ICSE & CBSE Board Question Based on Linear Equations in two Variables of RS Aggarwal

Here the denominator of a fraction is greater than its numerator by the given number. if a number is added to both its numerator and denominator, it becomes 3/4.

You have to find the fraction.

This is the Question Number 21 Of Exercise 3E of RS Aggarwal Solution.

Deepak BoraNewbie

# The denominator of a fraction is greater than its numerator by 11. if 8 is added to both its numerator and denominator, it becomes 3/4. Find the fraction.

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Let the required fraction be x/y .

According to the Question

1st Equation

⇒ y = x + 11

⇒ y – x = 11 ……(i)

2nd Equation

⇒ x+8/y+8 = 3/4

⇒ 4(x + 8) = 3(y + 8)

⇒ 4x+ 32 = 3y + 24

⇒ 4x – 3y = -8 ……(ii)

On multiplying Equation(i) by 4

4y – 4x = 44

On adding Eq (ii) and (iii)

⇒ y = (-8 + 44)

⇒ y = 36

on substituting y = 36 in (i), we get:

⇒ y – x = 11

⇒ 36 – x = 11

⇒ x = (36 – 11)

⇒ x = 25

Hence, the required fraction is 25/36