# Can someone solve this for me?

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

- 4 years, 4 months ago

Perfect solution! Thanks...!

- 4 years, 4 months ago

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

- 3 years, 9 months ago