BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

A contractor combined x tons of a gravel . . .

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 2276
Joined: Sat Oct 14, 2017 6:10 am
Followed by:3 members

A contractor combined x tons of a gravel . . .

by VJesus12 » Thu Nov 23, 2017 6:38 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

The OA is D .

How can I find the value of x with each statement alone? Experts, may you show me how to find x? Thanks.
Top
Source: — Data Sufficiency |

GMAT/MBA Expert

User avatar
Legendary Member
Posts: 503
Joined: Thu Jul 20, 2017 9:03 am
Thanked: 86 times
Followed by:15 members
GMAT Score:770

by ErikaPrepScholar » Mon Nov 27, 2017 9:04 am
We have a mixtures problem here. With mixtures problem, our best bet is often to build a system of equations. Using the information given in the question stem, we can create two equations:

x + y = z (the total weight of the first gravel mixture plus the total weight of the second gravel mixture equals the total weight of the combined gravel mixture)

and

0.1x + 0.02y = 0.05z (multiplying the percent gravel in each gravel mixture by the total weight of each gravel mixture gives the weight of just gravel in each mixture - so the weight of just gravel in the first mixture plus the weight of just gravel in the second mixture equals the weight of just gravel in the combined mixture)

So we see here that we have two distinct equations with three unknown variables. A rule we should know: to solve for all of our variables, we need the same number of distinct equations as the number of variables. So if we have one variable, we need one equation to solve; if we have two variables, we need two distinct (not the same) equations to solve; etc. So right now we have one too many variables. If we either get one new equation or lose one of our variables, we should have the same number of variables as equations, which should allow us to solve for all of our variables.

Statement 1 gives us y. This means that we now only have two variables (x and z) and two distinct equations. So we should be able to solve for x. We could plug in our value for y, then use substitution to solve for x, but we don't actually need to do that since we know that we can. Sufficient.

Statement 2 gives us z. Again, we now only have two variables (x and y) and two distinct equations, allowing us to solve for x. Sufficient.
Image

Erika John - Content Manager/Lead Instructor
https://gmat.prepscholar.com/gmat/s/

Get tutoring from me or another PrepScholar GMAT expert: https://gmat.prepscholar.com/gmat/s/tutoring/

Learn about our exclusive savings for BTG members (up to 25% off) and our 5 day free trial

Check out our PrepScholar GMAT YouTube channel, and read our expert guides on the PrepScholar GMAT blog
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Mon Nov 27, 2017 9:30 am
A contractor combined x tons of a gravel mixture that contained 10% gravel G, by weight, with y tons of a mixture that contained 2 percent of the 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
x = 10% gravel
y = 2% gravel.
z = the MIXTURE of x and y = 5% gravel.

To determine the required ratio of x to y, use ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the percentages for x and y on the ends and the percentage for mixture z in the middle.
x 10%-----------5%-----------2% y

Step 2: Calculate the distances between the percentages.
x 10%-----5-----5%----3-----2% y

Step 3: Determine the ratio in the mixture.
The required ratio of x to y is equal to the RECIPROCAL of the distances in red.
x:y = 3:5.

Since x:y = 3:5, and 3+5 = 8, every 8 tons of mixture z is composed of 3 tons of x and 5 tons of y.

Statement 1: y=10
Since x:y = 3:5 = 6:10, x=6.
SUFFICIENT.

Statement 2: z=16
Since x:y = 3:5 = 6:10, and 6+10 = 16, the 16 tons of mixture z must be composed of 6 tons of x and 10 tons of y.
SUFFICIENT.

The correct answer is D.

Check here for a similar problem:
https://www.beatthegmat.com/weighted-ave ... 74015.html
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pm
Location: Vancouver, BC
Thanked: 5254 times
Followed by:1268 members
GMAT Score:770

Re: A contractor combined x tons of a gravel . . .

by Brent@GMATPrepNow » Wed Jan 22, 2020 4:18 pm
VJesus12 wrote: ↑
Thu Nov 23, 2017 6:38 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

The OA is D .

How can I find the value of x with each statement alone? Experts, may you show me how to find x? Thanks.
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

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Post new topic Post Reply
• Page 1 of 1