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**Q1. A social-welfare-society donates a certain monthly amount towards education of students from economically weaker section with a fixed increment every year. The donation after 4 years was â‚¹ 15000 and â‚¹ 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 â‚¹ x and the fixed annual increment be

â‚¹ y.

âˆ´ Contribution after 4 years from the starting contribution = â‚¹(x + 4y)

Contribution after 10 years from the starting contribution = â‚¹(x + 10y)

â‡’ We have,

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

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

Subtracting (1) from (2),

â‡’ y = 3000/6 = 500

Substituting y = 500 in (1), we have

x + 4(500) = 15000

â‡’ x + 2000 = 15000

â‡’ x = 15000 â€“ 2000 = 13000

Thus, the starting donation was â‚¹ 13000.

(ii) âˆµ y = 500

âˆ´ The annual increment is â‚¹ 500

(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 â‚¹ 200 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 first student be â‚¹ 7x and that of the second student be ` 9x.

Let the expenditure of first student be â‚¹ 3y and that of the second student be 4y.

âˆ´ Saving of first student = 7x â€“ 3y

Saving of second student = 9x â€“ 4y

Solving the set of above simultaneous equations, we have :

4 Ã— [7x â€“ 3y = 200]

â‡’ 28x â€“ 12y = 800 ...(1)

3 Ã— [9x â€“ 4y = 200]

â‡’ 27x â€“ 12y = 600 ... (2)

Subtracting (2) from (1),

Substituting

x = 200, in 7x â€“ 3y = 200, we get

7 Ã— 200 â€“ 3y = 200

â‡’ 1400 â€“ 3y = 200

â‡’ â€“3y = 200 â€“ 1400 = â€“1200

â‡’ y = -1200/-3 = 400

Now, 7x = 7 Ã— 200 = 1400; 9x = 9 Ã— 200 = 1800 and donation = â‚¹ 200 each

âˆ´ Pocket money of first student = â‚¹ 1400

Pocket money of second student = â‚¹ 1800

(ii) Pair of Linear Equations in two variable.

(iii) Charity**Q3. The progressive club help needy students by giving free text books and stationery items. The club purchased 5 books and 7 pens together cost â‚¹ 79 whereas 7 book and 5 pens together cost â‚¹ 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 â‚¹ x and that of one pen be â‚¹ y.

âˆ´ Cost of 5 books + cost of 7 pen = â‚¹ 79

â‡’ 5x + 7y = 79 ...(1)

Also cost of 7 books + cost of 5 pens

= â‚¹ 77

â‡’ 7x + 5y = 77 ...(2)

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

â‡’

â‡’ 35x + 49y = 553 .....(3)

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

Subtracting (4) from (3), we have

24y = 168 â‡’ y = 168/24 = 7

Substituting y = 7 in (1),

5x + (7 Ã— 7) = 79

5x + 49 = 79

5x = 79 â€“ 49 = 30

â‡’ x = 30/5 = 6

i.e. cost of 100 books = â‚¹ 7 Ã—100 = â‚¹ 700

cost of 100 pens = â‚¹ 6 Ã— 100 = â‚¹ 600

â‡’ cost of 100 books + cost of 100 pens = â‚¹ 700 + â‚¹ 600 = â‚¹ 1300

(ii) Pair of Linear Equations in two variables.

(iii) Charity.**Q4. Aniket saves some amount daily from his pocket money. He saved â‚¹ 200 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 â‚¹ 30 and a game of Hoopla costs â‚¹ 40, find the number of rides she had and the number of times she played Hoopla, provided she spent â‚¹ 200?****(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

âˆ´ According to the condition, we get

y = ....(1)

and ..(2)

Solving (1) and (2), we have

30x + 20x = 20

â‡’50x = 200

x = 200/50 = 4

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

Thus, the required number of rides is 2 and 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

âˆ´ 3x + 7y = 14

âˆ´ x + y = 40 ...(1)

According to the condition

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

â‡’ y = x + 4

or x â€“ y = â€“4 ...(2)

Solving (1) and (2),

From (1), x + y = 40

â‡’ 18 + y = 40

â‡’ y = 40 â€“ 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

âˆ´Total marks obtained = 3x â€“ y

3x â€“ y = 40 ...(1)

Marks deducted for 1 wrong answer = 2

âˆ´Total marks = 4x â€“ 2y

â‡’ 4x â€“ 2y = 50 ...(2)

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

â‡’ x = 30/2 = 15

From (1), we have

3x â€“ y = 40 â‡’ 3(15) â€“ y = 40

â‡’ 45 â€“ y = 40

â‡’ y = 45â€“40 = 5

Thus, number of questions attempted correct = 15

number of questions attempted incorrect = 5

â‡’ Total number of questions in the test = 15 + 5 = 20

(ii) Simultaneous equations in two variables.

(iii) Competitive spirit.

Let one man can finish the task in y days.

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

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

âˆ´ (2 womenâ€™s 1 day work) + (5 menâ€™s 1 day work) = 1/4

âˆµ They together finish the work in 4 days.

â‡’ ...(1)

Also (3 womenâ€™s 1 day work) + (6 menâ€™s 1 day work) = 1/3 ...(2)

...(2)

Solving (1) and (2), we have

â‡’

â‡’ 3/y = 1/12

â‡’ y = 3 Ã— 12 = 36 days

Substituting y = 36 in (2), we get

â‡’

â‡’

â‡’ x = 6 Ã— 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 â‚¹ 27 for a book kept for 7 days, while Susy paid â‚¹ 21 for the book she kept for 5-days.**

Let the charge for each extra day = â‚¹ y

For Saritha

[Amount for first 3 days] + [Amount for next 4 days] = â‚¹ 27

â‡’ x + 4y = 27 ...(1)

For Susy

[Amount for first 3 days] + [Amount for 2 extra days] = â‚¹ 21

â‡’ x + 2y = 21 ...(2)

Solving (1) and (2) we get

Substituting y = 3 in (1), we have

x + 4(3) = 27

â‡’ x + 12 = 27

â‡’ x = 27 â€“ 12 = 15

Thus, fixed charge = â‚¹ 15

Charges for each extra day = â‚¹ 3

(ii) Pair of Linear equations in two variables.

(iii) Charity.

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