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14 Sep 2021, 08:00
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Difficulty:
65%
(hard)
Question Stats:
62% (02:12) correct
38%
(02:21)
wrong
based on 212
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A solution of water and sugar is 20% sugar by weight. After several weeks, some of the water evaporates so that the solution is 60% sugar by weight. What is the ratio of the final weight of water to the initial weight of water in the mixture?
(A) 1:6
(B) 1:3
(C) 1:4
(D) 4:1
(E) 6:1
(A) 1:6
(B) 1:3
(C) 1:4
(D) 4:1
(E) 6:1
Updated on: 14 Sep 2021, 09:52
Originally posted by BrentGMATPrepNow on 14 Sep 2021, 09:36.
Last edited by BrentGMATPrepNow on 14 Sep 2021, 09:52, edited 1 time in total.
Last edited by BrentGMATPrepNow on 14 Sep 2021, 09:52, edited 1 time in total.
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Bunuel
The question is asking us to find a certain RATIO, let's start by assigning a "nice" value to the original volume of solution
A solution of water and sugar is 20% sugar by weight.
Let's say that we ORIGINALLY have 100 ml of solution.
20% of 100 = 20
So, the original solution contains 20 ml of sugar, and 80 ml of water
After several weeks, some of the water evaporates so that the solution is 60% sugar by weight.
Let x = the volume of water in the NEW solution.
Since no sugar was evaporated from the mixture, we know that the new solution still contains 20 ml of sugar.
So, the total volume of the NEW solution = x + 20
Since the new solution is 60% sugar by weight, we can write: 20/(x + 20) = 60/100
Simplify to get: 20/(x + 20) = 3/5
Cross multiply: 3(x + 20) = 100
Expand: 3x + 60 = 100
Subtract 60 from both sides: 3x = 40
Solve: x = 40/3
What is the ratio of the final weight of water to the initial weight of water in the mixture?
The ratio = (40/3)/80
= (40/3)(1/80)
= 1/6
= 1:6
Answer: A
Given: A solution of water and sugar is 20% sugar by weight. After several weeks, some of the water evaporates so that the solution is 60% sugar by weight.
Asked: What is the ratio of the final weight of water to the initial weight of water in the mixture?
Before water evaporation: -
Water = .8k
Sugar = .2k
Total = k
After water evaporation: -
Sugar = .2k = .6T; T = k/3
Water = T - .6T = .4T = .4k/3
Total = T = k/3
The ratio of the final weight of water to the initial weight of water in the mixture = (.4k/3)/.8k = 1/6 i.e. 1:6
IMO A
Asked: What is the ratio of the final weight of water to the initial weight of water in the mixture?
Before water evaporation: -
Water = .8k
Sugar = .2k
Total = k
After water evaporation: -
Sugar = .2k = .6T; T = k/3
Water = T - .6T = .4T = .4k/3
Total = T = k/3
The ratio of the final weight of water to the initial weight of water in the mixture = (.4k/3)/.8k = 1/6 i.e. 1:6
IMO A
