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The revenue increased by 10 percent from March to April and then the revenue decreased by 10 percent from April to May. By what percent is increased (or decreased) the revenue from March to May?

A. 10% less
B. 10% more
C. 1% less
D. 1% more
E. 0.9% less

* A solution will be posted in 2 days.

There is a shortcut formula for solving consecutive % increase -

\(x + y + \frac{xy}{100}\) ( With corresponding signs + for Increase and - for decrease)

So, \(+10 - 10 +\frac{(10*-10)}{100}\) = \(- 1\) %

Hence correct answer will be (C) \(- 1\) %
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==>If the revenue increases by 10% from March to April, R -->R(1+10%)=1.1R, and if this decreases by 10% from April to May, 1.1R  1.1R(1-10%)=1.1R(0.9)=0.99R. If so, the revenue from March to May is R  0.99R, hence 1% decrease. The answer is C.
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The revenue increased by 10 percent from March to April and then the revenue decreased by 10 percent from April to May. By what percent is increased (or decreased) the revenue from March to May?

A. 10% less B. 10% more C. 1% less D. 1% more E. 0.9% less
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MathRevolution
The revenue increased by 10 percent from March to April and then the revenue decreased by 10 percent from April to May. By what percent is increased (or decreased) the revenue from March to May?

A. 10% less B. 10% more C. 1% less D. 1% more E. 0.9% less


Let the initial revenue be 100 units i.e., for the month of March
It increased by 10% in April so the new revenue is 110 units.
It decreased by 10% in May so the new revenue is 110-11= 99 units.
Percentage change in revenue from March to May = 99-100/100 = 1% decrease
Option C
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==>If the revenue increases by 10% from March to April, R -->R(1+10%)=1.1R, and if this decreases by 10% from April to May, 1.1R  1.1R(1-10%)=1.1R(0.9)=0.99R. If so, the revenue from March to May is R  0.99R, hence 1% decrease. The answer is C.

Answer: C
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