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Mixtures

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by KItuz » Sun Jun 26, 2011 8:21 pm
A contractor combind x tons of gravel mixture that contained 10% gravel G, by weight, with y tons of a mixture that contained 2 percent gravel G, by weight, to produce z tons of a mixture that was 5% gravel G, by weight. What is the value of x?
[1] y = 10
[2] z = 16

On solving you get 10x+2y=5z. You need 1 and 2 to solve.

Ans is D. Can someone explain how?
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Source: — Data Sufficiency |
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by smackmartine » Sun Jun 26, 2011 8:47 pm
IMO D
This is tricky!!!

So know the trick!!! WHENEVER YOU SEE THAT ALL THE COMPOSITION IS GIVEN IN TERMS OF % (PERCENTAGES) APPLY ALLIGATION CONCEPT RATHER THAN WRITING AN EQUATION.

X ---CommonMixture--Y

10 ------------5-------------2

(5-2)=3------|------(10-5)=5 (Sides of the results are reversed while applying Alligation method)*

This means that 3 parts of X tons and 5 parts of Y tons of mixtures are present in the final mixture.
So, 3P + 5P = 8P where P is the common factor.

1) y=10

y is nothing but 5P, so 5P=10 --> P=2 --> x= 3P= 3*2=6 (sufficient), no need to calculate though.

2) z= 16

or 8P=16 , P=2 , x= 3P=3*2=6 (sufficient), no need to calculate though.
Smack is Back ...
It takes time and effort to explain, so if my comment helped you please press Thanks button :)
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by Frankenstein » Sun Jun 26, 2011 8:56 pm
KItuz wrote: On solving you get 10x+2y=5z. You need 1 and 2 to solve.

Ans is D. Can someone explain how?
Hi,
You also have another equation x+y = z. You have probably ignored that.
Cheers!

Things are not what they appear to be... nor are they otherwise
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