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Manhattan Challenge Problem

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by shankar.ashwin » Wed Nov 02, 2011 2:12 am
The ratio, by weight, of the four ingredients A, B, C, and D of a certain mixture is 4:7:8:12. The mixture will be changed so that the ratio of A to C is quadrupled and the ratio of A to D is decreased. The ratio of A to B will be held constant. If B will constitute 20% of the weight of the new mixture, by approximately what percent will the ratio of A to D be decreased?

A) 15%
B) 25%
C) 35%
D) 45%
E) 55%
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by shankar.ashwin » Wed Nov 02, 2011 2:14 am
My solution; not sure of the OA.

A|B|C|D = 4|7|8|12

Now, A|C is quadrupled - 16|8
A|B remains the same - 4|7 = 16|28
Let D be x; so the new ratio is
A|B|C|D = 16|28|8|x

B constitutes 20% of the weight or 1/5th of the weight;

So, 28/(16+28+8+x) = 1/5
x = 88

Ratio of A : D (New) = 16:88 = 2:11
Ratio of A : D (Old) = 4:12 = 1: 3

% Decrease = (1/3-2/11)/(1/3) * 100 = 45
D IMO
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by knight247 » Wed Nov 02, 2011 2:38 am
Mitch has solved this problem here
https://www.beatthegmat.com/manhattan-ra ... 94277.html
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by neelgandham » Wed Nov 02, 2011 2:40 am
Step by Step

The ratio, by weight, of the four ingredients A, B, C, and D of a certain mixture is 4:7:8:12.
A:B:C:D = 4:7:8:12

The mixture will be changed so that the ratio of A to C is quadrupled
A:B:C:D = 16:7:8:12

The ratio of A to D is decreased.
A:B:C:D = 16:7:8:D

The ratio of A to B will be held constant.
A:B:C:D = 16:28:8:D

If B will constitute 20% of the weight of the new mixture
28:(16+8+D)::20/80 => D=88, new ratio A:B:C:D = 16:28:8:88

By approximately what percent will the ratio of A to D be decreased?

(((4/12)-(16/88))/(16/88)) * 100
((88-48)/88))*100
(40/88)*100
~ 45.x

IMO D
Anil Gandham
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