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Question

A certain mixture of nuts consists of 4 parts almonds to 1 part walnuts, by weight. By removing some of the almonds, the remained nuts consist 75 percent of almonds, by weight. What percent of the almonds were removed?
(A) 10%
(B) 12.5%
(C) 20%
(D) 25%
(E) 30%

Anju@Gurome · Answer

shashwats wrote:A certain mixture of nuts consists of 4 parts almonds to 1 part walnuts, by weight. By removing some of the almonds, the remained nuts consist 75 percent of almonds, by weight. What percent of the almonds were removed?

Let us assume initially there was 100 gm of nuts.
So, there was 100*4/(4 + 1) = 80 gm almonds and (100 - 80) = 20 gm walnuts.

Let us assume x gm of walnut was removed.
So, in the remaining mixture, there is (80 - x) gm almonds and 20 gm walnuts.

So, 20/(100 - x) = 25/100 = 1/4
--> (100 - x) = 80
--> x = 100 - 80 = 20

So, 20 gm almonds was removed.
--> 20 = (20/80)*100 % of 80 = 25% of 80

The correct answer is D.

Last edited by Anju@Gurome on Mon Apr 29, 2013 5:35 am, edited 1 time in total.

Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

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srcc25anu · Answer

Let Total Mixture be 100
Almonds = 4/5 * 100 = 80
Nuts = 1/5 * 100 = 20
If we remove Y amount of nuts, Nuts / Total Mixture ratio becomes 75%
Lets plug in answer choices to see what results.

A. we remove 10% or 8 units of almonds.
Now ALMONDS to Total % = (80-8) or 72 / 92 NOT EQUAL TO 75%

C. we remove 20% or 16 units of almonds
Almonds / Total = 64 / 86 (NOT EQUAL TO 75%)

D. lets remove 25% of almonds or 20 units
Almonds / Total = 60 / 80 or 3/4 or 75% (CORRECT ANSWER)

fifafreak · Answer

Almonds = 80; Walnuts = 20 80 + 20 = 100....[1] 80-x + 20 = 100-x ...[2] [80-x]/[100-x]= 75/100 => x = 20 which is 25% of 80. Ans. D

GMATGuruNY · Answer

shashwats wrote:A certain mixture of nuts consists of 4 parts almonds to 1 part walnuts, by weight. By removing some of the almonds, the remained nuts consist 75 percent of almonds, by weight. What percent of the almonds were removed?

Let us assume initially there was 100 gm of nuts.
So, there was 100*4/(4 + 1) = 80 gm almonds and (100 - 80) = 20 gm walnuts.

Let us assume x gm of walnut was removed.
So, in the remaining mixture, there is (80 - x) gm almonds and 20 gm walnuts.

So, 20/(100 - x) = 25/100 = 1/4
--> (100 - x) = 80
--> x = 100 - 80 = 20

So, 20 gm almonds was removed.
--> 20 = (20/80)*100 % of 80 = 25% of 80

The correct answer is D.

Last edited by Anju@Gurome on Mon Apr 29, 2013 5:35 am, edited 1 time in total.

Anju Agarwal
Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

§ GMAT with Gurome § Admissions with Gurome § Career Advising with Gurome §

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