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?

Result

p =  12.5 %

Solution:

q1=1+25/100=54=1.25 q2=1−10/100=910=0.9  S1=ab S2=(a⋅ q1)⋅ (b⋅ q2)  p=100⋅ S2−S1S1  p=100⋅ (a⋅ q1)⋅ (b⋅ q2)−abab  p=100⋅ q1⋅ q2−11=100⋅ 1.25⋅ 0.9−11=252=12.5=12.5%q_{1}=1+25/100=\dfrac{ 5 }{ 4 }=1.25 \ \\ q_{2}=1-10/100=\dfrac{ 9 }{ 10 }=0.9 \ \\ \ \\ S_{1}=ab \ \\ S_{2}=(a \cdot \ q_{1}) \cdot \ (b \cdot \ q_{2}) \ \\ \ \\ p=100 \cdot \ \dfrac{ S_{2}-S_{1} }{ S_{1} } \ \\ \ \\ p=100 \cdot \ \dfrac{ (a \cdot \ q_{1}) \cdot \ (b \cdot \ q_{2}) - ab }{ ab } \ \\ \ \\ p=100 \cdot \ \dfrac{ q_{1} \cdot \ q_{2}-1 }{ 1 }=100 \cdot \ \dfrac{ 1.25 \cdot \ 0.9-1 }{ 1 }=\dfrac{ 25 }{ 2 }=12.5=12.5 \%

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