An Important question based on Linear Equations in two Variables Chapter of R.S Aggarwal book for ICSE & CBSE Board.

Here the sum of the numerator and denominator of a fraction is n more than twice the numerator. if the numerator and denominator are increased by the given number, they are in the ratio is given.

you have to determine the fraction.

This is the Question Number 23 Exercise 3E of RS Aggarwal Solution.

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# The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2:3. Determine the fraction.

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Let n be the numerator and d be the denominator

Given,

The sum of the given n and d is equal to twice the numerator plus 4,i.e

n+d=2n+4

n+d-2n-4=0

-n+d=4

n-d=-4———-(i)

From Given, the numerator n and the denominator d must be increased by 3 to get the ratio 2:3

(n+3)/(d+3)=2/3

3n+9=2d+6

3n-2d=6-9

3n-2d=-3————-(ii)

From equations (i) and (ii) we get,

(i)x3=3n-3d=-12

(ii)3n-2d=-3

Solving (i) and (ii) we get

-d=-9

d=9

Substituting d=9 in equation (i), we get

n-d=-4

n-9=-4

n=-4+9

n=5

Therefore, the given fraction is

n/d=5/9