### Chapter 3: Pair of Linear Equations in Two Variables

Q
##### Pair of linear equations in two variables CBSE NCERT Solutions

Question:

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 ½ if we only add 1 to the denominator. What is the fraction?

(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?

(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.

(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.

(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.

Solutions: (i) Let numerator is x and denominator is y

According to the given condition, we have

and

+ 1 = y – 1 and 2x = y + 1

– y = −2 … (1) and 2x – y = 1… (2)

So, we have equations (1) and (2), multiplying equation (1) by 2 we get.

2x − 2y = −4… (3)

2x – y = 1… (2)

Subtracting equation (2) from (3), we get

− y = − 5  = 5

Putting the value of y in (1), we get

– 5 = −2  = −2 + 5 = 3

Therefore, the fraction is:

(ii) Let the present age of Nuri is x years and the present age of Sonu is y years.

5 years ago, the age of Nuri is (x – 5) years.

5 years ago, the age of Sonu is (y – 5) years.

According to the given condition, we have

(− 5) = 3 (− 5)

– 5 = 3– 15

− 3= −10… (1)

10 years later from present, the age of Nuri is (+ 10) years

10 years later from the present, the age of Sonu is (+ 10) years

According to the given condition, we have

(+ 10) = 2 (+ 10)

+ 10 = 2+ 20

− 2= 10 … (2)

Subtracting equation (1) from (2), we get

= 10 − (−10) = 20 years

Putting the value of y in (1), we get

– 3 (20) = −10

– 60 = −10

= 50 years

Therefore, the present age of Nuri is 50 years and the present age of Sonu is 20 years.

(iii) Let digit at ten’s place is and digit at one’s place is y

According to the given condition, we have

= 9 … (1)

And 9 (10y) = 2 (10x)

90+ 9= 20+ 2x

88= 11y

8y

8– = 0 … (2)

Adding (1) and (2), we get

9= 9  = 1

Putting the value of x in (1), we get

1 + = 9

= 9 – 1 = 8

Therefore, number = 10= 10 (1) + 8 = 10 + 8 = 18

(iv) Let the number of Rs 100 notes is and the number of Rs 50 notes is y

According to given conditions, we have

= 25 … (1)

and 100+ 50= 2000

2= 40 … (2)

Subtracting (2) from (1), we get

= −15  = 15

Putting the value of x in (1), we get

15 + = 25

= 25 – 15 = 10

Therefore, the number of Rs 100 notes is 15 and the number of Rs 50 notes is 10.

(v) Let fixed charge for 3 days is Rs x
and the additional charge for each day thereafter is Rs y.
According to the given condition, we have
+ 4= 27 … (1)
+ 2= 21 … (2)
Subtracting (2) from (1), we get
2= 6
= 3
Putting the value of y in (1), we get
x + 4 (3) = 27

= 27 – 12 = 15
Therefore, a fixed charge for 3 days is Rs 15 and the additional charge for each day after 3 days is Rs 3.

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