Let's break the problem down:
Let's denote:
1. \(S\) as the cost of the small box.
2. \(L\) as the cost of the large box. So, \(L = 2S\).
3. \(w_s\) as the weight of chocolate in the small box.
4. \(w_l\) as the weight of chocolate in the large box.
5. Price per gram of chocolate in the small box = \(\frac{S}{w_s}\).
6. Price per gram of chocolate in the large box = \(0.88(\frac{S}{w_s})\) (since it's 12% less than that in the small box).
Now, from the large box pricing, \(\frac{L}{w_l} = 0.88(\frac{S}{w_s})\)
Substitute \(L\) in the above equation:
\(\frac{2S}{w_l} = 0.88 \times \frac{S}{w_s}\)
\(\frac{2}{w_l} = 0.88 \times \frac{1}{w_s}\)
\(w_l = 2.27w_s\)
This means the large box contains 2.27 times the weight of chocolate compared to the small box. Therefore, it exceeds the small box's weight by:
\(2.27 - 1 = 1.27\)
Or 127%.
The percentage by which the weight of chocolate in the large box exceeds that in the small box is approximately 127%
Let's assume the following:
Selling price of the large boxes be Rs.200
Selling price of the small boxes be Rs.100
Now. let's consider the following:
Large box has L grams of Chocolate
Small box has S grams of Chocolate
Now, the relation between the selling price per gram of chocolate can be represented as :
\(\frac{200}{L}=0.88\times\frac{100}{S}\)
By solving the above relation , we get the ratio of the amount of chocolates in each box :
\(\frac{L}{S}=\frac{25}{11}\)
Now , the percentage by which weight of chocolate in the large box exceeds that in the small box :
\((\frac{25}{11}-1)\times100\)
≈ 127%
Therefore, the correct option is (A) : 127.