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# A bowl was filled with 10 ounces of water, and 0.008 ounce of the wate

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Math Expert
Joined: 02 Sep 2009
Posts: 54696
A bowl was filled with 10 ounces of water, and 0.008 ounce of the wate  [#permalink]

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25 May 2016, 04:25
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Difficulty:

25% (medium)

Question Stats:

74% (01:08) correct 26% (01:22) wrong based on 118 sessions

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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%

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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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25 May 2016, 04:44
Total amount of water evaporated each day during a 50-day period = .008 * 50
=.008 * 100/2
= .8/2
= .4

percent of the original amount of water evaporated during this period = (.4/10) * 100%
= 4 %
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Joined: 05 Mar 2015
Posts: 999
Re: A bowl was filled with 10 ounces of water, and 0.008 ounce of the wate  [#permalink]

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25 May 2016, 12:46
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%

water evapoated during 50 day period=50 *.008=0.4 ounces
now 0.4 is What percent of the original amount
let required % of original amount=X
0.4=X% of 10
X=4
Ans D
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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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31 Aug 2017, 10:20
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%

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Joined: 04 Jun 2017
Posts: 7
Re: A bowl was filled with 10 ounces of water, and 0.008 ounce of the wate  [#permalink]

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02 Sep 2017, 11:55
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%
Re: A bowl was filled with 10 ounces of water, and 0.008 ounce of the wate   [#permalink] 02 Sep 2017, 11:55
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