Define the following variables:
The price per gram of chocolate in the small box is calculated as: \[ \frac{S}{w_s} \] The price per gram of chocolate in the large box is 12% less than in the small box, which can be expressed as: \[ 0.88 \times \frac{S}{w_s} \]
Based on the large box pricing, we establish the equation: \[ \frac{L}{w_l} = 0.88 \times \frac{S}{w_s} \] Substitute \(L = 2S\) into the equation: \[ \frac{2S}{w_l} = 0.88 \times \frac{S}{w_s} \] Simplify the equation: \[ \frac{2}{w_l} = 0.88 \times \frac{1}{w_s} \] Solve for \(w_l\): \[ w_l = 2.27w_s \]
This result indicates that the large box contains 2.27 times the weight of chocolate compared to the small box. The excess weight of chocolate in the large box over the small box is: \[ 2.27 - 1 = 1.27 \] This equates to 127%.
The percentage increase in the weight of chocolate in the large box compared to the small box is approximately \( \boxed{127\%} \).
A positive integer $m$ is increased by 20% and the resulting number is 1080. Then the integer $m$ is
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