Question

if numerator of a fraction is a decreased by 20% and its denominator is increased by 10% then the value of a fraction becomes 16 by 55 find the original fraction​

Answer 1

Answer

Let the numerator be x.

Let the denominator be y.

According to the question,

${\sf \dashrightarrow \dfrac{x - 20\% \: of \: x}{y + 10\% \: of \: y} = \dfrac{16}{55}}$

${\sf \dashrightarrow \dfrac{x - \dfrac{20}{100} \times x}{y + \dfrac{10}{100} \times y} = \dfrac{16}{55}}$

${\sf \dashrightarrow \dfrac{x - \dfrac{1}{5} \times x}{y + \dfrac{1}{10} \times y} = \dfrac{16}{55}}$

${\sf \dashrightarrow \dfrac{x - \dfrac{x}{5}}{y + \dfrac{y}{10}} = \dfrac{16}{55}}$

${\sf \dashrightarrow \dfrac{\dfrac{4x}{5}}{\dfrac{11y}{10}} = \dfrac{16}{55}}$

${\sf \dashrightarrow \dfrac{x}{y} = \dfrac{16}{55} \times \dfrac{11}{10} \times \dfrac{5}{4}}$

${\sf \dashrightarrow \dfrac{x}{y} = \dfrac{2}{55} \times \dfrac{11}{1} \times \dfrac{1}{1}}$

${\sf \dashrightarrow \dfrac{x}{y} = \dfrac{22}{55}}$

So, the original fraction is $\sf \dfrac{22}{55}$

Verification

${\sf \dashrightarrow \dfrac{x - 20\% \: of x}{y + 10\% \: of \: y} = \dfrac{16}{55}}$

${\sf \dashrightarrow \dfrac{22 - 20\% \: of 22}{55 + 10\% \: of \: 55} = \dfrac{16}{55}}$

${\sf \dashrightarrow \dfrac{22 - \dfrac{20}{100} \times 22}{55 + \dfrac{10}{100} \times 55} = \dfrac{16}{55}}$

${\sf \dashrightarrow \dfrac{22 - \dfrac{1}{5} \times 22}{55 + \dfrac{1}{10} \times 55} = \dfrac{16}{55}}$

${\sf \dashrightarrow \dfrac{22 - \dfrac{22}{5}}{55 + \dfrac{11}{2}} = \dfrac{16}{55}}$

${\sf \dashrightarrow \dfrac{\dfrac{110 - 22}{5}}{\dfrac{110 + 11}{2}} = \dfrac{16}{55}}$

${\sf \dashrightarrow \dfrac{\dfrac{88}{5}}{\dfrac{121}{2}} = \dfrac{16}{55}}$

${\sf \dashrightarrow \dfrac{88}{5} \times \dfrac{2}{121} = \dfrac{16}{55}}$

${\sf \dashrightarrow \dfrac{8}{5} \times \dfrac{2}{11} = \dfrac{16}{55}}$

${\sf \dashrightarrow \dfrac{16}{55} = \dfrac{16}{55}}$

${\sf \dashrightarrow LHS = RHS}$