## A contractor combined $$x$$ tons of a gravel mixture that contained $$10$$ percent gravel $$G,$$ by weight, with y tons

##### This topic has expert replies
Legendary Member
Posts: 1441
Joined: 01 Mar 2018
Followed by:2 members

### A contractor combined $$x$$ tons of a gravel mixture that contained $$10$$ percent gravel $$G,$$ by weight, with y tons

by Gmat_mission » Sun Sep 12, 2021 8:33 am

00:00

A

B

C

D

E

## Global Stats

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 mixture that was $$5$$ percent gravel $$G,$$ by weight. What is the value of $$x?$$

(1) $$y = 10$$

(2) $$z = 16$$

Source: GMAT Prep

### GMAT/MBA Expert

GMAT Instructor
Posts: 15793
Joined: 08 Dec 2008
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1267 members
GMAT Score:770

### Re: A contractor combined $$x$$ tons of a gravel mixture that contained $$10$$ percent gravel $$G,$$ by weight, with y t

by [email protected] » Sun Sep 12, 2021 12:15 pm

00:00

A

B

C

D

E

## Global Stats

Gmat_mission wrote:
Sun Sep 12, 2021 8:33 am
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 mixture that was $$5$$ percent gravel $$G,$$ by weight. What is the value of $$x?$$

(1) $$y = 10$$

(2) $$z = 16$$

Source: GMAT Prep
Let's use some weighted averages to solve this question
Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

Target question: What is the value of x ?

Given: 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 mixture that was 5 percent gravel G, by weight.
First, we can write: x + y = z

Also, the total weight of the mixture = z (aka x + y)
So, when we apply the above formula, we get: 5% = (x/z)(10%) + (y/z)(2%)
Ignore the % symbols: 5 = (x/z)(10) + (y/z)(2)
Multiply both sides by z to get: 5z = 10x + 2y
Since x + y = z, we can rewrite the above equation as: 5(x +y) = 10x + 2y
Expand: 5x + 5y = 10x + 2y
Simplify to get: 5x - 3y = 0

Now onto the statements!!!!!

Statement 1: y = 10
Replace y with 10 to get: 5x - 3(10) = 0
Solve to get, x = 6
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: z = 16
In other words, x + y = 16

So, we have:
5x - 3y = 0 and x + y = 16
Since we have 2 linear equations with 2 variables, we COULD solve the system for x, which means we COULD answer the target question
So, statement 2 is SUFFICIENT