One fourth

[BTGmoderatorDC]

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%

What is the easiest solution in this problem?

OA A

[GMATGuruNY]

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%

Let the total volume = 100 pounds.

Resulting solution:
Since the resulting solution is 16% sugar, the amount of sugar in the resulting solution = 16% of 100 = 16 pounds.

Original solution:
3/4 of the original 10% solution -- in other words, 75 pounds of the original 10% solution -- is NOT replaced.
Amount of sugar in 75 pounds of original 10% solution = 10% of 75 = 7.5 pounds.

Replacement solution:
Since 1/4 of the total weight is replaced, the weight of the replacement solution = (1/4)(100) = 25 pounds.
Amount of sugar in the replacement solution = (total sugar in the resulting solution) - (sugar in the remaining 75 pounds of original solution) = 16 - 7.5 = 8.5 pounds.
Sugar percentage in the replacement solution = (8.5)/25 * 100 = 34.

The correct answer is A.

[DrMaths]

Originally: 1/4 @10% + 3/4 @ 10%
Later: 1/4 @ p% + 3/4 @ 10% = 16%
Multiply all by 4:
p% + 30% = 64%
p = 64-30 = 34
Answer = A

[Scott@TargetTestPrep]

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