Data Sufficiency

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?

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

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

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