Mixture Problems - GMAT Quantitative
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If Audrey is currently 
years old, and Matt’s age is 2 years more than 
of Audrey’s age, what will be Matt’s age in 5 years?
If Audrey is currently
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of Audrey’s age is 
.
2 years more than 
of Audrey’s age is
.
However, be careful because the question is asking for Matt’s age in 5 years, so you need to add 5 years to Audrey’s current age:
2 years more than
However, be careful because the question is asking for Matt’s age in 5 years, so you need to add 5 years to Audrey’s current age:
Charlie, the barrista at Sunday Coffee, has a problem. He must mix together Licorice Dream tea, which costs $20/lb, and Strawberry Explosion tea, which costs $12/lb to make sixty pounds of a new flavor of tea. Unfortunately, he forgot the correct proportions. He does know that the new tea costs $14/lb.
How many pounds of the Licorice Dream tea go into the mix?
(You may assume that the tea is to sell for the same amount of money as if it had been sold separately.)
Charlie, the barrista at Sunday Coffee, has a problem. He must mix together Licorice Dream tea, which costs $20/lb, and Strawberry Explosion tea, which costs $12/lb to make sixty pounds of a new flavor of tea. Unfortunately, he forgot the correct proportions. He does know that the new tea costs $14/lb.
How many pounds of the Licorice Dream tea go into the mix?
(You may assume that the tea is to sell for the same amount of money as if it had been sold separately.)
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Let 
be the number of pounds of Licorice Dream tea; then 
is the number of pounds of Strawberry Explosion. Then the total value of each ingredient tea, as well as the total value of the new tea, will be:
Licorice: 
or 
Strawberry: 
or 
New tea: 
Add the individual costs of the ingredient teas to get the total cost of the new tea.
Charlie will use 15 pounds of the Licorice Dream; as this is not one of the given answers, the correct answer is "None of the answers are correct".
Let
Licorice:
Strawberry:
New tea:
Add the individual costs of the ingredient teas to get the total cost of the new tea.
Charlie will use 15 pounds of the Licorice Dream; as this is not one of the given answers, the correct answer is "None of the answers are correct".
A scientist needs a 10% saline solution for an experiment. In his closet he finds a 20 ounce bottle of 25% saline solution. How many ounces of pure water should he add to the mixture to produce the correct saline solution?
A scientist needs a 10% saline solution for an experiment. In his closet he finds a 20 ounce bottle of 25% saline solution. How many ounces of pure water should he add to the mixture to produce the correct saline solution?
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The solution he needs has only a 10% salt level. Currently we know that his solution has 20 ounces at 25% salt. We can calculate the amount of salt in the 20 ounce container by utilizing the given information. (20)*(.25) = 5 ounces of salt. Let x be the volume of pure water (in ounces) added.
Therefore, we know the total volume of our new solution will be 20+x. We know we want our solution to have 10% salt, so our salt amount in the new solution will have to be (20+x) *(.10). Since we are not adding any salt when we add our "pure water", we know the total salt in the solution will not change. Therefore, we can write the equation. (20+x)*(.10) = 5
Solve for x and you get x = 30 ounces of pure water.
The solution he needs has only a 10% salt level. Currently we know that his solution has 20 ounces at 25% salt. We can calculate the amount of salt in the 20 ounce container by utilizing the given information. (20)*(.25) = 5 ounces of salt. Let x be the volume of pure water (in ounces) added.
Therefore, we know the total volume of our new solution will be 20+x. We know we want our solution to have 10% salt, so our salt amount in the new solution will have to be (20+x) *(.10). Since we are not adding any salt when we add our "pure water", we know the total salt in the solution will not change. Therefore, we can write the equation. (20+x)*(.10) = 5
Solve for x and you get x = 30 ounces of pure water.
Mike, the barista at Moose Jaw Coffee, has to mix together two kinds of coffee beans - Mocha Madness, which costs $12 a pound, and Sumatra Sweetness, which costs $20 a pound - to produce forty pounds of a coffee that costs $14 a pound. The beans in the mixture sell for the same price as they would separately.
How many pounds of Mocha Madness coffee will Mike put into the mixture?
Mike, the barista at Moose Jaw Coffee, has to mix together two kinds of coffee beans - Mocha Madness, which costs $12 a pound, and Sumatra Sweetness, which costs $20 a pound - to produce forty pounds of a coffee that costs $14 a pound. The beans in the mixture sell for the same price as they would separately.
How many pounds of Mocha Madness coffee will Mike put into the mixture?
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Let 
be the number of pounds of Mocha Madness coffee beans Mike uses. Then he will use 
pounds of Sumatra Sweetness coffee beans.
The Mocha Madness coffee costs $12 a pound times 
pounds, or 
dollars; similarly, the Sumatra Sweetness coffee costs 
. The total cost of the coffee will be 
. Since the beans will sell for the same price as unmixed, we can add the prices to obtain and solve this equation:
30 pounds of the Mocha Madness will go into the mixture.
Let
The Mocha Madness coffee costs $12 a pound times
30 pounds of the Mocha Madness will go into the mixture.
24 coins, all nickels and dimes, are together worth $1.75.
How many of the coins are nickels?
24 coins, all nickels and dimes, are together worth $1.75.
How many of the coins are nickels?
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Let 
be the number of nickels. Then there are 
dimes.
The amount of money is defined by the expression 
. Set this equal to 1.75 and solve:
The mix includes 13 nickels.
Let
The amount of money is defined by the expression
The mix includes 13 nickels.
Some dimes and some quarters are together worth $8.95.
Which of the following is a possible number of dimes in this mixture?
Some dimes and some quarters are together worth $8.95.
Which of the following is a possible number of dimes in this mixture?
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If the dimes are removed, the amount of money remaining, being only quarters, must be a multiple of $0.25. We can test each choice accordingly.
Of the choices given, only the removal of 17 dimes leaves an amount that coud possibly be made up only of quarters.
If the dimes are removed, the amount of money remaining, being only quarters, must be a multiple of $0.25. We can test each choice accordingly.
Of the choices given, only the removal of 17 dimes leaves an amount that coud possibly be made up only of quarters.
How much of a 20% acid solution would a chemist have to mix with one liter of a 40% acid solution to yield a 36% acid solution?
How much of a 20% acid solution would a chemist have to mix with one liter of a 40% acid solution to yield a 36% acid solution?
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Let 
be the amount of 20% acid solution used in milliliters. Then the amount of total solution will be 
(one liter being 1,000 milliliters). The amount of acid in each solution will be as follows:
In the 20% solution: 
In the 40% solution: 
In the 36% solution: 
Add the acid in the two source solutions to get the acid in the resulting solution; then solve for 
:
The chemist must add 250 milliliters of the 20% acid solution.
Let
In the 20% solution:
In the 40% solution:
In the 36% solution:
Add the acid in the two source solutions to get the acid in the resulting solution; then solve for
The chemist must add 250 milliliters of the 20% acid solution.
A certain solution is 
glycol by volume. How many liters of this solution must be added to 
liters of a 
glycol solution in order to produce a solution with an overall concentration of 
?
A certain solution is
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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:
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:
A chemist wants to make one liter of 
hydrochloric acid solution. He only has two concentrations on hand - 
and 
. If 
is the amount in milliliters of the 
solution he is to use, then which equation can be used to solve for 
?
A chemist wants to make one liter of
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If the chemist is to use 
milliliters of 12% to make the solution, then he will mix it with 
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: 
30% solution: 
15% solution (result): 
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
If the chemist is to use
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:
30% solution:
15% solution (result):
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
A chemist has 
milliliters of a solution of 
alcohol on hand, and he wants to mix it with enough pure alcohol to turn it into a 
alcohol solution. If 
is the amount of pure alcohol he needs, which equation can he use to solve for 
?
A chemist has
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If the chemist adds 300 milliliters of a solution of 25% alcohol to 
milliliters of pure alcohol, he will have 
milliliters of the 40% alcohol 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:
25% solution: 
40% solution (result): 
Since the first solution is being added to 
milliliters of pure alcohol, the following equation can be set up:
If the chemist adds 300 milliliters of a solution of 25% alcohol to
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:
25% solution:
40% solution (result):
Since the first solution is being added to
A chemist has 
milliliters of a solution of 
alcohol on hand, and she wants to mix it with enough 
alcohol solution to turn it into a 
alcohol solution. If 
is the amount of 
alcohol she needs, which equation can she use to solve for 
?
A chemist has
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The chemist will add 400 milliliters of 20% solution to 
milliliters of 60% solution to make 
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: 
60% solution: 
30% solution (result): 
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
The chemist will add 400 milliliters of 20% solution to
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:
60% solution:
30% solution (result):
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
A chemist has 400 milliliters of a solution of 20% alcohol on hand, and he wants to mix it with enough pure alcohol to turn it into a 30% alcohol solution. How much pure alcohol will this require?
Select the response that is closest to the correct answer.
A chemist has 400 milliliters of a solution of 20% alcohol on hand, and he wants to mix it with enough pure alcohol to turn it into a 30% alcohol solution. How much pure alcohol will this require?
Select the response that is closest to the correct answer.
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If the chemist adds 400 milliliters of a solution of 20% alcohol to milliliters of pure alcohol, he will have 
milliliters of the 30% alcohol 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: 
30% solution (result): 
Since the alcohol in the first solution is being added to milliliters of pure alcohol to yield the alcohol in the second, the following equation can be set up:
milliliters.
The closest response is 60 milliliters.
If the chemist adds 400 milliliters of a solution of 20% alcohol to
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:
30% solution (result):
Since the alcohol in the first solution is being added to
The closest response is 60 milliliters.
A chemist has 600 milliliters of a solution of 20% alcohol on hand, and she wants to mix it with enough 50% alcohol solution to turn it into a 30% alcohol solution. How much of the 50% solution will she need?
A chemist has 600 milliliters of a solution of 20% alcohol on hand, and she wants to mix it with enough 50% alcohol solution to turn it into a 30% alcohol solution. How much of the 50% solution will she need?
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The chemist will add 600 milliliters of a solution of 20% alcohol to milliliters of 50% alcohol solution to make 
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: 
50% solution: 
30% solution (result): 
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
milliliters, the correct response.
The chemist will add 600 milliliters of a solution of 20% alcohol to
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:
50% solution:
30% solution (result):
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
A chemist wants to make one liter of 
hydrochloric acid solution. He only has two concentrations on hand - 
and 
. How much of the 
solution will the chemist use?
A chemist wants to make one liter of
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The choices are given in milliliters, so we will convert one liter to 1,000 milliliters.
If the chemist is to use 
milliliters of 10% to make the solution, then he will mix it with 
milliliters of the 50% solution. The result will be 1,000 milliliters of 25% 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:
10% solution: 
50% solution: 
25% solution (result): 
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:
milliliters of the weaker solution.
The choices are given in milliliters, so we will convert one liter to 1,000 milliliters.
If the chemist is to use
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:
10% solution:
50% solution:
25% solution (result):
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:
A nut mixture consists of peanuts, pistachios, and macadamia nuts in the ratio 
, respectively, by weight. What amount of pistachios will be in 40 pounds of the mixture?
A nut mixture consists of peanuts, pistachios, and macadamia nuts in the ratio
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Using the ratio, we can write the following equation:
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
pounds of peanuts
pounds of pistachios
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.
Using the ratio, we can write the following equation:
There are
Since the question asked about pistachios, the correct answer is that there are
A chemist is diluting 
of a pure acid solution with water in order to obtain a solution that is only 
acid. What is the total amount of the resulting solution?
A chemist is diluting
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Let 
be the amount of water used to dilute the acid. The resulting solution will then have 
. The amount of acid in the final solution is: 
. That amount of acid in the final solution is equal to the original 
of acid, since there is no acid in water.
The amount of the final solution is 
, and the amount of water used to dilute the original acid solution is 
.
Let
The amount of the final solution is
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?
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
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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
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 
and 
; add these to get the amount of acid in 1,000 milliliters of a 20% solution, or 
milliliters:
The correct system is:
Since
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
The correct system is:
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.
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.
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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 
liters of solution. the amount of acid in the solution will be 25% of this, or 
.
The amount of acid in the solutions remains constant, so we set up and solve this equation:
The scientist will need to add 0.6 liters of water.
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
The amount of acid in the solutions remains constant, so we set up and solve this equation:
The scientist will need to add 0.6 liters of water.
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?
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?
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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 
liters of solution and 
liters of acid, and the concentration will be
percent. Solve for 
:
The scientist will need to add 0.2 liters of pure acid to the solution.
40% of one liter of the original solution is 0.4 liters of acid.
Suppose the scientist adds
The scientist will need to add 0.2 liters of pure acid to the solution.
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?
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?
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Since a normal batch of candies is one-sixth green candies, then it contains 
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 
pounds of green candies in a batch of 
candies. One sixth of the candies are green, so set and solve the equation:
Six pounds of green candies are added.
Since a normal batch of candies is one-sixth green candies, then it contains
Six pounds of green candies are added.