Class 10 Mathematics

Linear Equations

Exercise 3.4 (NCERT) Part 2

Question 2: Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method.

(a) 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?

Solution: Let us assume that numerator is x and denominator is y. Then as per question;

`(x+1)/(y-1)=1`

Or, `x+1=y-1`
Or, `y=x+2`
Or, `y-x=2`


The second condition can give following equation;

`(x)/(y+1)=1/2`

Or, `2x=y+1`
Or, `y=2x-1`
Or, `2x-y=1`
Or, `-y+2x=1`

Adding first and second equations, we get;

`y-x-y+2x=2+1`
Or, `x=3`

Substituting the value of x in first equation, we get;

`y – x = 2`
Or, `y – 3 = 2`
Or, `y = 5`

Hence, the required fraction is; 3/5

(b) 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?

Solution: Let us assume Nuri’s age = x and Sonu’s age = y. Then we can get following equations;

Five years ago:

Nuri’s age `= x – 5` and Sonu’s age `= y – 5`

`x – 5 = 3(y – 5)`
Or, `x – 5 = 3y – 15`
Or, `x = 3y - 15 + 5`
Or, `x = 3y - 10` ………(1)

Ten years from now;

Nuri’s age `= x + 10` and Sonu’s age `= y + 10`

`x + 10 = 2(y + 10)`
Or, `x + 10 = 2y + 20`
Or, `x = 2y + 20 - 10`
Or, `x = 2y + 10` …….(2)

From both equations, we get;

Or, `3y - 10 = 2y + 10`
Or, `3y = 2y + 10 + 10 = 2y + 20`
Or, `y = 20`
So, `x = 2y + 10 = 50`


(c) 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.

Solution: Let us assume, one of the digits is x and another digit is y.

`x + y = 9` ……..(2)

The number can be given as;

`10x + y`

And the number obtained after reversing the digits is;

`10y + x`

As per question;

`9(10x + y) = 2(10y + x)`
Or, `90x + 9y = 20y + 2x`
Or, `90x – 2x = 20y – 9y`
Or, `88x = 11y`
Or, `88x – 11y = 0` …….. (2)


Multiplying the first equation by 11, we get;

`11x + 11y = 99`

Adding this to the second equation, we get;

`88x – 11y + 11x + 11y = 99`
Or, `99x = 99`
Or, `x = 1`

Substituting the value of x in first equation, we get;

`x + y = 9`
Or, `1 + y = 9`
Or, `y = 8`

Hence, the required number = 18

(d) 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 out how many notes of Rs. 50 and Rs. 100 she received.

Solution: Let us assume the number of Rs. 50 notes is x and that of Rs. 100 notes is y.

Total number of notes `= x + y = 25` ……..(1)

Total amount `= 50x + 100y = 2000` …….. (2)

Multiplying the first equation by 50 we get;

`50x + 50y = 1250`

Subtracting this equation from second equation, we get;

`50x + 100y – 50x – 50y = 2000 – 1250`

Or, `50y = 750`

Or, `y = 15`

Substituting the value of y in first equation, we get;

`x + y = 25`

Or, `x + 15 = 25`

Or, `x = 10`

Hence, number or Rs. 50 notes = 10 and number of Rs. 100 notes = 15

(e) 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.

Solution: Let us assume fixed charge is x and per day charge after three days is y.

Amount paid by Saritha:

`x + 4y = 27` …… (1)

Amount paid by Susy:

`x + 2y = 21` …….. (2)

Subtracting the second equation from first equation, we get;

`x + 4y – x – 2y = 27 – 21`
Or, `2y = 6`
Or, `y = 3`

Substituting the value of y in second equation, we get;

`x + 2y = 21`
Or, `x + 2 x 3 = 21`
Or, `x = 21 – 6 = 15`

Hence, fixed charge for three days = Rs. 15 and per day charge after three days = Rs. 3


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Introduction

Exercise:3.1

Exercise:3.2(Part 1)

Exercise:3.2(Part 2)

Exercise:3.3(Part 1)

Exercise:3.3 (part 2)

Exercise:3.4 (part 1)

Exercise:3.5 (part 1)

Exercise:3.5 (part 2)

Exercise:3.6 (part 1)

Exercise:3.6 (part 2)

Exercise:3.6 (part 3)