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How many liters of a solution that is 15 percent salt must be added to 5 liters of a solution that is 8 percent salt so

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How many liters of a solution that is 15 percent salt must be added to 5 liters of a solution that is 8 percent salt so

by BTGModeratorVI » Tue Feb 16, 2021 8:22 am

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How many liters of a solution that is 15 percent salt must be added to 5 liters of a solution that is 8 percent salt so that the resulting solution is 10 percent salt?

A. 1
B. 1.5
C. 2
D. 2.5
E. 3

Answer: C
Source: official guide
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Re: How many liters of a solution that is 15 percent salt must be added to 5 liters of a solution that is 8 percent salt

by Brent@GMATPrepNow » Tue Feb 16, 2021 8:34 am
BTGModeratorVI wrote: ↑
Tue Feb 16, 2021 8:22 am
 How many liters of a solution that is 15 percent salt must be added to 5 liters of a solution that is 8 percent salt so that the resulting solution is 10 percent salt?

A. 1
B. 1.5
C. 2
D. 2.5
E. 3

Answer: C
Source: official guide
When solving mixture questions, it can be useful to sketch the solutions with the ingredients SEPARATED.
8% of 5 liters = 0.4 liters. So, the original mixture contains 0.4 liters of salt

Let x = the volume (in liters) of 15% solution required
So, this solution contains 0.15x liters of salt

We get:
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We want the resulting solution to be 10% salt.

In other words, we want: [m][fraction]0.4 + 0.15x/5+x[/fraction]=[fraction]10/100[/fraction][/m] (aka 10%)

Simplify: [m][fraction]0.4 + 0.15x/5+x[/fraction]=[fraction]1/10[/fraction][/m]

Cross multiply: [m](10)(0.4 + 0.15x) = 5+x[/m]

Expand: [m]4 + 1.5x = 5+x[/m]

Solve: [m]x = 2[/m]

Answer: C

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Brent
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Re: How many liters of a solution that is 15 percent salt must be added to 5 liters of a solution that is 8 percent salt

by Scott@TargetTestPrep » Tue Feb 23, 2021 3:52 pm
BTGModeratorVI wrote: ↑
Tue Feb 16, 2021 8:22 am
 How many liters of a solution that is 15 percent salt must be added to 5 liters of a solution that is 8 percent salt so that the resulting solution is 10 percent salt?

A. 1
B. 1.5
C. 2
D. 2.5
E. 3

Answer: C
Source: official guide
Solution:

We see that currently there are 5 x 0.08 = 0.4 liters of salt in the 5-gallon solution. Let x be the number of liters of a 15% salt solution needed to add to the 8% salt solution to yield a 10% salt solution. So 0.15x liters of salt from the new solution will be added. We can create the equation:

(0.4 + 0.15x)/(5 + x) = 10/100

(0.4 + 0.15x)/(5 + x) = 1/10

4 + 1.5x = 5 + x

0.5x = 1

x = 2

Alternate Solution:

To a solution of 5 liters that is 8% salt we will add x liters of a solution that is 15% salt, and the result will be (5 + x) liters of a solution that is 10% salt. We can create the following equation:

5(0.08) + x(0.15) = (5 + x)(0.10)

0.4 + 0.15x = 0.5 + 0.10x

0.05x = 0.1

5x = 10

x = 2

Answer: C

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