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 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
- 127
- 135
- 124
- 144
The Correct Option is A
Approach Solution - 1
Let's define the following variables:
- \(S\) as the cost of the small box.
- \(L\) as the cost of the large box. So, \(L = 2S\).
- \(w_s\) as the weight of chocolate in the small box.
- \(w_l\) as the weight of chocolate in the large box.
Step 1: Price per gram of chocolate
The price per gram of chocolate in the small box is: \[ \frac{S}{w_s} \] The price per gram of chocolate in the large box is 12% less than in the small box, so: \[ 0.88 \times \frac{S}{w_s} \]
Step 2: Setting up the equation for the large box
From the large box pricing, we have: \[ \frac{L}{w_l} = 0.88 \times \frac{S}{w_s} \] Substituting \(L = 2S\) into the above equation: \[ \frac{2S}{w_l} = 0.88 \times \frac{S}{w_s} \] Simplifying: \[ \frac{2}{w_l} = 0.88 \times \frac{1}{w_s} \] Solving for \(w_l\): \[ w_l = 2.27w_s \]
Step 3: Conclusion
This means the large box contains 2.27 times the weight of chocolate compared to the small box. Therefore, the weight of chocolate in the large box exceeds that in the small box by: \[ 2.27 - 1 = 1.27 \] Or 127%.
Final Answer:
The percentage by which the weight of chocolate in the large box exceeds that in the small box is approximately \( \boxed{127\%} \).
Approach Solution -2
Let’s assume the following:
- The selling price of the large box is Rs. 200
- The selling price of the small box is Rs. 100
Let:
- Large box has \( L \) grams of chocolate
- Small box has \( S \) grams of chocolate
Step 1: Setting up the relation between the selling price per gram of chocolate
The relation between the selling price per gram of chocolate can be represented as: \[ \frac{200}{L} = 0.88 \times \frac{100}{S} \]
Step 2: Solving the relation
By solving the above relation, we get the ratio of the amount of chocolates in each box: \[ \frac{L}{S} = \frac{25}{11} \]
Step 3: Calculating the percentage by which the weight of chocolate in the large box exceeds that in the small box
The percentage by which the weight of chocolate in the large box exceeds that in the small box is: \[ \left( \frac{25}{11} - 1 \right) \times 100 \] Simplifying: \[ \left( \frac{25}{11} - \frac{11}{11} \right) \times 100 = \frac{14}{11} \times 100 \approx 127\% \]
Final Answer:
The percentage by which the weight of chocolate in the large box exceeds that in the small box is approximately \( \boxed{127} \% \).
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