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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

  1. $135$
  2. $127$
  3. $144$
  4. $124$
in Quantitative Aptitude retagged by
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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 $= \left[\dfrac{\left(\frac{25}{11}\right)\frac{x}{k}-\frac{x}{k}}{\frac{x}{k}}\right] \times 100\%$

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


$\textbf{Shortcut Method:}$

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 $:\text{B}$

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