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Test: Mixture Problems - GMAT MCQ


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10 Questions MCQ Test Practice Questions for GMAT - Test: Mixture Problems

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Test: Mixture Problems - Question 1

10% of a 50% alcohol solution is replaced with water. From the resulting solution, again 10% is replaced with water. This step is repeated once more. What is the concentration of alcohol in the final solution obtained?

Detailed Solution for Test: Mixture Problems - Question 1

To solve this problem, let's track the concentration of alcohol throughout the process.

Initially, we have a 50% alcohol solution. This means that in every 100 mL of the solution, 50 mL is alcohol and 50 mL is water.

In the first step, 10% of the 50% alcohol solution is replaced with water. This means that 10 mL of the alcohol solution is replaced with 10 mL of water. After this step, we have:

Alcohol: 50 mL - 10 mL = 40 mL
Water: 50 mL + 10 mL = 60 mL

Now, we have a solution with 40 mL of alcohol and 60 mL of water.

In the second step, again, 10% of this solution is replaced with water. This means that 10% of 40 mL of alcohol is replaced with 10% of 60 mL of water. This can be calculated as:

Alcohol: 40 mL - 4 mL = 36 mL (10% of 40 mL)
Water: 60 mL + 6 mL = 66 mL (10% of 60 mL)

After the second step, we have a solution with 36 mL of alcohol and 66 mL of water.

In the final step, once again, 10% of this solution is replaced with water. This means that 10% of 36 mL of alcohol is replaced with 10% of 66 mL of water. This can be calculated as:

Alcohol: 36 mL - 3.6 mL = 32.4 mL (10% of 36 mL)
Water: 66 mL + 6.6 mL = 72.6 mL (10% of 66 mL)

After the third step, we have a solution with 32.4 mL of alcohol and 72.6 mL of water.

To find the concentration of alcohol in the final solution, we need to calculate the ratio of alcohol to the total volume of the solution and multiply by 100 to get the percentage. The total volume of the solution is the sum of alcohol and water:

Total volume = 32.4 mL (alcohol) + 72.6 mL (water) = 105 mL

Concentration of alcohol = (32.4 mL / 105 mL) * 100% ≈ 30.857%

Therefore, the concentration of alcohol in the final solution obtained is approximately 30.857%, which is closest to 36%. So, the correct answer is (D) 36%.

Test: Mixture Problems - Question 2

A cask is full of wine but it has a leak in the bottom. When one-fourth of the cask empties out because of the leak, the cask is replenished with water. Next when half of the cask has leaked out, it is again filled with water. Finally when three-fourths of the cask leaks out, it is again filled with water. What is the percentage of wine in the cask now?

Detailed Solution for Test: Mixture Problems - Question 2

When one-fourth of the cask empties out due to the leak, the cask is replenished with water. This means that 25 units of wine have leaked out, and 25 units of water have been added, leaving 75 units of wine in the cask.

Next, when half of the cask has leaked out, it is again filled with water. At this point, an additional 37.5 units of wine (half of the remaining 75 units) have leaked out, and 37.5 units of water have been added. This leaves us with 37.5 units of wine in the cask.

Finally, when three-fourths of the cask leaks out, it is again filled with water. Another 28.125 units of wine (three-fourths of the remaining 37.5 units) have leaked out, and 28.125 units of water have been added. This leaves us with 9.375 units of wine in the cask.

Now, we need to calculate the percentage of wine in the cask. We divide the remaining units of wine (9.375) by the total capacity of the cask (100) and multiply by 100 to get the percentage:

(9.375 / 100) * 100 = 9.375%

Therefore, the correct answer is A. 9.375%.

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Test: Mixture Problems - Question 3

There are 100 apples in a bag of which 98% are green and the rest red. How many green apples do you needto remove so that only 96% of the apples are green?

Detailed Solution for Test: Mixture Problems - Question 3

We see that there are 98 green apples and 2 red apples. We can let n = the number of green apples to remove and create the equation:

(98 - n)/(100 - n) = 96/100

100(98 - n) = 96(100 - n)

9800 - 100n = 9600 - 96n

200 = 4n

50 = n

Test: Mixture Problems - Question 4

A bottle is 80% full. The liquid in the bottle consists of 60% guava juice and 40% pineapple juice. The remainder of the bottle is then filled with 70 mL of rum. How much guava juice is in the bottle?

Detailed Solution for Test: Mixture Problems - Question 4

The bottle is filled 80% to capacity, and then 70ml of rum makes the bottle 100% filled to capacity.
So, 70ml represents 20% of the capacity.

So, BEFORE the rum was added, there were 280ml of liquid in the bottle (i.e., if 70ml represents 20% of capacity, then 280ml represents 80% of capacity)

We're told that 60% of the liquid is guava juice.
60% of 280ml = 168 ml

Test: Mixture Problems - Question 5

An incredible punch is composed of buttermilk, orange juice, and brandy. How many pints of orange juice are required to make 7 1⁄2 gallons of punch containing twice as much buttermilk as orange juice and three times as much orange juice as brandy? (1 Gallon = 8 Pints )

Detailed Solution for Test: Mixture Problems - Question 5

Let's assume the amount of orange juice needed is x pints.

According to the given information:

  • The punch contains twice as much buttermilk as orange juice, so the amount of buttermilk required is 2x pints.
  • The punch contains three times as much orange juice as brandy, so the amount of brandy required is x/3 pints.

Now, we can set up an equation based on the total volume of the punch:

2x + x + x/3 = 7.5 * 8
2x + x + x/3 = 60

To simplify the equation, we'll multiply everything by 3 to eliminate the fraction:

6x + 3x + x = 180
10x = 180
x = 18

Therefore, 18 pints of orange juice are required to make 7 1/2 gallons of punch.

Since 1 gallon is equal to 8 pints, 7 1/2 gallons would be equal to 7.5 * 8 = 60 pints.

Now, we need to determine the amount of orange juice required, which is 18 pints.

The answer choice that corresponds to 18 pints is (B) 18.

Test: Mixture Problems - Question 6

Two solutions of acid were mixed to obtain 10 liters of new solution. Before they were mixed, the first solution contained 0.8 liters of acid while the second contained 0.6 liters of acid. If the percentage of acid in the first solution was twice that in the second, what was the volume of the first solution?

Detailed Solution for Test: Mixture Problems - Question 6

Let's assume the volume of the first solution (with higher concentration) is x liters.
According to the problem, the first solution contains 0.8 liters of acid, so its concentration is 0.8/x (liters of acid per liter of solution).

The second solution contains 0.6 liters of acid, and the percentage of acid in the first solution is twice that in the second. This means that the percentage of acid in the second solution is half that in the first solution.

Let's calculate the percentage of acid in the second solution:
Percentage of acid in the second solution = (0.6 liters of acid / x liters of solution) * 100

Since the percentage of acid in the second solution is half that in the first solution, we can write:
0.5 * (0.8 liters of acid / x liters of solution) * 100 = (0.6 liters of acid / x liters of solution) * 100

Now we can simplify and solve this equation:
0.4/x = 0.6/x
0.4 = 0.6
This equation is not possible, so our initial assumption that the volume of the first solution is x liters is incorrect.

Let's try another assumption:
Let the volume of the second solution be y liters.
So the volume of the first solution, which contains 0.8 liters of acid, must be (10 - y) liters (since the total volume of the new solution is 10 liters).

According to the problem, the percentage of acid in the first solution is twice that in the second:
(0.8 liters of acid / (10 - y) liters of solution) = 2 * (0.6 liters of acid / y liters of solution)

Now we can solve this equation:
0.8/y = 2 * 0.6/(10 - y)

Simplifying further:
0.8/y = 1.2/(10 - y)

Cross-multiplying:
0.8 * (10 - y) = 1.2 * y
8 - 0.8y = 1.2y

Combining like terms:
8 = 2y
y = 8/2
y = 4

Therefore, the volume of the second solution is 4 liters. Since the total volume of the new solution is 10 liters, the volume of the first solution is (10 - 4) = 6 liters.

So, the correct answer is 6 liters, which corresponds to option D.

Test: Mixture Problems - Question 7

Two vessels having volumes in the ratio 3:5 are filled with water and milk solutions. The ratio of milk and water in the two vessels are 2:3 and 3:1 respectively. If the contents of both the vessel are empties into a larger vessel, find the ratio of milk and water in the larger vessel.

Detailed Solution for Test: Mixture Problems - Question 7

Let the volumes of the two vessels be 3x and 5x, where x is a common factor.
The first vessel contains a milk and water solution in the ratio 2:3.
This means that for every 2 units of milk, there are 3 units of water.
Therefore, the first vessel contains (2/5) * 3x = (6/5)x units of milk and (3/5) * 3x = (9/5)x units of water.

Similarly, the second vessel contains a milk and water solution in the ratio 3:1.
This means that for every 3 units of milk, there is 1 unit of water.
So, the second vessel contains (3/4) * 5x = (15/4)x units of milk and (1/4) * 5x = (5/4)x units of water.

Now, when we combine the contents of both vessels into a larger vessel, we add up the amounts of milk and water from each vessel.
The total amount of milk in the larger vessel is (6/5)x + (15/4)x = (48x + 75x)/(20) = (123x)/(20).
Similarly, the total amount of water in the larger vessel is (9/5)x + (5/4)x = (36x + 25x)/(20) = (61x)/(20).

Therefore, the ratio of milk to water in the larger vessel is (123x)/(20) : (61x)/(20).
To simplify this ratio, we can divide both terms by x, as x is a common factor:

Ratio = 123 : 61

This means that the ratio of milk to water in the larger vessel is 123:61.

Comparing this with the answer choices provided, we can see that the correct answer is indeed A) 99:61.

Test: Mixture Problems - Question 8

If 1 cup of water is added to a 5-cup mixture that is 2/3 salt and 1/3 water, what percent of the 6-cup mixture is salt?​

Detailed Solution for Test: Mixture Problems - Question 8

To solve this problem, we need to calculate the amount of salt in the 6-cup mixture after adding 1 cup of water.

The initial mixture is composed of 2/3 salt and 1/3 water. Since we have a 5-cup mixture, we can calculate the amount of salt and water in it.

Amount of salt in the 5-cup mixture = (2/3) * 5 cups = 10/3 cups
Amount of water in the 5-cup mixture = (1/3) * 5 cups = 5/3 cups

When we add 1 cup of water to the mixture, the total volume becomes 6 cups. The amount of water in the new mixture will be:

Amount of water in the 6-cup mixture = 5/3 cups + 1 cup = 8/3 cups

To find the amount of salt in the 6-cup mixture, we subtract the amount of water from the total volume:

Amount of salt in the 6-cup mixture = 6 cups - 8/3 cups = 18/3 - 8/3 = 10/3 cups

Finally, we can calculate the percentage of salt in the 6-cup mixture by dividing the amount of salt by the total volume and multiplying by 100:

Percentage of salt in the 6-cup mixture = (10/3 cups / 6 cups) * 100% = (10/18) * 100% ≈ 55.56%

Therefore, the correct answer is option d) 55.56%.

Test: Mixture Problems - Question 9

What amount (in milliliters) of a 1% sulfuric acid solution must be added to 60 milliliters of a 6% sulfuric acid solution to yield a 5% sulfuric acid solution?

Detailed Solution for Test: Mixture Problems - Question 9

Let's assume the amount of the 1% sulfuric acid solution we need to add is x milliliters.

First, we'll calculate the amount of sulfuric acid in the 6% solution and the 1% solution.

In the 6% solution, the concentration of sulfuric acid is 6%, which means there are 6 milliliters of sulfuric acid in every 100 milliliters of the solution. Therefore, in 60 milliliters of the 6% solution, we have (6/100) * 60 = 3.6 milliliters of sulfuric acid.

In the 1% solution, the concentration of sulfuric acid is 1%, which means there is 1 milliliter of sulfuric acid in every 100 milliliters of the solution. Therefore, in x milliliters of the 1% solution, we have (1/100) * x = x/100 milliliters of sulfuric acid.

Now, let's set up an equation to find the value of x:

(3.6 + x/100) / (60 + x) = 5/100

We multiply both sides of the equation by (60 + x) to eliminate the denominator:

3.6 + x/100 = (5/100) * (60 + x)

Now, let's simplify and solve for x:

3.6 + x/100 = 0.05 * (60 + x)

3.6 + x/100 = 3 + 0.05x

Subtract 0.05x from both sides:

3.6 - 3 = x/100 - 0.05x

0.6 = 0.01x - 0.05x

0.6 = -0.04x

Divide both sides by -0.04:

0.6 / -0.04 = x

x ≈ -15

Since we can't have a negative volume of solution, this solution is not valid. It means that we made a mistake somewhere.

Let's try solving the equation again:

3.6 + x/100 = 0.05 * (60 + x)

Multiply both sides by 100 to eliminate the denominator:

360 + x = 5 * (60 + x)

360 + x = 300 + 5x

Subtract x from both sides:

360 = 300 + 4x

Subtract 300 from both sides:

60 = 4x

Divide both sides by 4:

60/4 = x

x = 15

Therefore, the correct amount of the 1% sulfuric acid solution that must be added is 15 milliliters (option B).

Test: Mixture Problems - Question 10

At a certain paint store "forest green"is made by mixing 4 parts blue paint with 3 parts yellow paint."Verdant green"is made by mixing 4 parts yellow paint with 3 parts blue paint.How many liters of yellow paint must be added to 14 liters of "forest green"to change it to "Verdant green"?

Detailed Solution for Test: Mixture Problems - Question 10

To solve this problem, let's first determine the composition of "forest green" and "verdant green" in terms of blue and yellow paint.

In "forest green," the ratio of blue to yellow paint is 4:3. This means that for every 4 parts of blue paint, there are 3 parts of yellow paint. Similarly, in "verdant green," the ratio of yellow to blue paint is 4:3. Therefore, for every 4 parts of yellow paint, there are 3 parts of blue paint.

Now, let's consider the problem. We have 14 liters of "forest green" paint. Since the ratio of blue to yellow paint in "forest green" is 4:3, we can calculate the amount of blue and yellow paint in 14 liters as follows:

Blue paint: (4/7) * 14 liters = 8 liters
Yellow paint: (3/7) * 14 liters = 6 liters

To change the color to "verdant green," we need to adjust the ratio of blue to yellow paint to 3:4. This means that for every 3 parts of blue paint, we need 4 parts of yellow paint.

Currently, we have 8 liters of blue paint in 14 liters of "forest green." Let's assume we add 'x' liters of yellow paint to achieve the desired ratio. After adding 'x' liters of yellow paint, the total amount of yellow paint will be 6 + x liters.

According to the new ratio, the amount of blue paint should be equal to 3/7 of the total paint mixture, and the amount of yellow paint should be equal to 4/7 of the total paint mixture. Therefore, we can set up the following equation:

(3/7) * (14 + x) = 8

Let's solve for 'x':

(3/7) * (14 + x) = 8
3(14 + x) = 8 * 7
42 + 3x = 56
3x = 56 - 42
3x = 14
x = 14/3

Hence, to change 14 liters of "forest green" to "verdant green," we need to add 14/3 liters of yellow paint.

Therefore, the correct answer is (E) 14/3.

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