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23 Apr 2025, 10:00
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D
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Difficulty:
45%
(medium)
Question Stats:
70% (02:02) correct
30%
(02:29)
wrong
based on 423
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Liquids A, B, and C contain 20%, 50%, and 80% alcohol respectively. The three liquids are mixed in a ratio of 1:2:2 and the resultant solution is mixed with pure alcohol in a ratio of 1:1. What will be the percentage of alcohol in this final solution?
A. 56%
B. 64%
C. 72%
D. 78%
E. 84%
A. 56%
B. 64%
C. 72%
D. 78%
E. 84%
23 Apr 2025, 13:48
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Alcohol in (1 + 2 + 2) ml => 5 ml mixed solution will be (0.2 + 1 + 1.6) ml => 2.8 ml
Now it is mixed with 5 ml of pure alcohol solution, the total alcohol content of this solution will be 2.8 + 5 = 7.8 ml => 7.8/10 = 78%
Answer D
Now it is mixed with 5 ml of pure alcohol solution, the total alcohol content of this solution will be 2.8 + 5 = 7.8 ml => 7.8/10 = 78%
Answer D
General Discussion
23 Apr 2025, 14:08
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Let's first realize that the 100% alcohol is also just another ratio in the overall mixture.
We're mixing 5 parts (1:2:2) and then mixing those 1:1 with 100% alcohol.
Because of this, the total ration can be seen as 1:2:2:5 (5 because the first mixture adds up to 5 parts)
Now multiply each percentage by it's number of parts and divide these by the total number of parts (1+2+2+5=10)
1(20) + 2(50) + 2(80) + 5(100) = 780
780 / 10 = 78
So 78% or D is your answer.
We're mixing 5 parts (1:2:2) and then mixing those 1:1 with 100% alcohol.
Because of this, the total ration can be seen as 1:2:2:5 (5 because the first mixture adds up to 5 parts)
Now multiply each percentage by it's number of parts and divide these by the total number of parts (1+2+2+5=10)
1(20) + 2(50) + 2(80) + 5(100) = 780
780 / 10 = 78
So 78% or D is your answer.
