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If a 30%-solution of alcohol was mixed with

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by DavidG@VeritasPrep » Tue Sep 05, 2017 8:19 am
NandishSS wrote:If a 30%-solution of alcohol was mixed with a 50%-solution of alcohol to obtain 10 liters of a 45%-solution of alcohol, how much of the 30%-solution was used?

A. 2.0 liters
B. 2.5 liters
C. 2.7 liters
D. 3.0 liters
E. 3.2 liters

OA : B
If we plot the solution percentages on a number line, we'll get:

30----------------45-----50
Gap:------15--------5------

Notice that we have a 15:5 or 3:1 ratio of the two components, so we can call them 3x and x. Notice also that we must have more of the 50% solution than the 30% solution as the overall average is closer to 50%.

50% solution: 3x
30% solution: x
Total: 10

3x + x = 10 ---> 4x = 10 ---> x = 2.5. The answer is B.
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by Brent@GMATPrepNow » Tue Sep 05, 2017 8:29 am
NandishSS wrote:If a 30%-solution of alcohol was mixed with a 50%-solution of alcohol to obtain 10 liters of a 45%-solution of alcohol, how much of the 30%-solution was used?

A. 2.0 liters
B. 2.5 liters
C. 2.7 liters
D. 3.0 liters
E. 3.2 liters
Let's keep track of the volume of pure alcohol
The combined solution has a volume of 10 liters, and it's 45% pure alcohol
45% of 10 liters = (0.45)(10) = 4.5 liters
So, the combined solutions contains 4.5 liters of PURE alcohol

Let x = the volume of 30% solution
So, 10-x = the volume of 50% solution (since we have 10 liters of resulting combined solution

This means 0.3x = the volume of PURE alcohol in the 30% solution
And 0.5(10 - x) = the volume of PURE alcohol in the 50% solution

We can write: 0.3x + 0.5(10 - x) = 4.5
Expand: 0.3 + 5 - 0.5x = 4.5
Simplify: 5 - 0.2x = 4.5
Subtract 5 from both sides: -0.2x = -0.5
Divide both sides by -0.2 to get: x = 2.5

Answer: B

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by ErikaPrepScholar » Tue Sep 05, 2017 8:53 am
Right off the bat, the word "mixed" should indicate that this is a mixture problem. This means that we'll be setting up equations using what we know from the problem.

The problem tells us that we combine some amount of a 30% alcohol solution with some amount of a 50% alcohol solution. This gives us 10 liters of a 45% alcohol solution. Let's say we have X liters of 30% alcohol solution and Y liters of 50% alcohol solution. We want to solve for X. We can build one equation for the total volume of the solution:

X + Y = 10

and one equation for the total alcohol content of the solution:

0.3X + 0.5Y = 0.45 (X+Y)

Simplifying gives:

0.3X + 0.5Y = 0.45X + 0.45Y
0.05Y = 0.15X
Y = 3X

Plugging that into our initial equation gives:

X + 3X = 10
4X = 10
X = 10/4
X = 2.5

Or answer choice B.

Alternatively, you can solve by visualizing where the percents fall on a number line.

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50% is 3 times closer to 45% than 30% is. This means there must be 3 times as much of the 50% solution in the final solution than there is of the 30% solution. So Y = 3X. Knowing that X + Y = 10, we can solve for X using the math shown above.
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by GMATGuruNY » Tue Sep 05, 2017 9:31 am
NandishSS wrote:If a 30%-solution of alcohol was mixed with a 50%-solution of alcohol to obtain 10 liters of a 45%-solution of alcohol, how much of the 30%-solution was used?

A. 2.0 liters
B. 2.5 liters
C. 2.7 liters
D. 3.0 liters
E. 3.2 liters
An alternate approach is to PLUG IN THE ANSWERS, which represent the amount of 30% solution.
When the correct answer is plugged in, the average alcohol percentage per 10 liters = 45%.

D: 3 liters 30% solution, implying 7 liters 50% solution
Average alcohol percentage per 10 liters = (3*30 + 7*50)/10 = (90+350)/10 = 440/10 = 44%.
The resulting alcohol percentage is TOO LOW.
To increase the resulting alcohol percentage, we must use MORE 50% SOLUTION (since it has a higher percentage of alcohol) and LESS 30% SOLUTION (since it has a lower percentage of alcohol).
Eliminate D and E.

B: 2.5 liters 30% solution, implying 7.5 liters 50% solution
Average alcohol percentage per 10 liters = (2.5*30 + 7.5*50)/10 = (75 + 375)/10 = 450/10 = 45%.
Success!

The correct answer is B.
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If a 30%-solution of alcohol was mixed with

by Admin1 » Tue Sep 05, 2017 1:04 pm
If a 30%-solution of alcohol was mixed with a 50%-solution of alcohol to obtain 10 liters of a 45%-solution of alcohol, how much of the 30%-solution was used?
A. 2.0 liters
B. 2.5 liters
C. 2.7 liters
D. 3.0 liters
E. 3.2 liters

Solution:
Let 30%-solution of alcohol be 'x' litres.
Then 50%-solution of alcohol will be '(10-x)' litres.

The situation described in the problem statement can be described in equation form as,
'x' litres of 30%-solution + '(10-x)' litres of 50%-solution = 10 litres of 45%-solution
x(30%) + (10 - x)(50%) = 10(45%)

Converting percentages to decimals
x (30/100) +(10-x)(50/100) = 10(45/100)

Simplifying
0.3x + 0.5(10 - x) = 10(0.45)

Using Distributive Property
0.3x + 5 - 0.5x = 4.5

Combining like terms
0.3x -0.5x = 4.5 - 5

Simplifying
-0.2x = -0.5
0.2x = 0.5
x = 0.5/0.2
x = 2.5 litres

Hence, 2.5 litres of 30%-solution was used.
Option B is the correct answer.
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by Scott@TargetTestPrep » Mon Sep 11, 2017 10:20 am
NandishSS wrote:If a 30%-solution of alcohol was mixed with a 50%-solution of alcohol to obtain 10 liters of a 45%-solution of alcohol, how much of the 30%-solution was used?

A. 2.0 liters
B. 2.5 liters
C. 2.7 liters
D. 3.0 liters
E. 3.2 liters

OA : B
We can let x = the amount of 30% solution and y = the amount of 50% solution. Thus:

x + y = 10

y = 10 - x

and

0.3x + 0.5y = 0.45(x + y)

30x + 50y = 45x + 45y

5y = 15x

y = 3x

Substituting 3x for y in y = 10 - x, we have:

3x = 10 - x

4x = 10

x = 10/4 = 2.5

Alternate Solution:

We mix x liters of 30% solution with (10 - x) liters of 50% solution to obtain 10 liters of 45% solution. We can express this information in the following equation:

x(0.30) + (10 - x)(0.50) = 10(0.45)

0.3x + 5 - 0.5x = 4.5

3x + 50 - 5x = 45

-2x = -5

x = 5/2, or 2.5

Answer: B

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