Question

(i) The sum of numerator and denominator of a fraction is 4 more than
twice the numerator. If 3 is added to each of the numerator and denominator their ratio becomes 2:3. Find the fraction 5
[CBSE 2010] ​

Answer 1

Answer:

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Answer 2

Answer:

The given fraction will be 5/9.

Solution:

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.

$\frac{n + 3}{d + 3} = \frac{2}{3} \\ = > 3n + 9 = 2d + 6 \\ = > 3n - 2d = 6 - 9 \\ = > 3n - 2d = - 3......(ii)$

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

$(i) \times 3 = > 3n - 3d = - 12 \\ \\ (ii) = > 3n - 2d = - 3$

Solving (i) and (ii) we get,

$- d = - 9 \\ = > d = 9$

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

$n - d = - 4 \\ = > n - 9 = - 4 \\ = > n = - 4 + 9 \\ = > n = 5$

∴ The given fraction is

$\frac{n}{d} = \frac{5}{9}$