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## GMAT - Problem Solving - Percentage

### If x > 0, x/50 + x/25 is what percent of x? (more)

If x > 0, x/50 x/25 is what percent of x?

A. 6%
B. 25%
C. 37 1/2%
D. 60%
E. 75 %

x/50 + x/25 = (x + 2x) /50 = 3x/50

% =  (3x/50) * 100 / x

= 6%

A is correct

### Mary's income is 60 percent more than Tim's income, and Tim's income is 40 percent less than Juan's income. What percent of Juan's income is Mary's income? (more)

Mary's income is 60 percent more than Tim's income, and Tim's income is 40 percent less than Juan's income. What percent of Juan's income is Mary's income?

(A) 124%
(B) 120%
(C) 96%
(D) 80%
(E) 64%

Assume Tim's income be x

Mary's income = x + 60% of x = 1.6x

Assume Juan's income be y

Tim's income = x = y - 40% of y = 0.6y

y = x/0.6

Mary's income * 100/ Juan's income

1.6x * 100 / x/0.6

= 16*6 = 96%

### A tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume. If 2,500 gallons of water evaporate from the tank, the remaining solution will be approximately what percent sodium chloride? (more)

A tank contains 10,000 gallons of a solution that is 5 percent sodium chloride by volume. If 2,500 gallons of water evaporate from the tank, the remaining solution will be approximately what percent sodium chloride?

(A) 1.25%
(B) 3.75%
(C) 6.25%
(D) 6.67%
(E) 11.7%

Initial volume of sodium chloride = 5% of 10000 = 500

new volume of the solution after evaporation of 2500 gallons of water = 10000 - 2500 = 7500

New percentage of sodium chloride in the solution = 500/7500 * 100 = 6.67%

D is correct

### Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 % fescue. If a mixture of X and Y contains 30% ryegrass, what percent of the weight of the mixture is X? (more)

Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 % fescue. If a mixture of X and Y contains 30% ryegrass, what percent of the weight of the mixture is X?

A. 10%
B. 33 1/3%
C. 40%
D. 50%
E. 66 2/3%

Let's assume that the weight of X in the mixture of X and Y be x and that of Y be y

Total weight of the mixture = x + y

Amount of rye grass in the mixture = 40% of x + 25% of y = 0.4x + 0.25y

% of ryegrass in the mixture = (0.4x + 0.25y) / (x + y)

0.3 = (0.4x + 0.25y) / (x + y)

0.3x + 0.3y = 0.4x + 0.25y

0.05y = 0.1x

y = 2x

Percentage of X in the mixture = 100*x / (x + y) = 100*x / (x + 2x) = 100*1/3 = 33.33%

B is correct.

### Company P had 15 percent more employees in December than it had in January. If Company P had 460 employees in December, how many employees did it have in January? (more)

Company P had 15 percent more employees in December than it had in January. If Company P had 460 employees in December, how many employees did it have in January?

(A) 391
(B) 400
(C) 410
(D) 423
(E) 445

Assume number of employees in January be x.

Number of employees in december = x + 15% of x = 1.15x

460 = 1.15x

x = 400