:
Subhas Chandra Bose was the leader who thought of fighting the brute force of the British by armed force. He was a revolutionary freedom fighter and leader with indomitable courage and vigour with continuous effort he moulded himself as one of the most promising leaders during the Indian freedom struggle.

He cast his influence on several Indians spread all over the world, he moved ahead with his meticulous plans to wage a war against the British. His political views were different from those of other important leaders. He differed in their opinions and ultimately it lead to disassociation with Congress.

Bose was of the opinion that India should take advantage of the second World War in which Britain also participated. He thought that British should grant independence in return to the help Indians will give Britain in the war. But prominent leaders like Gandhiji and Nehru differed with him.

Question 2.
Subhas was a wonderful organiser. Give three examples of this.
Answer:
Subhas was a wonderful organiser. One of the example is he raised a modem army of about 75,000 men and women. He inspired them with intense patriotic fervour and a burning desire to free their enslaved motherland.

Vivekananda taught him the way to spiritual development. He also taught yogic exercise has to be supplemented by social service. Bose joined a group students who used to go to villages to bring relief to victims of epidemics like plague, cholera and smallpox.

In the Presidency college, Calcutta, he organised a students strike to demonstrate against an English professor who had insultingly pushed away an Indian student for this, Bose was expelled from college.

Question 3.
Subhas was an inspirational figure. Give three examples for this.
Answer:
Nethaji organised Indian National Army (I.N.A) to fight against the British. The Indians soldiers and civilians at once declared their allegiance to him and began to call him ‘Nethaji’. He gave them the inspiring call of ‘ Jai hind’ and “Dilli chalo”.

Bose has strong views and advocated total freedom. He traveled throughout the country to organize public opinion against the was effort. Bose went on hunger strike against the ill-treatment of prisoners inspiring Indian to wage was against British.

Bose preferred a life of trials and tribulations as a fighter for his country’s freedom. He was a doer as well as thinker, and a fighter who never submitted to defeat. In life, he was a natural leader of men who inspired immense devotion and loyalty.

Question 4.
Subhas was broad-minded and spiritual. He did social work even as a young man. He worked for national integration and he was a man of vision. Give examples.
Answer:
He was broad-minded and looked at India in an international perspective. He had a special regard for his Muslim countrymen. He had very broad-minded views on freedom and other social issues and his hungry soul was not satisfied with textbooks alone.

He had a deeply spiritual nature. He went to Shri Ramakrishna Paramahamsa and Swami Vivekananda and learned the practice of yoga. Vivekananda taught him the way to spiritual development.

He learned yoga exercises and involved in social service. He joined a group of students and visited villages to give relief to victims of epidemics like plague, cholera, and smallpox.

Question 5.
Give an account of the sacrifices Subhas made.
Answer:
Subhas promised his father .to take up the ICS examination, and after a year he came out passing in merit. But his conscience did not allow him to serve the alien rulers even as a hight executive he resigned the job to sacrifice his life to serve the motherland.

When Britain was engaged in a world war, Bose traveled throughout the country to organize public opinion against the war effort. So he was arrested and went on a hunger strike to protest against the ill-treatment of prisoners. He was ready to undergo sacrifice until he got justice.

Question 6.
Give an account of Subhas’ adventurous life.
Answer:
One of the most adventurous- was the formation of in a to fight against the British the Indian soldiers and civilians at once declared to call him ‘Nethaji’. The army fought so well, soon liberated about 3,200 kms of Indian territory from the well-entrenched British army. But when the Japanese army was withdrawn, they were disarmed and court-martialled. As Bose boarded an aircraft to Tokyo, it caught fire while taking off. Nethaji was badly burnt and died in hospital the same night. This is one of the most adventurous act of Nethaji in his life.

Question 7.
Subhas worked in many different fields – administration, journalism, education, statesmanship and war. Give examples.
Answer:
Administrative field: Subhas made an impact on Congress that he was elected twice as President of INC in 1938 and 1939. In his presedential address, he talked not only of freedom but also of reconstruction, the need of planning by setting up planning commission and of gradual socialization of the entire agriculture and industrial system. He urged strongly to give an ultimate to the British government for complete “swaraj”.

Journalism: Bose’s intention was to bring awareness of polities of British as well as every indian to urge for the freedom. He found the latest media to reach the people through journals, newspapers After being released from the prison, he was appointed as the editor of party’s journal ‘Forward’. Bose became executive officer of Calcutta corporation and gained valuable experience of planning and practical administration.

Education field: Subhash joined Scottish Church College in Calcutta where he secured first class in B. A exam. Then he left for England to appear at ICS, he came fourth in order of merit. After having long discussions with Gandhi and C.R. Das he settled down as Principal which was later on resigned by him.

Statesmanship and war: Bose Believed that Gandhiji tactics of non-violence would .never be sufficient to secure India’s independence and advocated violent resistance He established a separate political party. All India forward Bolo and continued to fight for the full and immediate independence from British rule. This famous motto was “Give me blood and I will give you freedom At the outset of the war, he left India, traveling to Soviet Union, Nazi Germany and Imperial Japan seeking an alliance with each to attack the British government in India.

Word Formation:

a. Write three verbs and make the verbs adjectives by using the suffix ‘able’
Example: eat – eatable.

b. Make six sentence using the verbs and the adjectives you have written
Example: We gave the fruit to the monkey, The monkey ate the fruit. So we thought the fruit was eatable.

1. enjoy – enjoyable
Rita learnt photography. Though she was amateur photographer, she felt it is enjoyable.

2. move – moveable.
Shyam trying to move the big stone. After putting great effort it was moveable.

3. Watch – watchable
Girls saw the interesting movie which was even watchable by children.

4. Note – notable
Judged noted the point which was a notable point.
series

5. Walk – Walkable
After going to a temple they came to know temple was at walkable distance from their house.

6. Drink – Drinkable
Even though the drink was cold, it was drinkable.

Exercise

Combine these pairs of sentences into single sentences

a. The farmers loaded the paddy bags on their carts.
b. They drove back to the village.
c. Loading the paddy bags on their carts, farmers drove back to the village.

a. The weather was windy.
b. The boys came to the field to fly their kites.
c. Weather being windy, boys came to the field to fly their kites.

a. Raju had missed the last bus.
b. He walked all the way home.
c. Having missed the last bus. Raiu walked all the way home.

Writing:

Write a short paragraph about him/ her. You can use the title “The most unforgettable person I have met.”

The most unforgettable person I have met is my teacher who is always entrusted in educating children. She led her childhood miserably by which her parents couldn’t give her higher education But our teacher determined to educate her younger brother and sister by sacrificing her life to serve them. Even whenever she found a student Ms to pay the fees she used to pay for them and helped them to continue, their studies. Our teacher is very good inspiration to me. She always made our minds to think positively and be studying as long as we can Our teacher is the most unforgettable person I have met.

Question 1.
Write a brief family background of Nethaji.
Answer:
Subhas was born at Cuttak, Odisha on January 23, 1897. His father JanakinathBose was a famous lawyer. Bose was a brilliant student. Some of his ancestors like Mahipati Bose and Gopalnath Bose held the offices of minister and naval commander under the Muslim CPC Honest Success series rulers of Bengal. He was brought up in touch with English people, English education and English culture.

Question 2.
What was his opinion about muslims.
Answer:
The quarters in which we lived says Bose in his autobiography was prominently a Muslim one and our neighbours were mostly Muslims. I cannot remember ever to have looked upon muslims as different from ourselves in any way, except that they go to pray in a mosque.

Question 3.
How did Nethaji Seek the blessings of Gandhiji? What was it? Quote his words.
Answer:
While Launching the war against, the Imperial British army, Netaji went on air on July 6th, 1944 and sought Gandhiji’s blessings in the following words. ‘Father of our Nation, in this Holy war of India’s Liberation, we ask for your blessings and good wishes.”

A Tribute to Netaji Summary in English

Subhash Chandra Bose was a great freedom fighter. He inspired Indians to give their blood and he would give him freedom But Gandhiji promised to give freedom without violence. Hindu, Muslims and Sikhs adored Subhas and called him ‘Netaji’.

As he was a good organiser he organized an army of about 75,000 men and women. He inspired them with intense patriotic fervour and desire to free their enslaved motherland. Subhash had a special regard for his Muslim countrymen and he said they are not different from us in any way, except that they go to pray in a mosque.

Bose was rebellious, as a student he organised a student strike to demonstrate against an English professor who had insultingly pushed away an Indian student. Though he was appointed as High executive his conscience would not allow him to serve the alien rulers. So he resigned and returned to India.

In his presidential address, he talked not only of freedom but also of reconstructions need of planning by setting up of planning commission and of gradual socialization of the entire agriculture and industrial system.

Subhash went to south-east Asia and organized the I.N. A to fight against the British. He also took Gandhiji’s blessings and in a fought well, but when Japanese forces were withdrawn war went badly for them Nethaji was badly, burnt in a plane crash he was rushed to a hospital where died the same night. People refuse to believe that Nethaji is dead they are right, for martyrs never die.

A Tribute to Netaji Summary in Kannada

The above furnished information regarding KSEEB Solutions for Class 7 English Chapter 7 A Tribute to Netaji Questions and Answers is true as far as our knowledge is concerned. If you have any doubts feel free to reach us via the comment section and we will reach you at the soonest possible.

KSEEB Solutions for Class 7 English Supplementary Chapter 3 A Day’s Wait is available Online for students. Revise all the concepts of Chapter 3 A Day’s Wait Questions and Answers easily taking help from the KSEEB Solutions. Download the KSEEB Solutions for Class 7 English PDF free of cost and get good scores in the board exams. English Chapter 3 A Day’s Wait Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 7 English Karnataka State Board Solutions help you to revise complete Syllabus.

Karnataka State Board Class 7 English Supplementary Chapter 3 A Day’s Wait

Enhance your subject knowledge through the KSEEB Solutions for Class 7 English A Day’s Wait Questions and Answers lay a stronger foundation of your basics. Verify your answers with the Karnataka State Board Class 7 English Chapter 3 A Day’s Wait with Answers provided and know where you went wrong.

A Day’s Wait Questions and Answers, Summary, Notes

I. Answer these questions:

Question 1.
How old was Schatz? What disease was he suffering from?
Answer:
Schatz was nine years old. He was suffering from fever.

Question 2.
What is the relationship of the author with Schatz? Which line gives you that information?
Answer:
After a while he said to me, ‘You don’t have to stay in here with me, papa, if it bothers you gives us the information that the author and the boy Schatz are father and son.

Question 3.
How many capsules did the doctor prescribe? State the purpose of each one?
Answer:
Doctor gave three capsules, one was to bring down the fever, another a purgative, the third to overcome an acid condition.

Question 4.
How does the author describe nature?
Answer:
The day was bright and cold. The ground covered with sleet that had frozen. The hare trees, the bushes, the cut brush and all the grass and bare ground looked as if it had been varnished with ice. The ground was glassy and slippery. There was a covey of quail under a high clay bank. Some covey lit in trees. There was a frozen creek also.

Question 5.
“I took the young Irish setter up the road and along a frozen creek”. Who does the young Irish Setter refer to?
Answer:
Irish setter referred to a red coloured dog with long hair.

Question 6.
Why did the writer go out of the house? What did he hunt?
Answer:
Writer was a little puzzled by the son’s behaviour as he was trying to get him out of his room. So the writer thought his son is light-headed due to the temperature, went out for a walk. He hunted only two birds.

Question 7.
What happened at the house in the absence of the writer?
Answer:
In the absence of a writer, the father found his son exactly in the same position as he had left him. His face was white but the tops of his cheeks flushed by the fever, and he never allowed anyone in the house to enter the room as he did not want them to get what he had.

Question 8.
What caused Schatz to panic?
Answer:
The doctor had said that Schatz had a temp of 102°. In his school in France, the boys had said that no one could live with a temperature of 44°. He thought that his temperature was much higher so he was surely going to die so he was in a panic.

Question 9.
How could the author convince Schatz that nothing was wrong with him?
Answer:
The author convinced Schatz that nothing was wrong with him. There are different thermometer to read the body’s temperature. In one thermometer 37° is normal and the thermometer the doctor used to read his temperature is a different one. In this thermometer 98° is normal and nothing happens to a person if it does not cross 104° to 105°.

His temperature is 102 and it is very much safe. He gave an example of miles and kms. How many kilometers we make when we do seventy miles in the car. This clarification made Schatz think that there is nothing wrong with him and he was not going to die.

Question 10.
What was the effect of the clarification of Schatz? Which lines tell us about the effect?
Answer:
The father’s clarification made Schatz himself free from the fear of death. His gaze at the foot of the bed relaxed slowly. The holdover himself also relaxed. The next day he started crying early at little things that were of no importance.

The father spoke like this – It is like miles and kilometers. You aren’t going to die. That’s a different thermometer on that thermometer thirty- seven is normal. On this kind, it’s ninety – eight. The above lines made a great effect on Schatz.

Question 11.
Are these statements true or false:
a. Schatz was suffering from Pneumonia.
b. Schatz liked his father reading to him
c. The father shot down a fox when he was hunting.
d. The cat in the house accompanied his father when he went out hunting.
e. The normal temperature of the human body is 100 Degrees centigrade.
Answers:
a-false, b-false, c-false, d-false, e-true.

Think and answer

Question 1.
How was Schatz really cured? Was it by medicine or by the clarification given by his father?
Answer:
Schatz was really cured by the clarification given by his father.

Question 2.
Why didn’t the boy allow anyone into his room?
Answer:
The boy didn’t allow anybody because he thought he was affected by some dangerous disease. If anyone comes in contact with him he was afraid even they may get it and they will die and he wanted them to be safe.

Question 3.
Why do you think the boy liked to stay awake instead of going to sleep?
Answer:
Quite possibly, he was waiting for the hour to come when he would die, since he had the notion that with 102 degrees temperature he would not survive.

Question 4.
What was the boy waiting for? How did the wait end?
Answer:
The boy was waiting all day to die In France the boys had told him that if a patient had 44 degrees, he was sure to die. He heard the doctor say that he had 102, which was more than twice 44 he was sure to die, father laughed and explained, in France, they had a different kind of thermometer in which 37° is normal, but in our country 98 is normal So his temperature was just a little above normal. The boy relaxed immediately and the long wait ended.

A Day’s Wait by Ernest Hemingway About the Author:

Ernest Hemingway, the son of a doctor, was born in Illinois, USA. He began his literary career as a war correspondent during world war I -and went on to become a highly successful writer of prose fiction, both novels, and short stories. His best-known novels are ‘A farewell to Arms’ ‘For whom the bell tolls’ and ‘The old man and the sea’ for which he won the Pulitzer prize in 1953. Hemingway influenced a whole generation of writers by his distinctive narrative techniques, his depiction of human conflicts, and his economy of style. His themes include war, sports, hunting, violence, and married life. He was awarded the Nobel prize for literature in 1954 for his contribution to literature. Hemingway died in 1965.

A Day’s Wait Summary in English

The story opens as a father discovers that his 9-year-old boy, Schatz has a fever. The father sends for the doctor and he diagnoses a mild case of influenza. The doctor says that the temperature is 102° and also prescribes three kinds of medicines and said the boy will be well if they avoid pneumonia and the fever did not go above 104°.

When a father gives Schatz his medication, Schatz asks if he thinks the medication will help and the father answers affirmatively. After attempting to interest Schatz in the pirate book and foiling the father stopped reading. Then Schatz ask his father when the father thinks Schatz will die. From this father came to know that Schatz had heard at school in France that no one can live with a temperature above 44, so Schatz thinks he is sure to die with a temperature of 102. He has been waiting to die all day.

After the father explains the difference between Fahrenheit and Celsius, Schatz relaxes, letting go of his iron self-control, and the next day he allows himself to get upset over little things.

A Day’s Wait Summary In Kannada

The above furnished information regarding KSEEB Solutions for Class 7 English Chapter 3 A Day’s Wait Questions and Answers is true as far as our knowledge is concerned. If you have any doubts feel free to reach us via the comment section and we will reach you at the soonest possible.

Students can Download Chapter 8 Comparing Quantities Ex 8.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

Question 1.
Convert the given fractional numbers to percents.
a) \(\frac{1}{8}\)
Solution:
Percent means per hundred. So multi-plied by 100.

b) \(\frac{5}{4}\)
Solution:

c) \(\frac{3}{40}\)
Solution:

d) \(\frac{2}{7}\)
Solution:

Question 2.
Convert the given decimal fractions to per cents.
a) 0.65
Solution:

b) 2.1
Solution:

c) 0.02
Solution:

d) 12.35
Solution:

Question 3.
Estimate what part of the figures is coloured and hence find the per cent which is coloured.

Solution:

∴ In the first figure 25% is coloured.

∴ in the second figur e 50% Is coloured,.

∴ In the third figure 37.5 % is coloured.

Question 4.
Find :
a) 15% of 250
Solution:

b) 1% of 1 hour
Solution:

c) 20% of ₹ 2500
Solution:

d) 75% of ₹ 1 kg
Solution:

Question 5.
Find the whole quantity if
a) 5% of it is 600.
Solution:
Let the whole quantity be ‘m’
∴ 5% of m = 600

∴ The whole quantity is 12,000.

b) 12% of its 1080.
Solution:
Let the whole quantity be ‘m’
∴ 12% of m = Rs. 1080

∴ The whole amount be Rs. 9.000

c) 40 % of it is 500 km
Solution:
Let the whole quantity be ‘m’
∴ 40% of m = 500 km

∴ The whole amount is 1250 kms

d) 70% of it is 14 minutes
Let the whole quantity be ‘m’

∴ The whole amount is 20 minutes

e) 8% of it is 40 litres.
Let the whole quantity be ‘m’

∴ The whole amount be 500 liters

Question 6.
Convert given per cents to decimal fractions and also to fractions in simplest forms:
Solution:

Question 7.
In a city, 30% are females, 40% are males and the remaining are children. What percent are children?
Solution:
females = 30%
males = 40%
children = 100 – (30 + 40)
remaining are children = 100 – 70 = 30%
∴ Percentage of children = 30%

Question 8.
Out of 15,000 votes in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?
Solution:
Total no. of voters in a constituency } = 15,000
Percentage of voted = 60%
∴ Percentage of voters who did not vote
= 100 – 60 = 40%
Actual number of voters who did not vote} = 40% of 15,000

6000 voters who did not vote.

Question 9.
Meeta saves ₹ 400 from her salary. If this is 10% of her salary. What is her salary?
Solution:
Let Meeta’s salary be ‘M’
10% of M is = Rs. 400
∴ M =?

∴ Her salary be Rs. 4,000/-

Question 10.
A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?
Let the total matches won by them be M 25% of 20 = M

∴ They win 5 matches.

Students can Download Chapter 12 Algebraic Expressions Ex 12.4, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4

Question 1.
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.
If the number of digits framed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern.
How many segments are required to form 5, 10, 100 digits of the kind 6, 4, 8.

Solution:
i) The number of segments required to form
5 digits of this kind
5n + 1 = 5 × 5 + 1 = 25 + 1 = 26

ii) The number of segments required to form 10 digits of this kind
5n + 1 = 5 × 10 + 1 = 50 + 1 = 51

iii) The number of segments required to form 100 digits of this kind
5n + 1 = 5 × 100 + 1 = 501

Let the number of digits formed be ‘n’.
Therefore the number of segments required to form ‘n’ digits is given by the algebraic expression 3n + 1.
i) The number of segments required to form 5 digits of this kind.
3n + 1 = 3 × 5 + 1 = 15 + 1 = 16

ii) The number of segments required to form 10 digits of this kind
3n + 1 = 3 × 10 + 1 = 30 + 1 = 31

iii) The number of segments required to form 100 digits of this kind
3n + 1 = 3 × 100 + 1 = 300 + 1 = 301

Let the number of digits formed be ‘n’. Therefore the number of segments required to form ‘n’ digits is given by the algebraic expression 5n + 2.
i) The number of segments required to form 5 digits of this kind.
5n + 2 = 5 × 5 + 2 = 25 + 2 = 27
ii) The number of segments required to form 10 digits of this kind
5n + 2 = 5 × 10 + 2 = 50 + 2 = 52

iii) The number of segments required to form 100 digits of this kind
5n + 2 = 5 × 100 + 2 = 500 + 2 = 502

Question 2.
Use the given algebraic expression to complete the table of number patterns.
Solution:

Students can Download Chapter 12 Algebraic Expressions Ex 12.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3

Question 1.
If m = 2, find the value of:
i) m – 2
2 – 2 = 0

ii) 3m – 5
3 × 2 – 5 = 6 – 5 = 1

iii) 9 – 5m
9 – 5 × 2 ÷ 9 – 10 = -1

iv) 3m² – 2m – 7
= 3 × 2² – 2 × 2 – 7
= 3 × 4 – 4 – 7
= 12 – 11 = 1

v)

Solution:

Question 2.
If p = -2, find the value of :
Solution:
i) 4p + 7
= 4x – 2 + 7
= -8 + 7 = -1

ii) -3p² + 4p + 7
= -3 (-2)² + 4 (-2) + 7
= 3 × 4 – 8 + 7
= 12 – 8 + 7
= -13

iii) -2p³ – 3p² – 4p + 7
= -2 (-2)³ – 3(2)² + 4 (-2) + 7
= -2(-8) – 3(4) – 8 + 7
= 16 – 12 – 8 + 7
= 16 + 7 – 12 – 8
= 23 – 20 = 3

Question 3.
Find the value of the following expressions, when x = -1:

i) 2x – 7
= 2(-1) – 7
= -2 – 7 = -9

ii) -x + 2
= -1 + 2 = 1

iii) x² + 2x + 1
= 1 – 2 + 1 = 0

iv) 2x² – x – 2
= 2(-1)² – (-1) – 2
= 2 × 1 + 1 – 2
= 2 + 1 – 2 = 1

Question 4.
If a = 2, b = -2, find the value of:
i) a² + b²
= 2² + (-2)²
= 4 + 4 = 8

ii) a² + ab + b²
= 2² + 2 × -2 + (-2)
= 4 – 4 + 4 = 4

iii) a² – b²
= (2)² – (-2)² = 4 – 4 = 0

Question 5.
When a = 0, b = -1, find the value of the given expressions :
i) 2a + 2b
= 2 × 0 + 2 × -1
= o – 2 = -2

ii) 2a² + 2b² + 1
= 2(0)² + (-1)² + 1
= 0 + 1 + 1 = 2

iii) 2a²b + 2ab² + ab
= 2 × 0² × -1 + 2 × 0 × (-1)² + 0 × -1
= 0 + 0 + 0 = 0

iv) a² + ab + 2
= (0)² + 0 × -1 + 2
= 0 + 0 + 2 = 2

Question 6.
Simplify the expressions and find the value if x is equal to 2
Solution:
i) x + 7 + 4 (x – 5)
= x + 7 + 4x – 20 (re-arranging the terms)
= 5x – 13 = 5(2) – 12
= 10 – 13 = -3

ii) 3(x + 2) + 5x – 7
= 3x + 6 + 5x – 1
= 8x – 1 = 8(2) – 1
= 16 – 1 = 15

iii) 6x + 5 + 5(x – 2)
= 6x+ 5x – 10
= 11x – 10 = 11(2) – 10
= 22 – 10 = 12

iv) 4(2x – 1) + 3x + 11
= 8x – 4 + 3x + 11
= 8x + 3x – 4 + 11 (re-arranging the terms)
= 11x + 7 = 11(2) + 7 = 22 + 7 = 29

Question 7.
Simplify these expressions and find their values if x = 3, a = -1, b = -2.
i) 3x – 5 – x + 9
= 3x – x + 9 – 5
= 2x + 4 = 2(3) + 4 = 6 + 4 = 10

ii) 2 – 8x + 4x + 4
= 2 + 4 – 4x
= -4x + 6
= -4 × 3 + 6
= -12 + 6 = -6

iii) 3a + 5 – 8a + 1
= 3a – 8a + 5 + 1
= -5a + 6
= -5(-1) + 6
= 5 + 6 = 11

iv) 10 – 3b – 4 – 5b
= -3b – 5b + 10 – 4
= -8b + 6
= -8 (-2) + 6
= 16 + 16 = 22

v) 2a – 2b – 4 – 5 + a
= 2a – 2b – 9
= 3a – 2b – 9
= 3(-1) – 2 (-2) – 9
= -3 + 4 – 9
= -12 + 4 = -8

Question 8.
i) If z = 10, find the value of z³ – 3(z – 10)
= (10)³ – 3 (10 – 10)
= 1000 – 3(0)
= 1000 – 0 = 1000

ii) If p = -10, find the value of p² – 2p – 100
= (10)² – 2(-10) – 100
= 100 + 20 – 100
= 120 – 100
= 20

Question 9.
What should be the value of a if the value of 2x² + x – a equals to 5, when x = 0 ?
Solution:
= 2x² + x – a
= 2(0)² + 0 – a
= 5
= 0 + 0 – a = 5
∴ a = -5

Question 10.
Simplify the expression and find its value when a = 5 and b = -3, 2(a² + ab) + 3 – ab
= 2a² + 2ab + 3 – ab
= 2(5)² + 2 × 5 × -3 + 3 – 5 × -3
= 2 × 25 – 30 + 3 + 15
= 50 – 30 + 3 + 15
= 50 + 3 + 15 – 30
= 68 – 30 = 38

Students can Download Chapter 8 Comparing Quantities Ex 8.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1

Question 1.
Find the ratio of :
a) 5 to 5o paise
Solution:
Rs. 5 : 50 paise
5 × 100 : 50 paise
500 paise : 50 paise (Dividing by 50)
10 : 1

b) 15 kg to 210 g
15 kg: 210 g
15 × 1000: 210 g
15000 g : 210 g (Dividing by 30) 500 : 7

c) 9 m to 27 cm
9 m : 27 cm
9 × 100 : 27 cm
900 cm : 27 cm (Dividing by 9)
100 : 3

d) 30 days to 36 hours
30 days : 36 hours
30 × 24 : 36 hours
720 hours : 36 hours (Dividing by 36)
20 : 1

Question 2.
In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?
Solution:
Let ‘x’ computers to be needed for 24 students. Then the ratio will be 3 : x = 6 : 24

∴ 24 students need 12 computers.

Question 3.
Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs. Area of Rajasthan = 3 lakh km² and area of UP= 2 lakh km².
Solution:
Population of Rajasthan = 570 lakhs
Population of UP = 1660 lakhs
Area of Rajasthan = 3 lakh sq. kms
Area of U.P = 2 lakh sq. kms

i) How many people are there per km² in both these States ?
Solution:

ii) Which State is less populated?
Solution:
Rajasthan State is less populated.

Students can Download Chapter 7 Congruence of Triangles Ex 7.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2

Question 1.
Which congruence criterion do you use in the following ?
a) Given : AC = D
AB = DE
BC = EF

So, ∆ ABC = ∆ DEF
SSS congruence criterion

b) Given : ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ∆ PQR ≅ ∆ XYZ

SAS congruence criterion

c) Given: ∠MLN = ∠FGH
∠NML = ∠GFH
ML = FG
So, ∆LMN ≅ ∆GFH

ASA congruence criterion.

d) Given : EB = DB
AE = BC
∠A = ∠C = 90°
So, ∆ ABC ≅ ∆ CDB

RHS congruence criterion

Question 2.
You want to show that ∆ ART ≅ ∆ PEN,

a) If you have to use SSS criterion, then you need to show
i) AR =
AR = PE,
ii) RT =
RT = EN,
iii) AT =
AT = PN

b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have

i) RT =
RT = EN and
ii) PN =
PN = AT
c) If it is given that AT = PN and you to use ASA criterion, you need to have
i) ?
∠ATR = ∠PNE and
ii) ?
∠TAR = NPE.

Question 3.
You have to show that ∆ AMP ≅ ∆ AMQ.
In the following proof, supply the missing reasons

Question 4.
In ∆ ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ∆ PQR,

A student says that ∆ ABC ≅ ∆ PQR by AAA congruence criterion. Is he justified? Why or why not?
Solution:
No, he is not justified because AAA is not a criterion for the congruence of triangles.

Question 5.
In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ∆ RAT ≅ ?
Solution:

Question 6.
Complete the congruence statement:
Solution:

∆ BCA ≅ ? ∆ QRS ≅ ?
∆ BCA ≅ ? ∆ BTA ≅ ?
∆ QRS ≅ ? ∆ TPQ

Question 7.
In a squared sheet, draw two triangles of equal areas such that

i) The triangles are congruent.
Solution:
Consider the ∆^S PQS and SQR

In ∆ PQS and ∆ SQR
PS = QR = 6 cms
∠SPQ = ∠QRS = 90°
QS = QS = common
By RHS congruence criterion
∆PQS ≅ ∆SQR.
Perimeter of the ∆ PQS = PQ + QS + PS
Perimeter of the ∆ SQR = SR + QS + QR
∴ Perimeter of the ∆ PQS = Perimeter of the ∆ SQR (∵ PQ = SR & PS = QR)

ii) the triangles are not congruent. What can you say about their perimeters?
Solution:

∴ area of ∆ PQS = Area of ∆ PQM.
By seeing the figure the ∆ PQS = PQ + PS + SQ = 8 + 6 + 10 = 24 cms.
Perimeter of the ∆ PQM = PQ + PM + QM
= 8 + 7.2 + 7.2
= 22.4 cms.
∴ Their perimeters are not equal.
PM = QM
(PM)² = PN² + MN²
= 4² + 6²
= 16 + 36 = 52
PM = QM = \(\sqrt{52}\) = 7.2

Question 8.
Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.
Solution:

Considering the two triangles PQR and XYZ
In ∆ PQR and XYZ
PQ = XZ = 7 cms
PR = YZ = 6 cms
RQ = XY = 5 cms
∠PRQ = ∠XYZ
∠PQR = ∠XZY
In the above 2 triangles 5 pairs of the congruent present. Still, they are not congruent.

Question 9.
If ∆ ABC and ∆ PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

The additional corresponding part is BC = RQ by ASA congruence rules

Question 10.
Explain, why ∆ ABC ≅ ∆ FED

Solution:
∠ABC = ∠DEF = 90°
BC = DE ∠ACD = ∠EDF
( ∵ The sum of the measures of three angles of a triangle is 180°.)
∴ ∆ ABC ≅ ∆ DEF
(By ASA congruence criterion.)

Students can Download Chapter 12 Algebraic Expressions Ex 12.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

Question 1.
Simplify combining like terms :
i) 21b – 32 + 7b – 20b
(re-arranging the terms)
= 21b + 7b – 20b – 32
= 28b – 20b – 32
= 8b – 32

ii) z² + 13z² – 5z + 7Z² – 15z
(re-arranging the terms)
= -z² + 13z² – 5z – 15z + 7z³
= 12z² – 20z + 7z³

iii) p – (p – q) – q – (q – p)
= p – p + q – q – q + p
(re-arranging the terms)

= p – q

iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b – a
= 3a – a – a – 2b + b + b – ab – ab + 3ab

= a + ab

v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
(re-arranging the terms)
= 5x²y + 3yx² – 5x² + x² – 3y² – y² – 3y² + y²
= 8x²y – 4x² – 7y² + 8xy²

vi) (3y² + 5y – 4) – (8y – y² – 4) .
3y² + 5y – 4 – 8y – y² – 4
(re-arranging the terms)
3y² + y² + 5y – 8y – 4 + 4
= 14y² – 3y

Question 2.
Add:
i) 3mn, -5mn, 8mn, -4mn
3mn + 8nm – 5mn – 4mn
(re-arranging the terms)
= 11mn – 9mn = 2mn

ii) t – 8tz, 3tz – z, z – t

(re-arranging the terms)
= -8tz + 3tz = -5tz

iii) -7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
-7mn + 12 mn + 9mn – 2mn + 5 + 2 – 8 – 3
(re-arranging the terms)
= -7mn – 2mn + 12mn + 9mn + 7 – 11
= -9mn + 21 mn -4
= 12mn – 4

iv) a + b -3, b – a +3, a – b + 3

(re-arranging the terms)
= (2a – a + 2b – b + 6 – 3)
= a + b + 3

v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(re-arranging the terms)

= 17x + 51

vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(re-arranging the terms)
= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5
= 7m – 4m – 7n + 3n – 3mn – 3
= 13m – 4n – 3mn – 3

vii) 4x²y, – 3xy², 5y², 5x²y
(Re-arranging the terms)
4x²y + 5x²y – 3xy² – 5xy²
= 9x²y – 8xy²

viii) 3p²q² – 4pq + 5, 10p²q², 15 + 9pq + 7p²q²
3p²q² – 10p²q² + 7p²q² – 4pq + 9pq + 5+ 15
(Re-arranging the terms)
= 3p²q² + 7p²q² – 10p²q² – 4pq + 9pq + 5 + 15

= 5pq + 20

ix) ab – 4a, 4b – ab, 4a – 4b

(Re-arranging the terms)
(All terms are cancelling)
= 0

x) x² – y² – 1, y² – 1 – x², 1 – x² – y²
(Re arranging the terms)

= -x² – y² – 1

Question 3.
Subtract :
i) -5y² from y²
= y² – (-5y²)
= y² + 5y² = 6y²

ii) 6xy from – 12xy
= -12xy – (6xy)
= -12xy – 6xy = -18xy

iii) (a – b) from (a + b)
= a + b – (a – b)

= 2b

iv) a(b – 5) from b(5 – a)
= ab – 5a from 5b – ab
= (5b – ab) – (ab – 5a)
= 5b – ab – ab + 5a
= 5a – 5b – 2ab

v) -m² + 5mn from 4m² – 3mn + 8
= (4m² – 3mn + 8) – (-m² + 5mn)
= 4m² – 3mn + 8 + m² – 5mn
= 4m² + m² – 3mn – 5mn + 8
= 5m² – 8mn + 8

vi) -x² + 10x – 5 from 5x – 10
= (5x – 10) – (-x² + 10x – 5)
= 5x – 10 + x² – 10x + 5
= 5x – 10x + x² – 10 + 5
= x² – 5x – 5

vii) 5a² – 7 ab + 5b² from 3ab – 2a² – 2b²
= (3ab – 2a² – 2b²) – (5a² – 7ab + 5b²)
= 3ab – 2a² – 2b² – 5a² + 7ab – 5b
= -2a² – 5a² – 2b² – 5b² + 7ab + 3ab
= -7a² – 7b² + 10ab

viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq
= (5p² + 3q² – pq) – (4pq – 5q² – 3p²)
= 5p² + 3q² – pq² – 4pq + 5q² + 3p²
= 5p² + 3p² + 3q² + 5q² – 4pq – pq
= 8p² + 8q² – 5pq

Question 4.
a) What should be added to x² + xy + y² to obtain 2x² + 3xy ?
Solution:
= 2x² + 3xy – (x² + xy + y²)
= 2x² + 3xy – x² – xy – y² – 2x² – x² + 3xy – xy – y²
(re-arranging the terms)
∴ x² + 2xy – y² should be added to x² + xy + y² to get 2x² + 3xy

b) What should be subtracted from 2a + 8b + 10 to get -3a + 7b + 16 ?
Solution:
= (2a + 8b + 10) – (-3a + 7b + 16)
= 2a + 8b+ 10 + 3a – 7b – 16
= 2a + 3a + 8b – 7b + 10 – 16
(re-arranging the terms)
= 5a + b – 6
∴ (5a + b – 6) should be subtracted from 2a + 8b +10 to get -3a + 7b + 16.

Question 5.
What should be taken away from – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20?
Solution:
= (3x² – 4y² + 5xy + 20) – (-x² – y² + 6xy + 20)
= 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20
= 3x² + x² – 4y² + y² + 5xy – 6xy + 20 – 20
(re – arranging the terms)
= 4x² – 3y – xy
∴ 4x² + 3y² – xy should be taken away from 3x² – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20

Question 6.
From the sum of 3x – y + 11 and -y – 11, subtract 3x – y – 11.
Solution:

b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and -x² + 2x + 5.
Solution:

or
[(4 + 3x) + (5 – 4x + 2x²)] – [(3x² – 5x) + (x² + 2x + 5)]
= 4 + 3x + 5 – 4x + 2x² – 3x² + 5x + x² – 2x – 5 = 4 + 3x – 4x + 2x² + 5 – 3x² + x² + 5x – 2x – 5
(re-arranging the terms)

= 2x + 4

Students can Download Chapter 10 Practical Geometry Ex 10.5, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 10 Practical Geometry Ex 10.5

Question 1.
Construct the right angled ∆ PQR? where m∠Q = 90°, QR = 8 cm and PR = 10 cm.
Solution:

Solution:
Steps of Construction

1. Draw a line segment of length 8 cm and named as QR. At Q draw QM ⊥ QR.
2. With R as centre, draw an arc of radius 10 cm and cut the ⊥^le line at P.
3. Join PR. Now we get the required PQR ∆

Question 2.
Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4
cm long.
Solution:

Steps of Construction

1. Draw a line segment of length 4 cm and named it AB. At A draw a ⊥ line AM.
2. With B as centre, draw an arc of 6 cms cut the x line at ‘C ’. Now we obtained the required triangle ABC.

Question 3.
Construct an isosceles right-angled triangle ABC, where m∠ACB = 90° and AC = 6 cm.
Solution:

Steps of Construction

1. Draw a line segment of 6 cm, ie., AC. At C draw CM ⊥ CA.
2. With ‘C’ as the centre draw an arc of radius 6 cm to intersect CM at ‘B’
3. Join AB. Now we get the required ∆ACB.