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A rectangle's length is decreased by 20% and its width increased by 20%. By how much percent does the area get affected?

Note by Muskaan Ohri 4 years, 4 months ago

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Let original length = \( l \) and original width be = \( w \)

Then Original Area = \( l \times w \) = \( lw \)

Now, New length = \( \frac{100 -20}{100} \times l \) = \( \frac{80}{100} \times l \) = \( \frac{8l}{10} \)

New Breadth = \( \frac{100 +20}{100} \times w \) = \( \frac{120}{100} \times w \) = \( \frac{12w}{10} \)

Then New Area = \( \frac{8l}{10} \times \frac{12w}{10} \) = \( \frac{96lw}{100} \)

Then Percent change = \( \dfrac{\text{New Area - Original Area}}{\text{Original Area}} \)

\(= \dfrac{\frac{96lw}{100} - lw}{lw} \)

\( = \dfrac{\frac{-4lw}{100}}{lw} \)

\(= \dfrac{-4}{100} \)

= \( - 4\)%

Therefore the percent change is -4% or a 4% decrease

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Perfect solution! Thanks...!

1-20/1001+20=.81.2=.96 so area is decreased by 4% so -4%, is the answer

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## Comments

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TopNewestLet original length = \( l \) and original width be = \( w \)

Then Original Area = \( l \times w \) = \( lw \)

Now, New length = \( \frac{100 -20}{100} \times l \) = \( \frac{80}{100} \times l \) = \( \frac{8l}{10} \)

New Breadth = \( \frac{100 +20}{100} \times w \) = \( \frac{120}{100} \times w \) = \( \frac{12w}{10} \)

Then New Area = \( \frac{8l}{10} \times \frac{12w}{10} \) = \( \frac{96lw}{100} \)

Then Percent change = \( \dfrac{\text{New Area - Original Area}}{\text{Original Area}} \)

\(= \dfrac{\frac{96lw}{100} - lw}{lw} \)

\( = \dfrac{\frac{-4lw}{100}}{lw} \)

\(= \dfrac{-4}{100} \)

= \( - 4\)%

Therefore the percent change is -4% or a 4% decrease

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Perfect solution! Thanks...!

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1-20/100

1+20=.81.2=.96 so area is decreased by 4% so -4%, is the answerLog in to reply