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07 Aug 2022, 09:59
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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:
65%
(hard)
Question Stats:
53% (02:27) correct
47%
(02:35)
wrong
based on 19
sessions
History
Date
Time
Result
Not Attempted Yet
Three metals A, B and C are mixed such that the quantity of A is x % more than that of C and the quantity of A is x % less than that of B. The quantity of C is increased by 66.67% by adding pure C to the alloy mixture. This results in the metal C contribution to the alloy becoming 33.33% of the total. Find the value of x.
A. 25
B. 33.33
C. 50
D. 66.67
E. 83.33
(adapted from gmatfree)
A. 25
B. 33.33
C. 50
D. 66.67
E. 83.33
(adapted from gmatfree)
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11 Sep 2022, 08:30
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pintukr
As per the question, original values of B and C w.r.t. to A are
B = 100A/(100 - x)
C = 100A/(100 + x)
Now after adding pure C we get
=> A+B+5C/3 = 3(C + 2C/3) = 5C
=> A+B = 5C - 5C/3
=> A+B = 10C/3
=> A + 100A/(100 - x) = 10/3 * 100A/(100 + x)
=> 1 + 100/(100 - x) = 10/3 * 100/(100 + x)
=> 3(200 + x)/(100 - x) = 10^3/(100+x)
=> (600 + 3x)(100 - x) = 1000(100 - x)
=> 60000 + 300x - 3x^2 = 100000 - 1000x
=> 3x^2 - 1300x + 40000 = 0
=> 3x^2 - 1200x - 100x + 40000 = 0
=> 3x(x - 400) - 100(x -400) = 0
=> (x-400)(3x-100) = 0
Now x can either be 400 or 100/3
Since A is x percentage less than B it can never be more than 100 hence x = 100/3 = 33.33
Option B
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