Question

If the numerator of a fraction is increased by 20% and the denominator is decreased by 10%, then the value of the fraction is $\dfrac{16}{21}$. What is the original fraction?
A. $\dfrac{3}{5}$
B. $\dfrac{3}{10}$
C. $\dfrac{4}{5}$
D. $\dfrac{4}{7}$

Answer 1

Hint: We have to use the concept of percentages to get the answer. If a number ‘k’ is increased by x%, it means that ‘k’ is increased by x% of ‘k’. Numerically, ‘k’ is increased by x% of ‘k’ can be written as $k+\left( \dfrac{x}{100}\times k \right)$, similarly ‘k’ is decreased by x% of ‘k’ can be written as $k-\left( \dfrac{x}{100}\times k \right)$. Let us assume that the required fraction be $\dfrac{a}{b}$. We apply the steps mentioned in the question about numerator and denominator and equal the final fraction to $\dfrac{16}{21}$ and get the required fraction.

Complete step-by-step answer:
Let us assume that the required fraction be $\dfrac{a}{b}$. Given that the numerator is increased by 20% and denominator is decreased by 10%.
The concept of percentages can be understood using unitary method. The different attributes in the unitary method are directly proportional to each other. A simple example is, if the cost of 1 book is Rs.10/- then the cost of 5 books is Rs.5*10 = Rs.50/-. Let ‘y’ be the value of x% of ‘k’. x% means ‘x’ out of 100. Numerically,
$\dfrac{x}{100}=\dfrac{y}{k}$
By cross-multiplying, we get
$y=\dfrac{x\times k}{100}$
So, ‘k’ is increased by x% of ‘k’ we can write it as as $k+\left( \dfrac{x}{100}\times k \right)$ and
‘k’ is decreased by x% of ‘k’ we can write it as $k-\left( \dfrac{x}{100}\times k \right)$.
Applying this property to the given fraction, we get
k=a, x=20 and numerator is increased.
Final numerator = $a+\left( \dfrac{20}{100}\times a \right)=a+0.2a=1.2a\to (1)$
Similarly for denominator k=b and x=10 and denominator is decreased.
Final denominator = $b-\left( \dfrac{10}{100}\times b \right)=b-0.1b=0.9b\to (2)$
From equations (1) and (2) we can write
Final fraction $=\dfrac{\text{Final numerator}}{\text{Final denominator}}=\dfrac{1.2a}{0.9b}=\dfrac{12a}{9b}=\dfrac{4a}{3b}\to \left( 3 \right)$
Equating this to $\dfrac{16}{21}$ gives,
$\dfrac{4a}{3b}=\dfrac{16}{21}$
Multiplying by $\dfrac{3}{4}$ on both sides, we get
$\begin{align}
  & \dfrac{3}{4}\times \dfrac{4a}{3b}=\dfrac{16}{21}\times \dfrac{3}{4} \\
 & \dfrac{a}{b}=\dfrac{4}{7} \\
\end{align}$
$\therefore $ Required fraction is $\dfrac{4}{7}$. The answer is option D.

Note:There is a chance of students not reading the question properly and students tend to apply the percentages to the whole fraction. That is when they see increase by 20%, they tend to write
$\dfrac{a}{b}\times \dfrac{120}{100}$ and after that decrease by 10% which will be $\dfrac{a}{b}\times \dfrac{120}{100}\times \dfrac{90}{100}$ which is a whole different problem and leads to a wrong solution.