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Re: A bowl was filled with 10 ounces of water, and 0.008 ounce of the wate [#permalink]
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Bunuel wrote:
A bowl was filled with 10 ounces of water, and 0.008 ounce of the water evaporated each day during a 50-day period. What percent of the original amount of water evaporated during this period?

A. 0.004%
B. 0.04%
C. 0.40%
D. 4%
E. 40%


We are given that 0.008 ounces of water evaporated each day. Furthermore, we know that this process happened over a 50-day period. To calculate the total amount of water that evaporated during this time, we need to multiply 0.008 by 50:

0.008 x 50 = 0.4 ounces

Finally, we are asked what percentage of the original amount of water evaporated during this period. To determine this percentage, we have to make sure we translate the expression correctly. We can translate it to:

(Amount Evaporated/Original Amount) x 100%

(0.4/10) x 100%

(4/100) x 100% = 4%

Answer: D
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Re: A bowl was filled with 10 ounces of water, and 0.008 ounce of the wate [#permalink]
0.008 oz evaporates each day for 50-day period => total evaporation is 0.4 oz....shift dec, multiply by 50, shift back.

Answer the question: 0.4 is what percent of 10? Simply multiple numerator and denominator by 10 to get 4/100. Answer is D=4%
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Re: A bowl was filled with 10 ounces of water, and 0.008 ounce of the wate [#permalink]
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Re: A bowl was filled with 10 ounces of water, and 0.008 ounce of the wate [#permalink]
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