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**Q1. A social welfare society donates a certain monthly amount towards the education of students from an economically weaker section with a fixed increment every year. The donation after 4 years was ₹ 15000 and ₹ 18000 after 10 years.****(i) What was the starting donation?****(ii) What is the annual increment?****(iii) Which mathematical concept is used in the above problem?****(iv) By contributing monthly financial help to students of EWS, which value is depicted by the social welfare society?****Sol.** (i) Let the starting contribution be ₹ x and the fixed annual increment be

₹ y.

∴ Contribution after 4 years from the starting contribution = ₹(x + 4y)

Contribution after 10 years from the starting contribution = ₹(x + 10y)

⇒ We have,

x + 4y = 15000 ...(1)

x + 10y = 18000 ...(2)

Subtracting (1) from (2),

⇒ y = 3000/6 = 500

Substituting y = 500 in (1), we have

x + 4(500) = 15000

⇒ x + 2000 = 15000

⇒ x = 15000 – 2000 = 13000

Thus, the starting donation was ₹ 13000.

(ii) ∵ y = 500

∴ The annual increment is ₹ 500

(iii) Pair of Linear Equations in two Variables

(iv) Charity**Q2. The ratio of pocket money of two students is 7: 9 and the ratio of their expenditures is 3: 4. If each of them saves ₹ 200 per month for donation to a Blind school. ****(i) Find their respective pocket money and amount of donation.****(ii) Which mathematical concept is used in the above problem?****(iii) By donating their saving from pocket money to Blind school, which value is depicted by the students?****Sol.** (i) Let the pocket money of the first student be ₹ 7x and that of the second student be ` 9x.

Let the expenditure of the first student be ₹ 3y and that of the second student be 4y.

∴ Saving of the first student = 7x – 3y

Saving of the second student = 9x – 4y

Solving the set of above simultaneous equations, we have :

4 × [7x – 3y = 200]

⇒ 28x – 12y = 800 ...(1)

3 × [9x – 4y = 200]

⇒ 27x – 12y = 600 ... (2)

Subtracting (2) from (1),

Substituting

x = 200, in 7x – 3y = 200, we get

7 × 200 – 3y = 200

⇒ 1400 – 3y = 200

⇒ –3y = 200 – 1400 = –1200

⇒ y = -1200/-3 = 400

Now, 7x = 7 × 200 = 1400; 9x = 9 × 200 = 1800 and donation = ₹ 200 each

∴ Pocket money of first student = ₹ 1400

Pocket money of second student = ₹ 1800

(ii) Pair of Linear Equations in two variable.

(iii) Charity**Q3. The progressive club helps needy students by giving free textbooks and stationery items. The club purchased 5 books and 7 pens together cost ₹ 79 whereas 7 book and 5 pens together cost ₹ 77.****(i) Find the total cost of 100 books and 100 pens.****(ii) Which mathematical concept is used in the above problem?****(iii) By donating books and stationery items to needy students which value is depicted by progressive-club?**

**Sol.** (i) Let the cost of one book be ₹ x and that of one pen be ₹ y.

∴ Cost of 5 books + cost of 7 pen = ₹ 79

⇒ 5x + 7y = 79 ...(1)

Also, cost of 7 books + cost of 5 pens

= ₹ 77

⇒ 7x + 5y = 77 ...(2)

To solve (1) and (2), we have

⇒

⇒ 35x + 49y = 553 .....(3)

35x + 25y = 385 ....(4)

Subtracting (4) from (3), we have

24y = 168 ⇒ y = 168/24 = 7

Substituting y = 7 in (1),

5x + (7 × 7) = 79

5x + 49 = 79

5x = 79 – 49 = 30

⇒ x = 30/5 = 6

i.e. cost of 100 books = ₹ 7 ×100 = ₹ 700

cost of 100 pens = ₹ 6 × 100 = ₹ 600

⇒ cost of 100 books + cost of 100 pens = ₹ 700 + ₹ 600 = ₹ 1300

(ii) Pair of Linear Equations in two variables.

(iii) Charity.**Q4. Aniket saves some amount daily from his pocket money. He saved ₹ 200 and gave to his younger sister Akhila, who wished to enjoy a fair. She wanted to enjoy rides on the Giant-Wheel and play Hoopla (a game in which you throw a ring on the items kept in a stall, and if the ring covers any object completely, you get it). The number of times she played Hoopla is half the number of rides she had on the Giant wheel.****(i) If each ride costs ₹ 30 and a game of Hoopla costs ₹ 40, find the number of rides she had and the number of times she played Hoopla, provided she spent ₹ 200?****(ii) Which mathematical concept is used in the above problem?****(iii) By saving and giving the amount from his pocket money, which value is depicted by Aniket?****Sol.** (i) Let the number of rides that Akhila had = x

And number of times Akhila played Hoopla = y

∴ According to the condition, we get

y = ....(1)

and ..(2)

Solving (1) and (2), we have

30x + 20x = 20

⇒50x = 200

x = 200/50 = 4

Again, substituting, x = 4 in (1), we get

Thus, the required number of rides is 2 and the number of Hoopa plays is 4.

(ii) Pair of Linear Equations in Two Variables.

(iii) Savings to give happiness to youngers.**Q5. Mr Ajay, the mathematics teacher encouraged 40 of his class X students to take part in a Mathematics quiz. If the number of girls is 4 more than the number of boys.****(i) Find the number of boys and girls who took part in the quiz.****(ii) Which mathematical concept is used in the above problem?****(iii) By encouraging the students to take part in quiz-contest, which value is depicted by Mr Ajay, the mathematics teacher?****Sol.** (i) Let the number of boys be x and the number of girls be y

∴ 3x + 7y = 14

∴ x + y = 40 ...(1)

According to the condition

[Number of girls] = [Number of boys] + 4

⇒ y = x + 4

or x – y = –4 ...(2)

Solving (1) and (2),

From (1), x + y = 40

⇒ 18 + y = 40

⇒ y = 40 – 18 = 22

Thus,

(ii) Pair of Linear Equations in two variables.

(iii) Competitive spirit.

**Q6. Mrs Sharma is a housewife, she arranges a mathematics test for her daughter every Sunday. Her daughter scored 40 marks in one of such test receiving 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then her daughter would have scored 50 marks.****(i) How many questions were there in the test?****(ii) Which mathematical concept is used in the above problem?****(iii) By giving a test to her daughter (with negative marking) every Sunday which value is depicted by Mrs Sharma? Sol.** (i) Let the number of questions attempted with correct answer = x.

Let the number of questions attempted with wrong answer = y.

Marks deducted for 1 wrong answer = 1

∴Total marks obtained = 3x – y

3x – y = 40 ...(1)

Marks deducted for 1 wrong answer = 2

∴Total marks = 4x – 2y

⇒ 4x – 2y = 50 ...(2)

To solve (1) and (2), we have

⇒ x = 30/2 = 15

From (1), we have

3x – y = 40 ⇒ 3(15) – y = 40

⇒ 45 – y = 40

⇒ y = 45–40 = 5

Thus, number of questions attempted correct = 15

number of questions attempted incorrect = 5

⇒ Total number of questions in the test = 15 + 5 = 20

(ii) Simultaneous equations in two variables.

(iii) Competitive spirit.

**Q7. A group of people take a voluntary decision to undertake a community task 2 women and 5 men of them can together finish the task in 4 days whereas 3 women and 6 men can finish the same task in 3 days.****(i) Find the time taken by 1 woman alone to finish the task.****(ii) Find the time taken by 1 man alone to finish the task****(iii) By taking a decision to do a community task, which value is depicted by the group of people.****Sol. **(i) Let one woman can finish the task in x days.

Let one man can finish the task in y days.

∴ Part of work finished by a woman in 1 day = 1/x.

Part of work finished by a man in 1 day = 1/y.

∴ (2 women’s 1-day work) + (5 men’s 1-day work) = 1/4

∵ They together finish the work in 4 days.

⇒ ...(1)

Also (3 women’s 1 day work) + (6 men’s 1 day work) = 1/3 ...(2)

...(2)

Solving (1) and (2), we have

⇒

⇒ 3/y = 1/12

⇒ y = 3 × 12 = 36 days

Substituting y = 36 in (2), we get

⇒

⇒

⇒ x = 6 × 3 = 18 days

Thus, one man can alone finish the task in 36 days.

(ii) And one woman can alone finish the task in 18 days.

(iii) Community service.**Q8. Mr Saini is a retired person. He has set up a library for lending books and donates the proceeds to an orphanage. His library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for 7 days, while Susy paid ₹ 21 for the book she kept for 5-days.**

Let the charge for each extra day = ₹ y

For Saritha

[Amount for first 3 days] + [Amount for next 4 days] = ₹ 27

⇒ x + 4y = 27 ...(1)

For Susy

[Amount for first 3 days] + [Amount for 2 extra days] = ₹ 21

⇒ x + 2y = 21 ...(2)

Solving (1) and (2) we get

Substituting y = 3 in (1), we have

x + 4(3) = 27

⇒ x + 12 = 27

⇒ x = 27 – 12 = 15

Thus, fixed charge = ₹ 15

Charges for each extra day = ₹ 3

(ii) Pair of Linear equations in two variables.

(iii) Charity.

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