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One fourth of a solution that was 10% sugar by weight was...

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One fourth of a solution that was 10% sugar by weight was...

by BTGmoderatorLU » Mon Mar 12, 2018 2:37 pm
One fourth of a solution that was 10% sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight?

A. 34%
B. 24%
C. 22%
D. 18%
E. 8.5%

The OA is A.

I get the solution as follow,

let 100 be the total volume.
total salt = 10
salt taken out = 10/4 = 2.5

to make solution 16%, total salt = 16
salt added = 16-7.5 = 8.5

solution had = 8.5/25*100 = 34% sugar.

I would like that any expert explains another way to solve this PS question, please! Thanks in advance!
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by ErikaPrepScholar » Tue Mar 13, 2018 5:46 am
Your solution is very efficient and works great - there's no reason to solve using a different method. However, if you want options that you can use on test day, we can also solve this problem by building an equation:

One fourth of a solution that was 10% sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight?

Our final solution is 3/4 the first solution and 1/4 the second solution, or 0.75 of the first solution and 0.25 the second solution.

The first solution is 10% sugar, so the amount of sugar contributed to the resulting solution by the first solution is 0.75(0.1).

We don't know the percent sugar in the second solution, so we can express the amount of sugar contributed to the resulting solution by the second solution is 0.25y, where y is the decimal form of the percent sugar in the second solution.

The resulting solution is 16% sugar. We can combine all of this information to add the sugar in solution 1 to the sugar in solution 2 to yield the sugar in the resulting solution: 0.75(0.1) + 0.25y = 0.16

We can then solve for y, which gives 0.34, or 34%.
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by Scott@TargetTestPrep » Wed Mar 14, 2018 3:30 pm
LUANDATO wrote:One fourth of a solution that was 10% sugar by weight was replaced by a second solution resulting in a solution that was 16 percent sugar by weight. The second solution was what percent sugar by weight?

A. 34%
B. 24%
C. 22%
D. 18%
E. 8.5%
We can let the weight of the original solution = n.

Thus, the original solution had 0.1n sugar.

If 1/4 of the original solution was removed, the new solution now has:

0.1n x 3/4 = 0.1n x 0.75 = 0.075n sugar

Some amount of solution was added back so that the overall weight was 16% sugar. We can let x = the amount of sugar in the second solution and create the following equation:

(0.075n + x)/n = 0.16

0.075n + x = 0.16n

x = 0.085n

We see that the amount of sugar in the second solution, in terms of n, is 0.085n. However, the total weight of the second solution is (¼)n or 0.25n since that is the amount being replaced. Thus, the percent of the second solution that is sugar is:

0.085n/0.25n = 85/250 = 340/1000 = 34/100 = 34%

Alternate Solution:

We start with x liters of a 10% sugar solution. From it, we take out x/4 liters (which is also 10% sugar). We then add back in x/4 liters of z% sugar solution, resulting in x liters of a 16% sugar solution.

Let's express this in an equation and solve for z:

x(0.10) - (x/4)(0.10) + (x/4)z = x(0.16)

Multiplying both sides by 4, we obtain:

0.40x - 0.10x + zx = 0.64x

0.30x + zx = 0.64x

zx = 0.34x

z = 0.34, or 34%.

Answer: A

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