CAT 2021 Set-1 | Quantitative Aptitude | Question: 22

Question

Identical chocolate pieces are sold in boxes of two sizes, small and large. The large box is sold for twice the price of the small box. If the selling price $\text{per gram}$ of chocolate in the large box is $12 \%$ less than that in the small box, then the percentage by which the weight of chocolate in the large box exceeds that in the small box is nearest to

• A. $135$
• B. $127$
• C. $144$
• D. $124$

Answer 1

We can draw the table for better understanding.

$$\begin{array}{l|cc} \hline  & \textbf{Small box} & \textbf{Large box} \\\hline \text{Selling price} & x & 2x \\ \text{Selling price per gram} & k & \left( \frac{88}{100}\right)k \\ \text{Weight} & \frac{x}{k} &  \dfrac{2x}{\left(\frac{88}{100}\right)k} = \left(\frac{25}{11}\right)\frac{x}{k} \\\hline \end{array}$$
$\therefore$ The weight of chocolate in the large box exceeds that in the small box $= \left[\dfrac{\left(\frac{25}{11}\right)\frac{x}{k}-\frac{x}{k}}{\frac{x}{k}}\right] \times 100\%$

$\qquad = \left(\frac{25-11}{11}\right) \times 100 \%= \left(\frac{14}{11}\right) \times 100 \%=127.2727\%\cong127\%.$

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$\textbf{Shortcut Method:}$

We can also do that in a more simple way.

$$\begin{array}{l|cc} \hline  & \textbf{Small box} & \textbf{Large box} \\\hline \text{Selling price} & 1000 & 2000 \\ \text{Selling price per gram} & 100 & 88 \\ \text{Weight} & 10 & \frac{2000}{88} = \frac{250}{11} \\\hline \end{array}$$
$\therefore$ The weight of chocolate in the large box exceeds that in small box $= \left(\frac{\frac{250}{11}-10}{10}\right) \times 100\%$

$\qquad = \left(\frac{250-110}{110}\right) \times 100\%= \left(\frac{140}{110}\right) \times 100\%= \left(\frac{14}{11}\right) \times 100\%= 127.2727 \cong 127\%.$

Correct Answer $:\text{B}$