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Percentage Problem

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Percentage Problem

by askaichin » Sun Jul 25, 2010 8:06 am
Concrete mixture A is composed of 35 percent cement, 40 pecent sand and 25 percent gravel; concrete mixture B is composed of 15 percent cement, 30 percent sand and 55 percent gravel. If a mixture of A and B contains 25 percent cement, what percent of the weight of the mixture is A?

A 20%
B 33 1/3%
C 45%
D 50%
E 55 1/3%
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by Patrick_GMATFix » Sun Jul 25, 2010 1:23 pm
Hi askaichin

To solve this, you can safely ignore the sand and gravel.

A is 35% cement. B is 15% cement. The mixture is 25% cement.

If the mix had only A but no B, then the mix would have been 35% cement. On the other hand, if the mix had only B but no A, then the mix would have been 15% cement. Essentially, the % cement in the mix is determined by the ratio of A:B in the mix. Since the % of cement in the whole mix is 25 (exactly halfway between 35 and 15), so there must be equal quantities of A and B. in other words the mix is 50% A and 50% B.

Pick answer D

In general, in weighted average questions between 2 things, the ratio of the distances between each datapoint and the weighted average is the inverse of the ratio of the two quantities being averaged. For example, A is 35% cement and B is 15% cement. If the mix had been 30% cement instead, then the ratio of distances between each datapoint (A=35, B=15) and the weighted average (30) would have been 5:15. This would mean that the ratio of quantities (inverse of ratio of distances) would have been A:B >> 15:5 >> 3:1. In other words, if the mix had 30% cement, then the mixture would have been 75% A and 25% B.

To practice weighted avg questions in timed drills, set topic='Weighted Averages' and difficulty='600-700 & 700+' in the Drill Generator

Good luck,
-Patrick
Last edited by Patrick_GMATFix on Mon Jul 26, 2010 4:45 am, edited 1 time in total.
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by indiantiger » Sun Jul 25, 2010 1:56 pm
given in the mixture the weight of Cement is 25 % of the weight of mixture

Let wm is the weight of the mixture wm this gives us

the weight of cement mixture = 25wm/100-----(1)

let wA be weight of concrete mixture A
let wB be weight of concrete mixture B

now

cement weight in mixture in A is = 35wA/100----(2)
cement weight in mixture in A is = 15wB/100----(3)

adding (2) and (3) = (1)
25wm/100 = 35wA/100 + 15wB/100
25wm = 35wA + 15wB
5wm = 7wA + 3wB
now we know wm = wA + wB
5wA+5wB = 7wA + 3wB
2wA = 2wB
wA = wB

% of A in combined mixture = wA/(wm) = wA/(wA+wB) = 1/2 = 50 %
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by dinesh19aug » Sun Jul 25, 2010 8:20 pm
Take a look at this:

https://www.beatthegmat.com/mba/2009/12/ ... art-2-of-2
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by gmatmachoman » Sun Jul 25, 2010 8:53 pm
Alligation principle :

Wa/Wb = ( Pa-Pm)/(Pm-Pb)
(35-25)/(25-15)
1:1
Wa=Wb
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by outreach » Sun Jul 25, 2010 10:00 pm
(.35A+.15B)/(A+B)=25/100
solving we Get A=B

hence D
askaichin wrote:Concrete mixture A is composed of 35 percent cement, 40 pecent sand and 25 percent gravel; concrete mixture B is composed of 15 percent cement, 30 percent sand and 55 percent gravel. If a mixture of A and B contains 25 percent cement, what percent of the weight of the mixture is A?

A 20%
B 33 1/3%
C 45%
D 50%
E 55 1/3%
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by thebigkats » Tue Jan 04, 2011 5:21 pm
say x tons of A and y tons of B is mixed to give us z tons of mixture ==> z = x+y ---------(1)
==> x contains x*(35/100) tons of cement
==> y contains y*(15/100) tons of cement
==> z contains (35x+15y)/100 tons of cement which is 25% of z = 25 *z/100
==> 35x+15y=25z ------------ (2)

?given (1) - 15x + 15y = 15z (multiplying (1) with 15) --------- (3)
subtracting (3) from (2)
20x = 10z
x = z/2

So A is 50% of mixture
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by GMATGuruNY » Tue Jan 04, 2011 7:46 pm
askaichin wrote:Concrete mixture A is composed of 35 percent cement, 40 pecent sand and 25 percent gravel; concrete mixture B is composed of 15 percent cement, 30 percent sand and 55 percent gravel. If a mixture of A and B contains 25 percent cement, what percent of the weight of the mixture is A?

A 20%
B 33 1/3%
C 45%
D 50%
E 55 1/3%
Almost no math is needed here. Since the resulting percentage of cement (25) is the average of the two original percentages of cement (35 and 15), equal amounts of A and B must have been used. Thus, the resulting mixture will be 50% A and 50% B.

The correct answer is D.
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by lordzilla » Wed Apr 27, 2011 4:08 am
This question is quite simple. Ignore all the details about sand and gravel and focus on cement.

Now subtract 35% cement content of mixture A from cement content of mixture A and B. Then subtract 15% cement content of mixture B from mixture A and B. Note: Use just absolute values. Now find the ratio of the results.

35-25=10

15-25=10

ratio= 1:1
so 50%
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by vishnuchaithanya » Mon May 30, 2011 2:38 am
it is 50%
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by sushantgupta » Mon Jul 04, 2011 10:42 pm
50%
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by sushantgupta » Mon Jul 04, 2011 10:42 pm
50%
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by prashant misra » Sat Oct 29, 2011 10:48 am
the answer to this question is D.i solved this question in following manner 35%of A + 15% of B=25% of (A+B) that gives us A/B=1/1 ratio
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by Rastis » Tue Nov 29, 2011 12:44 pm
I'm having a difficult time understanding the logic behind the answer explanations people are providing. Can someone help please?
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by GMATGuruNY » Tue Nov 29, 2011 1:19 pm
Rastis wrote:I'm having a difficult time understanding the logic behind the answer explanations people are providing. Can someone help please?
A = 15% cement.
B = 35% cement.

If we combine EQUAL amounts of A and B, the percentage of cement in the mixture will be HALFWAY between 15 and 35.
To illustrate:
In 100 pounds of A, the amount of cement = 15 pounds.
In 100 pounds of B, the amount of cement = 35 pounds.
When the two are combined, the percentage of cement in the mixture = (15+35)/(100+100) = 50/200 = 25/100 = 25%.
25% is HALFWAY between 15% and 35%.

In the problem above, since the mixture is 25% cement -- HALFWAY between 15% and 35% -- EQUAL amounts of A and B must have been used.
Thus, the percentage of A in the mixture = 50%.
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