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08 Sep 2019, 02:26
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
64% (01:44) correct
36%
(01:52)
wrong
based on 347
sessions
History
Date
Time
Result
Not Attempted Yet
In what proportion must water be mixed with alcohol to gain \(12\frac{1}{2}\)% by selling it at the cost price?
A) 1:3
B) 3:1
C) 8:1
D) 1:8
E) 2:3
A) 1:3
B) 3:1
C) 8:1
D) 1:8
E) 2:3
09 Sep 2019, 11:49
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Solution:
Assume SP for 1 L of mixture as Re 1. Given Gain = 12.5% .
So CP of 1 L of mixture= 1/1.125=> 8/9
CP of 1L Water = 0
CP of 1L Pure Alcohol = Re 1
Using Weighted Avg, we can solve as follows:
Qty of W/Qty of Alcohol = (1-8/9) / (8/9-0) => 1:8
Answer: Option D
Hope it is clear.
Assume SP for 1 L of mixture as Re 1. Given Gain = 12.5% .
So CP of 1 L of mixture= 1/1.125=> 8/9
CP of 1L Water = 0
CP of 1L Pure Alcohol = Re 1
Using Weighted Avg, we can solve as follows:
Qty of W/Qty of Alcohol = (1-8/9) / (8/9-0) => 1:8
Answer: Option D
Hope it is clear.
General Discussion
31 Aug 2020, 18:58
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Quote:
I think the question stem could have been better stated to say, water to alcohol will be in what proportion? Or something like this.
But -- anyways -- the way to think about this is 1/4 = 25% and half is 12.5% or 1/8 -- so the selling price will be 9/8 -- because selling at a 12.5% gain
so -- if 9/8x -- then we are going to have 8/9 alcohol and 1/9 water, or 8:1 alcohol to water (or 1:8 water to alcohol -- per the correct answer).
