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Test: Mixture Problems - GMAT MCQ


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10 Questions MCQ Test - Test: Mixture Problems

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Test: Mixture Problems - Question 1

How many quarts of lemonade concentrate should be added to 10 quarts of a 12-percent concentrate and water mixture to create a 20-percent concentrate and water mixture?

Detailed Solution for Test: Mixture Problems - Question 1

Let's assume that x quarts of lemonade concentrate need to be added.

The amount of lemonade concentrate in the 10 quarts of 12% concentrate and water mixture is (10 * 12%) = 1.2 quarts.

After adding x quarts of lemonade concentrate, the total amount of concentrate in the mixture will be (1.2 + x) quarts.

The total amount of the mixture will be (10 + x) quarts.

To create a 20% concentrate and water mixture, we can set up the following equation:

(1.2 + x) / (10 + x) = 20%

To solve this equation, we can convert the percentages to decimals:

(1.2 + x) / (10 + x) = 0.20

Cross-multiplying, we get:

0.20(10 + x) = 1.2 + x

2 + 0.20x = 1.2 + x

0.20x - x = 1.2 - 2

-0.80x = -0.8

x = -0.8 / -0.80

x = 1

Therefore, 1 quart of lemonade concentrate should be added to the mixture.

The correct answer is B: 1.

Test: Mixture Problems - Question 2

Jim needs to mix a solution in the following ratio: 1 part bleach for every 4 parts water. When mixing the solution, Jim makes a mistake and mixes in half as much bleach as he ought to have. The total solution consists of 18 mL. How much did Jim put into the solution?

Detailed Solution for Test: Mixture Problems - Question 2


According to the given ratio, the correct mixture should contain 1 part bleach for every 4 parts water. However, Jim made a mistake and mixed in half as much bleach as he should have.

Let's assume Jim put x mL of bleach into the solution.

Since the ratio is 1 part bleach to 4 parts water, the amount of water in the mixture will be 4x mL.

The total solution consists of 18 mL, so the sum of the bleach and water must equal 18 mL:

x + 4x = 18

5x = 18

x = 18 / 5

x = 3.6

However, since Jim made a mistake and mixed in half as much bleach, we need to find half of x:

0.5 * 3.6 = 1.8

Jim put 1.8 mL of bleach into the solution.

The correct answer is B: 2 mL (rounded to the nearest whole number).

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Test: Mixture Problems - Question 3

In calvin’s chemistry class, her final grade consists of lab work and the final exam. If lab work is 2/3 of the final grade, and she scores 80%, what must she score on the final exam to earn a final grade of 75%?

Test: Mixture Problems - Question 4

A 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2. 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more. At the end of the two removal and replacement, what is the ratio of milk and water in the resultant mixture?

Test: Mixture Problems - Question 5

Solution X, which is 50% alcohol, is combined with solution Y, which is 30% alcohol, to form 16 liters of a new solution that is 35% alcohol. How much of solution Y is used?

Detailed Solution for Test: Mixture Problems - Question 5

Let's assume the amount of solution X used is x liters. Since the total volume of the new solution is 16 liters, the amount of solution Y used would be (16 - x) liters.

To find the amount of alcohol in the final solution, we need to calculate the total amount of alcohol from each solution.

Amount of alcohol in solution X = 50% of x liters = 0.5x liters
Amount of alcohol in solution Y = 30% of (16 - x) liters = 0.3(16 - x) liters

Since we are creating a 35% alcohol solution, the total amount of alcohol in the final solution should be 35% of 16 liters = 0.35 * 16 liters = 5.6 liters.

Setting up the equation for the total amount of alcohol in the final solution:

0.5x + 0.3(16 - x) = 5.6

0.5x + 4.8 - 0.3x = 5.6

0.2x = 0.8

x = 0.8 / 0.2

x = 4

Therefore, the amount of solution X used is 4 liters, and the amount of solution Y used is (16 - 4) = 12 liters.

The correct answer is E: 12 liters.

Test: Mixture Problems - Question 6

Salad dressing A is made up of 30% vinegar and 70% oil, and salad dressing B contains 10% vinegar and 90% oil. If the 2 dressing are combined to produce a salad dressing that is 15% vinegar, dressing A comprises what % of the new dressing?

Detailed Solution for Test: Mixture Problems - Question 6

Let's assume that the quantity of dressing A used is x and the quantity of dressing B used is y to create the new dressing.

Since we are combining dressing A and dressing B to produce a dressing with 15% vinegar, we can set up the following equation based on the vinegar content:

(0.30 * x + 0.10 * y) / (x + y) = 0.15

Simplifying the equation:

0.30x + 0.10y = 0.15(x + y)

0.30x + 0.10y = 0.15x + 0.15y

0.30x - 0.15x = 0.15y - 0.10y

0.15x = 0.05y

3x = y

This means that the quantity of dressing B used (y) is three times the quantity of dressing A used (x).

To find the percentage of dressing A in the new dressing, we need to calculate the ratio of dressing A to the total dressing (A + B).

Percentage of dressing A = (x / (x + y)) * 100

Substituting y = 3x:

Percentage of dressing A = (x / (x + 3x)) * 100

Percentage of dressing A = (x / 4x) * 100

Percentage of dressing A = (1/4) * 100

Percentage of dressing A = 25%

Therefore, dressing A comprises 25% of the new dressing.

The correct answer is C: 25%.

Test: Mixture Problems - Question 7

The ratio of alcohol and water in three mixtures of alcohol and water is 3:2, 4:1, and 7:3. If equal quantities of the mixture are drawn and mixed, the concentration of alcohol in the resulting mixture will be?

Detailed Solution for Test: Mixture Problems - Question 7

To find the concentration of alcohol in the resulting mixture, we need to consider the ratios and quantities of alcohol and water in each of the initial mixtures.

Let's assume we have equal quantities of the three mixtures: Mixture 1, Mixture 2, and Mixture 3.

The ratio of alcohol to water in Mixture 1 is 3:2, meaning for every 3 parts of alcohol, there are 2 parts of water.

The ratio of alcohol to water in Mixture 2 is 4:1, meaning for every 4 parts of alcohol, there is 1 part of water.

The ratio of alcohol to water in Mixture 3 is 7:3, meaning for every 7 parts of alcohol, there are 3 parts of water.

To find the resulting concentration of alcohol in the mixture, we need to calculate the total amount of alcohol and water.

Total alcohol = (3 + 4 + 7) parts
Total water = (2 + 1 + 3) parts

The resulting concentration of alcohol in the mixture can be calculated as:
Concentration of alcohol = (Total alcohol) / (Total alcohol + Total water)

Concentration of alcohol = (3 + 4 + 7) / ((3 + 4 + 7) + (2 + 1 + 3))

Concentration of alcohol = 14 / 20

Concentration of alcohol = 0.7 or 70%

Therefore, the resulting concentration of alcohol in the mixture is 70%.

The correct answer is B: 70%.

Test: Mixture Problems - Question 8

To dilute 300 quarts of a 25% solution of garlic to a 20% solution, how much water should a chef add?

Detailed Solution for Test: Mixture Problems - Question 8

To dilute a 25% solution of garlic to a 20% solution, we need to add water to decrease the concentration of garlic.

Let's calculate the amount of garlic in the 25% solution:

Amount of garlic in the solution = 25% of 300 quarts = 0.25 * 300 = 75 quarts

To dilute the solution to 20%, we need to calculate the amount of garlic we want in the final solution:

Amount of garlic in the final solution = 20% of (300 + x) quarts

where x represents the amount of water added.

Since the amount of garlic should remain the same, we can set up the following equation:

75 quarts = 20% of (300 + x) quarts

Converting percentages to decimals:

0.20(300 + x) = 75

60 + 0.20x = 75

0.20x = 75 - 60

0.20x = 15

x = 15 / 0.20

x = 75

Therefore, a chef should add 75 quarts of water to dilute the 25% garlic solution to a 20% solution.

The correct answer is C: 75.

Test: Mixture Problems - Question 9

A and B are two alloys of gold and copper prepared by mixing metals in proportions 5: 3 and 5: 11 respectively. If equal quantities of this alloys are melted to a third alloy C, what would be the proportion of gold and copper in the alloys thus formed?

Detailed Solution for Test: Mixture Problems - Question 9

To find the proportion of gold and copper in the third alloy C formed by melting equal quantities of alloys A and B, we need to consider the individual proportions of gold and copper in each alloy.

Let's calculate the proportion of gold and copper in alloy A:
Gold: 5 parts out of (5 + 3) = 5/8
Copper: 3 parts out of (5 + 3) = 3/8

Similarly, let's calculate the proportion of gold and copper in alloy B:
Gold: 5 parts out of (5 + 11) = 5/16
Copper: 11 parts out of (5 + 11) = 11/16

Since we are melting equal quantities of alloys A and B, the proportions will be added together:

Gold in alloy C = (5/8) + (5/16) = 10/16 + 5/16 = 15/16
Copper in alloy C = (3/8) + (11/16) = 6/16 + 11/16 = 17/16

Therefore, the proportion of gold to copper in alloy C is 15:17.

The correct answer is C: 15:17.

Test: Mixture Problems - Question 10

A Food and Drug lab has two new samples: a 240 gram cup of drip coffee, which contains 124 mg of caffeine, and a 60 gram cup of espresso, containing 160 mg of caffeine. If a technician were to create a new 120 gram cup sample that contained 50% coffee and 50% espresso, how many mg of caffeine would the new drink contain?

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