Percent change

If the length of a rectangle is increased by 25% and the width is decreased by 10%, the area of the rectangle is larger than the area of the original rectangle by what percent?

Correct result:

p =  12.5 %

Solution:

$$\begin{gathered} q_1=1+25\mathrm{/}100={\displaystyle \frac{5}{4}}=1.25 \\ {q}_2=1-10\mathrm{/}100={\displaystyle \frac{9}{10}}=0.9 \\ {S}_1=ab \\ {S}_2=(a\cdot q_1)\cdot (b\cdot q_2) \\ p=100\cdot {\displaystyle \frac{S_2-S_1}{S_1}} \\ p=100\cdot {\displaystyle \frac{(a\cdot q_1)\cdot (b\cdot q_2)-ab}{ab}} \\ p=100\cdot {\displaystyle \frac{q_1\cdot q_2-1}{1}}=100\cdot {\displaystyle \frac{1.25\cdot 0.9-1}{1}}={\displaystyle \frac{25}{2}}={\displaystyle \frac{25}{2}}\mathrm{\%}=12.5\mathrm{\%} \end{gathered}$$

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