CBSE Class 10 Mathematics Chapter 3 - Pairs of linear equations in two variables Exercise 3.3

 CBSE Class 10 Mathematics  Chapter 3 - Pairs of linear equations in two variables  Exercise 3.3

Q1.	Solve the following pair of linear equations by the elimination method and the substitution method :
(i)	x + y = 5 and 2x – 3y = 4
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(ii)

3x + 4y = 10 and 2x – 2y = 2

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Given equation: 3x + 4y = 10 ……………….. (i) 2x – 2y = 2……………. (ii) Multiply Equation (ii) by 2 to make coefficients of y equal: 2(2x − 2y) = 2(2) 4x − 4y = 4 …………… (iii) Add equation (i) and (iii) to eliminate y: (3x + 4y) + (4x − 4y) = 10 + 4 3x + 4x – 4y + 4y = 14 7x = 14 x = 2 Put value of x in equation (i) 3(2) + 4y = 10 6 + 4y = 10 4y = 10 – 6 4y = 4 y = 1 So, the solution of equation is x = 2 and y = 1

(iii)	3x – 5y – 4 = 0 and 9x = 2y + 7
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(iv)
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Q2.	Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :
(i)	If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
Ans.	Let fraction represent as fallow : x/y According to question- First condition: If we add 1 to the numerator and subtract 1 from the denominator, the fraction reduces to 1. (x + 1)/(y - 1) = 1 ………….. (i) Second condition: It becomes 1/2 if we only add 1 to the denominator. x/(y + 1) = ½ ………………….. (ii) From equation (ii) x = (1/2)(y + 1) Substitute from equation (i) [(1/2)(y + 1) + 1]/(y - 1) = 1 Multiply both sides by (y - 1) to eliminate the denominator (1/2)(y + 1) + 1 = y – 1 Multiply both sides by 2 to get rid of the fraction: y + 1 + 2 = 2(y - 1) y + 3 = 2y – 2 y = 5 put y value in equation (ii) x = (1/2)(y + 1) x = (1/2)(5 + 1) x = (1/2)(6) x = 3 So, the fraction is 3/5.

(ii)

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

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Let Nuri's current age is N and Sonu's current age is S According to question- First condition: Five years ago, Nuri was thrice as old as Sonu N – 5 = 3(S – 5) N – 5 = 3S – 15 N – 3S = – 10 ………….. (i) Second condition: Ten years later, Nuri will be twice as old as Sonu N + 10 = 2(S + 10) N + 10 = 2S + 20 N – 2S = 20 – 10 N – 2S = 10 ………………….. (ii) Subtract equation (ii) from equation (i) (N – 3S) – (N – 2S) = (– 10) – (10) N – 3S – N + 2S = – 10 – 10 After N term canceled  – 3S + 2S = – 20 – S  = – 20     (divide both side by –1) S = 10 Put S value in equation (ii) N – 2(20) = 10 N – 40 = 10 N – 40 = 10 N = 10 + 40 N = 50 So, Nuri (N) is 50 year old and Sonu (S) is 10 year old.

(iii)

The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

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Let, we can represent the two-digit number as 10a + b where "a" is the tens digit and "b" is the units digit. According to question- First condition: the sum of the digits is 9 a + b = 9  ………….. (i) Second condition: nine times this number is twice the number obtained by reversing the order of the digits. The number obtained by reversing the digits is 10b + a. so we can write as: 9(number) = 2(Reverse Number) 9(10a + b) = 2(10b + a) 90a + 9b = 20b + 2a 90a – 2a + 9b – 20b = 0 88a – 11b = 0 8a – b = 0 ……………. (ii) (a + b) + (8a – b) = (0) + (9) a + b + 8a – b = 0 + 9 9a = 9 a = 1 put value in equaton (i) 1 + b = 9 b = 9 – 1 b = 8 So, ten’s digit (a) is 1 and unit digit (b) is 8. Therefore required number is 18.

(iv)

Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received.

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Let number of Rs. 50 note is x and number of Rs. 100 note is y According to question- First condition: Meena received a total of 25 notes. x + y = 25                    (by Multiply both side by 50) 50x + 50y = 1250 ………….. (i) Second condition: The total amount she received is Rs. 2000. 50x + 100y = 2000  ………………….. (ii) Add equation (i) and (ii) (50x + 50y) + (50x + 100y) = 1250 + 2000 50x + 50y + 50x – 100y = 1250 + 2000 – 50y = – 750 y = 15 Put y value in equation (ii) x + 15 = 25 x  = 25 – 15 x  = 10 So, Meena gets Rs. 50 Note (x) is 10 and Rs. 100 note (y) is 15.

(v)

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

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Let Fix charge is x (in rupees) and additional charges for each day is y (in rupees) According to question- First condition: Saritha paid Rs. 27 for a book kept for seven days. Which can be written as: x (for the first 3 days) + 4y (for the remaining 4 days) = 27 x + 4y = 27 ………….. (i) Second condition: Susy paid Rs. 21 for a book kept for five days. Which can be written as: x (for the first 3 days) + 2y (for the remaining 2 days) = 21 x + 2y = 21  ………………….. (ii) subtract equation (ii) from equation (i) (x + 4y) – (x + 2y) = 27 – 21 x + 4y – x – 2y = 27 – 21  2y = 6                  (by x term cancel out) y = 3 Put y value in equation (ii) x + 2(3) = 21 x + 6 = 21 x  = 21 – 6 x  = 15 So, the fixed charge (x) for the first three days is Rs. 15 and the additional charge for each day is Rs. 3.