GMAT Math : Mixture Problems

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Example Questions

Example Question #11 : Mixture Problems

A chemist has 400 milliliters of a solution of $20\%$ alcohol on hand, and she wants to mix it with enough $60\%$ alcohol solution to turn it into a $30\%$ alcohol solution. If is the amount of $60\%$ alcohol she needs, which equation can she use to solve for ?

Possible Answers:

$80+ 0.60\left ( x-400 \right ) = 0.30x$

0.20 (400-x)+ 0.60 x = 120

$240+ 0.20\left ( x-400 \right ) = 0.30x$

80+ 0.60 x = 0.30 (x+400)

0.20 x +240= 0.30 (x+400)

Correct answer:

80+ 0.60 x = 0.30 (x+400)

Explanation:

The chemist will add 400 milliliters of 20% solution to milliliters of 60% solution to make x+400 milliliters of 30% solution.

The amount of pure alcohol in each solution will be the amount of solution multiplied by the concentration expressed as a decimal. Therefore, the amounts of alcohol in the three solutions, in milliliters, will be:

20% solution: $0.20 \cdot 400 = 80$

60% solution: 0.60 x

30% solution (result): 0.30 (x+400)

Since the first two solutions are added to yield the third, the amounts of alcohol are also being added, so the equation to solve is

80+ 0.60 x = 0.30 (x+400)

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Example Question #12 : Mixture Problems

A chemist wants to make one liter of $15\%$ hydrochloric acid solution. He only has two concentrations on hand - $12\%$ and $30\%$ . If is the amount in milliliters of the $12\%$ solution he is to use, then which equation can be used to solve for ?

Possible Answers:

$0.12 x + 0.30 \left (1,000-x \right ) = 1,000$

$0.12 x - 0.30 \left (1,000-x \right ) = 1,000$

$0.15 x + 300 =0 .12\left (x+1,000 \right )$

$0.12 x + 0.30 \left (1,000-x \right ) = 150$

$0.12 x + 300 = 0.15\left (x+1,000 \right )$

Correct answer:

$0.12 x + 0.30 \left (1,000-x \right ) = 150$

Explanation:

If the chemist is to use milliliters of 12% to make the solution, then he will mix it with 1,000-x milliliters (1,000 milliliters being equal to 1 liter) of the 30% solution. The result will be 1,000 milliliters of 12% solution.

The amount of pure acid in each solution will be the amount of solution multiplied by the concentration expressed as a decimal. Therefore, the amounts of acid in the three solutions, in milliliters, will be:

12% solution: 0.12 x

30% solution: $0.30 \left (1,000-x \right )$

15% solution (result): $0.15 \cdot 1,000 = 150$

Since the first two solutions are added to yield the third, the amounts of acid are also being added, so the equation to solve is

$0.12 x + 0.30 \left (1,000-x \right ) = 150$

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Example Question #11 : Understanding Mixture Problems

A nut mixture consists of peanuts, pistachios, and macadamia nuts in the ratio 5:3:2 , respectively, by weight. What amount of pistachios will be in 40 pounds of the mixture?

Possible Answers:

Correct answer:

Explanation:

Using the ratio, we can write the following equation:

5x+3x+2x=40

10x=40

x=4

There are pounds of peanuts, pounds of pistachios, and pounds of macadamia nuts in the mixture, so that means that in 40 pounds of the mixture, there are

$5x=5\cdot4=20$ pounds of peanuts

$3x=3\cdot4=12$ pounds of pistachios

$2x=2\cdot4=8$ pounds of macadamia nuts

Since the question asked about pistachios, the correct answer is that there are pounds of pistachios in pounds of the mixture.

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Example Question #11 : Mixture Problems

A chemist is diluting $30 \:ml$ of a pure acid solution with water in order to obtain a solution that is only $15\%$ acid. What is the total amount of the resulting solution?

Possible Answers:

$100\:ml$

$110\:ml$

$200\:ml$

$170\:ml$

$140\:ml$

Correct answer:

$200\:ml$

Explanation:

Let be the amount of water used to dilute the acid. The resulting solution will then have $(30+x) \:ml$ . The amount of acid in the final solution is: $0.15 \cdot (30+x)$ . That amount of acid in the final solution is equal to the original $30 \:ml$ of acid, since there is no acid in water.

Mixture

0.15(30+x)+0=30

$30+x=\frac{30}{0.15}=\frac{3000}{15}=200$

x=200-30=170

The amount of the final solution is $200 \:ml$ , and the amount of water used to dilute the original acid solution is $170 \:ml$ .

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Example Question #11 : Mixture Problems

A certain solution is $77\%$ glycol by volume. How many liters of this solution must be added to liters of a $53\%$ glycol solution in order to produce a solution with an overall concentration of $60\%$ ?

Possible Answers:

$13.8 \text{ liters}$

$17.3 \text{ liters}$

$23.2 \text{ liters}$

$42.0 \text{ liters}$

$27.4 \text{ liters}$

Correct answer:

$17.3 \text{ liters}$

Explanation:

Let x equal the number of liters of the 77% solution, the value we need to solve for. If the first solution is 77% glycol, then 0.77x will give us the number of liters of glycol in the first solution. Accordingly, 0.53(42 liters) will give us the number of liters of glycol in the second solution. If we are mixing these two solutions, then the number of liters of glycol in the final mixture will equal the sum of the number of liters of glycol in each solution. The amount of glycol in the final mixture will be its desired concentration times the volume of the total mixture, 0.60(x+42), so we can write the following equation and solve for x:

0.77x+0.53(42)=0.60(x+42)

0.77x+22.26=0.60x+25.2

0.17x=2.94

x=17.3 liters

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Example Question #174 : Word Problems

A scientist has one liter of a 40% sulfuric acid solution; he needs to dilute it to 25%. How much water will he need to dilute it to the appropriate strength?

Round your answer to the nearest tenth of a liter, if applicable.

Possible Answers:

$0.7 \textup{ L}$

$0.5 \textup{ L}$

$0.6 \textup{ L}$

$0.8 \textup{ L}$

$0.9 \textup{ L}$

Correct answer:

$0.6 \textup{ L}$

Explanation:

40% of one liter is 0.4 liters of acid, which is the amount of acid that will be in the solution at the end as well.

After liters of water is added, there will be w + 1 liters of solution. the amount of acid in the solution will be 25% of this, or 0.25(w + 1) .

The amount of acid in the solutions remains constant, so we set up and solve this equation:

0.25(w + 1) = 0.4

0.25w + 0.25 = 0.4

0.25w = 0.15

$w = 0.15 \div 0.25$

w = 0.6

The scientist will need to add 0.6 liters of water.

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Example Question #171 : Problem Solving Questions

A scientist has one liter of a 40% sulfuric acid solution; he needs to strengthen it to 50%. How much pure sulfuric acid will he need to strengthen it to the appropriate concentration?

Possible Answers:

$0.2\textup{ L}$

$0.25\textup{ L}$

$0.4 \textup{ L}$

$0.15\textup{ L}$

$0.1\textup{ L}$

Correct answer:

$0.2\textup{ L}$

Explanation:

40% of one liter of the original solution is 0.4 liters of acid.

Suppose the scientist adds liters of acid. Then the scientist will have A + 1 liters of solution and A + 0.4 liters of acid, and the concentration will be

$\frac{A + 0.4}{A+1} \cdot 100 = 50$ percent. Solve for :

$\frac{100A + 40}{A+1} = 50$

$\frac{100A + 40}{A+1} \cdot (A+1) = 50 (A+1)$

100A + 40 = 50A + 50

50A = 10

A = 0.2

The scientist will need to add 0.2 liters of pure acid to the solution.

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Example Question #181 : Gmat Quantitative Reasoning

Read the following problem:

A scientist needs one liter of 20% hydrochloric acid solution; however, he only has two one-liter bottles of solution on hand, one of which is 15% acid and one of which is 35% acid. How much of each solution must he mix together to get the solution he wants?

Let be the number of milliliters of 15% solution he will use and be the number of milliliters of 35% solution he will use. Which of the following systems of equations will yield the correct values of both?

Possible Answers:

x+y = 1,000

0.15x+0.35y = 200

y = x-1,000

0.15x+0.35y = 200

x+y = 1,000

0.15x+0.35y = 20

x+y = 1,000

15x+ 35y = 200

y = x-1,000

0.15x+0.35y = 20

Correct answer:

x+y = 1,000

0.15x+0.35y = 200

Explanation:

Since and are being set to numbers of milliliters, we convert one liter to 1,000 milliliters. The sum of the amounts of the two solutions and is 1,000, so

x+y= 1,000

is one equation of the system.

The other equation will add the acid. The amount of acid in an acid solution is the concentration, as a decimal, multiplied by the amount of solution. The amount of acid in the pre-mixture solutions are 0.15x and 0.35y ; add these to get the amount of acid in 1,000 milliliters of a 20% solution, or $0.20 \times 1,000 = 200$ milliliters:

0.15x+0.35y = 200

The correct system is:

x+y= 1,000

0.15x+0.35y = 200

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Example Question #182 : Word Problems

Pieces of Rainbow Candy come in six colors, including green; the machines are carefully set to make sure each batch of sixty pounds contains an equal amount of candy of each color.

One day, something went wrong, and one sixty-pound batch only contained half as much green candy as it was supposed to; the other five colors were distributed equally. How many pounds of green candies needed to be added to this batch in order to restore the correct distribution of colors?

Possible Answers:

$12\textup{ lbs}$

$10 \textup{ lbs}$

$5\textup{ lbs}$

$6 \textup{ lbs}$

$11\textup{ lbs}$

Correct answer:

$6 \textup{ lbs}$

Explanation:

Since a normal batch of candies is one-sixth green candies, then it contains $\frac{1}{6} \times 60 = 10$ pounds of green candies; the defective batch contained half this, or 5 pounds. If we let be the number of pounds of green candies that were added, then, after the addition, there were G+5 pounds of green candies in a batch of G+60 candies. One sixth of the candies are green, so set and solve the equation:

$\frac{G+5}{G+60} = \frac{1}{6}$

$6\left (G+5 \right )= 1 (G+60)$

6G+30= G+60

6G+30 - G - 30= G+60 - G - 30

5G = 30

G=6

Six pounds of green candies are added.

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Example Question #183 : Word Problems

Pieces of Rainbow Candy come in six colors, including white; the machines are carefully set to make sure each batch of forty-eight pounds contains an equal number of each color candy.

One day, something went wrong, and one forty-eight-pound batch only contained half as much white candy as it was supposed to; the other five colors were distributed equally. White candies were added to this batch until the correct color distribution was restored. To the nearest pound, how many pounds of candy ended up in this corrected batch?

Possible Answers:

$56\textup{ lbs}$

$54\textup{ lbs}$

$55\textup{ lbs}$

$52 \textup{ lbs}$

$53 \textup{ lbs}$

Correct answer:

$53 \textup{ lbs}$

Explanation:

Initially, there were 48 pounds of candies. In a normal batch, one sixth of them, or $\frac{1}{6} \times 48 = 8$ pounds, should be white, but in the defective batch, half this, or 4 pounds, were white. Subsequently, 44 pounds of candies in the defective batch were not white.

Let be the number of pounds of candy in the batch after the white candies were added. Since $\frac{5}{6}$ of a normal batch of candies are not white, $\frac{5}{6} C$ pounds of candies are not white; since no candies that weren't white were added,

$\frac{5}{6} C = 44$

$\frac{6} {5}\cdot \frac{5}{6} C =\frac{6} {5}\cdot 44$

$C= 52\frac{4}{5}$

The closest choice, and the correct one, is 53 pounds.

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GMAT Math : Mixture Problems

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #11 : Mixture Problems

Example Question #12 : Mixture Problems

Example Question #11 : Understanding Mixture Problems

Example Question #11 : Mixture Problems

Example Question #11 : Mixture Problems

Example Question #174 : Word Problems

Example Question #171 : Problem Solving Questions

Example Question #181 : Gmat Quantitative Reasoning

Example Question #182 : Word Problems

Example Question #183 : Word Problems

All GMAT Math Resources

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