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let x=total weight of first mixture
weight of peanuts in second mixture=(3x/4)-8
weight of cashews in second mixture=x/4
(3x/4)-8/(x/4)=7/3
x=48 pounds
3(48)/4=36 pounds of peanuts in first mixture
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The first mixture consists of its 3/4 is peanuts and the rest is cashews. The second mixture has the same quantity of cashews as the first mixture but the peanuts’ quantity of the second mixture is 8 pounds fewer than that of the first mixture. If the ratio of the peanuts to cashews is 7 to 3 in the second mixture, what is the amount of the peanuts quantity in the first mixture, in pounds?

A. 24
B. 28
C. 32
D. 36
E. 40

This is frequently given on GMAT MATH, which is “2 by 2” question.
Attachment:
mixture.jpg
mixture.jpg [ 10.63 KiB | Viewed 3924 times ]
On the table above, 3f-8:f=7:3 -> 3(3f-8)=7f, 9f-24=7f, f=12. Hence, in the first mixture, peanuts=3f=3(12)=36 is the answer. Therefore, D is the answer.
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MathRevolution
The first mixture consists of its 3/4 is peanuts and the rest is cashews. The second mixture has the same quantity of cashews as the first mixture but the peanuts’ quantity of the second mixture is 8 pounds fewer than that of the first mixture. If the ratio of the peanuts to cashews is 7 to 3 in the second mixture, what is the amount of the peanuts quantity in the first mixture, in pounds?

A. 24
B. 28
C. 32
D. 36
E. 40


* A solution will be posted in two days.

Its an old question but I found a slightly different way to approach it than the listed methods.

In the first mixture, Peanuts are 3/4 and consequently cashews are 1/4. Their ratio:

\(\frac{c}{p} = \frac{1}{3}\)

\(c = \frac{p}{3}\)

In mixture 2:

\(\frac{c}{p-8} = \frac{3}{7}\)

\(7c = 3(p-8)\)

from the first mixture equation, we know \(c = \frac{p}{3}\)

\(7*\frac{p}{3} = 9p - 72\)

simplifying

\(72 = 9p - 7p\)

\(2p = 72\)

\(p = 36\)
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Given:

M1: cashew:peanut = 3:1
M2: cashew:peanut= 7:3

M2: M1 = 7:3:9
removal of 8pounds=
2X = 8, then 9X = 36 and 7X = 28. Hence the answer is 36.D
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