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A beaker was filled with a mixture of 40 liters of water and a liquid
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26 Oct 2023, 00:36
26 Oct 2023, 00:36
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66% (02:55) correct
33% (04:22) wrong based on 9 sessions
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A beaker was filled with a mixture of 40 liters of water and a liquid chemical in the ratio of 3 : 5, respectively. If 2 percent of the initial quantity of water and 5 percent of the initial quantity of liquid chemical evaporated each day during a 10-day period, what percent of the original amount of mixture evaporated during this period?
(A) 22.22%
(B) 33.33%
(C) 38.75%
(D) 44.44%
(E) 58.33%
(A) 22.22%
(B) 33.33%
(C) 38.75%
(D) 44.44%
(E) 58.33%
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A beaker was filled with a mixture of 40 liters of water and a liquid
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02 Dec 2023, 15:17
02 Dec 2023, 15:17
1
Given that the mixture was in a ratio of 3:5, we can set up this fraction:
percent of original amount of mixture evaporated = \(10(\frac{3}{8}*40*0.02+\frac{5}{8}*40*0.05)/40=0.3875=38.75%\)
Note about solving this is that you don't need to know the actual volume of the initial mixture in order to find the solution, since it's asking for a percent of the original, not an actual value. If we weren't given 40 as the initial volume, you could still set up the fraction as:
\(\frac{10(3x*0.02+5x*0.05)}{8x}=0.3875\)
percent of original amount of mixture evaporated = \(10(\frac{3}{8}*40*0.02+\frac{5}{8}*40*0.05)/40=0.3875=38.75%\)
Note about solving this is that you don't need to know the actual volume of the initial mixture in order to find the solution, since it's asking for a percent of the original, not an actual value. If we weren't given 40 as the initial volume, you could still set up the fraction as:
\(\frac{10(3x*0.02+5x*0.05)}{8x}=0.3875\)
gmatclubot
A beaker was filled with a mixture of 40 liters of water and a liquid [#permalink]
02 Dec 2023, 15:17
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