by applying alligation method and mental math we know all percentages of mixture, so we figure average rate and we only need to know one of each mixture agent's weight
|5-10|=5 assigned to y
|2-5|=3 assigned to x
5+3=8; z/8=y/5=x/3
OR
st(1)10/5=x/3 Sufficient
st(2) 16/8=x/3 Sufficient

Akansha wrote:A contractor combined x tons of gravel mix that contained 10% gravel G
by weight, with y tons of a mixture that contained 2% gravel G by weight to
produce z tons of a mixture that was 5% gravel G by weight. What is the value of
x? D
a. y = 10
b. z = 16

OA is D

here x+y=z, the problem says that
(10x+2y)/(x+y)=5, after little solving
5x=3y, and we need to find x-?
(1)if y=10 and 5x=3y, then it possible to find x, x=6 suff
(2) z=16. and it comes that x+y=z=16. solving with 5x=3y also possible to find x, suff

The statement provides us with informations that can be translated into two simple equations:

First one: x+y=z and second one: 0.05z= 0.1x + 0.02y or 5z= 10x + 2y

From (1): y=10 gives us a system of two equations with two unknowns (x and z), therefore x can be found. SUFFICIENT

From (2): z=16 gives us a system of two equations with two unknowns (x and y), therefore x can be found. SUFFICIENT

Both statements are sufficient, therefore the answer is D.

What exactly is the alligation method? And is it ever beneficial to use, as the last poster broke it down pretty simply and the other two I can't say I followed completely.

z/8=y/5=x/3

I have intentionally derived these ratios for us to see the legitimacy of alligation method.
Certainty comes with definitions.
The ways you or me would define a problem most probably are followed by
-assigning variables, the first
-finding the relationship, the second
-solving for variables, the third

with alligation method you've all three set - just need to solve one-two practice problems beforehand for applying alligation when needed.

uslalas22 wrote:What exactly is the alligation method? And is it ever beneficial to use, as the last poster broke it down pretty simply and the other two I can't say I followed completely.

I think we could have a simple approach:
from statement, we can have 2 equations straight:
1. x+y=z
2. x/10 + y/50 = z/20

From A, y= 10, which brings down above equation to "2 equations, 2 unknowns". A is sufficient.
From B, z=16, again "2 equations, 2 unknowns". Without even going ahead for the complete solution we can say that 'D' is the answer.