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Mixtures question

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by tahitiboy » Thu Apr 08, 2010 10:00 pm
119. A contractor combined x tons of a gravel mixture that contained 10 percent gravel G, by weight, with y tons of a mixture that contained 2 percent gravel G, by weight, to produce z tons of a muxture that was 5 percent gravel G, by weight. What is the value of x?
1. y=10
2. z=16

Can someone help?
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Source: — Data Sufficiency |
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by outreach » Thu Apr 08, 2010 10:46 pm
tahitiboy wrote:119. A contractor combined x tons of a gravel mixture that contained 10 percent gravel G, by weight, with y tons of a mixture that contained 2 percent gravel G, by weight, to produce z tons of a muxture that was 5 percent gravel G, by weight. What is the value of x?
1. y=10
2. z=16

Can someone help?
option D

we know that
x + y =z - (1)

also we know that gravel G
.1x+.02y=0.05z - (2)

from option 1

if we substitute y value in eq 1 and eq 2, we will two equations with x and z unknown variables. Solve the 2 eq to get the answer

from option 2
if we substitute z value in eq 1 and eq 2, we will two equations with x and y unknown variables. Solve the 2 eq to get the answer
Last edited by outreach on Thu Apr 08, 2010 10:47 pm, edited 1 time in total.
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by thephoenix » Thu Apr 08, 2010 10:46 pm
eqn is 0.1x+0.02y=0.05z
in order to get x we need bth y and z
hence C
a big mistake i forgot x+y=z
so Now D is the ans
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by eaakbari » Fri Apr 09, 2010 12:46 am
Ill tell you a neat trick to approach all mixtures questions.
Just remember the formula

% of higher-% of mean :% of mean - %of lower :: lower weight : higher weight

Stem:
Here our higher is x tons of 10%G
lower is y tons of 2%G
mean is z tons of 5%G


Substitute

5:3 :: y: x

Statement 1 gives us y , we can easily find x

Statement 2 gives us z which is x + y we can easily find x

Hence D
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