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15 Aug 2019, 00:21
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
74% (01:28) correct
26%
(01:54)
wrong
based on 62
sessions
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In a water tank, 3/4 of the volume is occupied by water and the remainder of the volume is occupied by air. If a contaminant occupies 1/3 of the water, by volume, and 20 percent of the air, what percent of the volume of the water tank is contaminated?
A. 25%
B. 30%
C. 32%
D. 33%
E. 34%
A. 25%
B. 30%
C. 32%
D. 33%
E. 34%
Archit3110
Major Poster
15 Aug 2019, 09:21
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Bunuel
let total = 12
so water ; 9
and air ; 3
so contaminated water ; 3 and air ; 0.6
total 3.6
% ; 30 % IMO B
15 Aug 2019, 09:39
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Bunuel
Given: In a water tank, 3/4 of the volume is occupied by water and the remainder of the volume is occupied by air. A contaminant occupies 1/3 of the water, by volume, and 20 percent of the air.
Asked: What percent of the volume of the water tank is contaminated?
Contaminated volume %\(= \frac{1}{3}* \frac{3}{4} + 20\% * \frac{1}{4} = \frac{1}{4} + \frac{1}{20} = \frac{6}{20} = \frac{30}{100} = 30\%\)
IMO B
