Test: Mixture Problems - GMAT MCQ

Let the total volume of the solution be 100l
The initial concentration of alcohol is 100%.
Thus, the volume of alcohol is 100l.

After first dilution, the volume of the alcohol removed = 10% of 100l = 100*10/100 = 10l
Thus, the volume of alcohol left = 100 - 10 = 90l or 0.9*100

After the second dilution, volume of the alcohol left = [(100-10)/100]*90l = 0.9*90 = 81l

Similarly, after the third dilution, the volume of alcohol left = 0.9*81 = 72.9l.

Thus, after nth dilution, the volume of alcohol left = (0.9)ⁿ*100

Since the percentage, if alcohol should be below 65% or 65*100/100 = 65l.

(0.9)ⁿ*100 < 65l
(0.9)ⁿ < 0.65

When n = 5, (0.9)⁵ = 0.5905

Thus, after 5 dilutions, the concentration of alcohol will be below 65%.

Thus, the correct answer is C.

Given:

• Solution A
  • Volume of water = x liters (say)
  • So, volume of alcohol = x liters (Volume of alcohol and water is equal on the solution)

Reduced volume of water in solution A = 0.5x liters, as 50% of the water evaporates from the original solution having x litres of water

• Solution B
  • Volume of solution B = 0.5x
    • (Volume of solution B is equal to reduced volume of water in solution A, as calculated above)
  • Also,(Volume of alcohol): (Volume of water) = 2: 3
• Solution A + Solution B (also called as ‘Resultant Solution’)
  • Volume of alcohol = 12 liters

To Find: Amount of water to be added to the resultant solution to increase the concentration of water to 50%?

• Let the amount of water to be added to the resultant solution be y liters.

Approach:

1. We know the volume of water and alchohol in the Solution A in terms of x. We will find the volume of water and alcohol in the Solution B in terms of x, by using the ratio of volume of alcohol to volume of water i.e. 2:3.
2. After knowing the volume of water and volume of alcohol in terms of x in both Solution A and Solution B, we can find the volume of water and alcohol in the resulting solution in terms of x, by adding the constituents of the Solutions A and B.
3. Knowing the volume of alcohol in the resulting solution in terms of x, we will be able to find the value of x, by equating volume of alcohol in resulting solution to 12 liters.
4. After knowing the value of x, we will be able to get the values of volume of alcohol and water in the resulting solution.
5. Now to make the concentration of alcohol to be 50%, we are adding y liters of water. We can finally calculate the value of y, by using the relation

• New Volume of water in solution after adding water= Volume of water in resultant solution + y
• Total volume of solution after adding water=Volume of resultant solution+y

Working out:

1. Volume of solution B = 0.5x
   • Volume of alcohol:Volume of water =2:3,

2. Volume of Alcohol  in the resultant solution = Volume  of alcohol in solution A + volume of Alcohol in solution B

• Volume of alcohol in resultant solution = x + 0.2x = 1.2x (See  diagram above)
• Now we know that volume of alcohol in the resultant solution = 12  liters
• Thus, 1.2x = 12 , x = 10 liters

3. Volume of Water  in the resultant solution = Volume  of Water  in solution A + volume of Water in solution B

•  Volume of water in resultant solution = 0.5x + 0.3x = 0.8x (See  diagram above)
• So, Volume of water in the resultant solution =0.8*10=8liters (As x=10, already calculated above)

4. Now the amount of water to be added is y liters

• So, New Volume of water in solution after adding water = 8 + y
• Total volume of solution after adding water= 12 + (8+y) = 20 + y
• Now we know the concentration of water is to be made 50%, so,

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Let the average runs of the team before adding E be x.

Total runs scored by A and B = 13x + 12x = 25x
Total runs scored by C and D = 2x (since average runs decreases by 2 when E is added)

Total runs scored by the team before adding E = 25x + 2x = 27x

After adding E, the average runs of the team decreases by 2. So, the new average runs = x - 2.

Total runs scored by the team after adding E = (x - 2) * 5

Since E scored 45 runs, we can write the equation:

(x - 2) * 5 = 45
x - 2 = 9
x = 11

So, the average runs before adding E = 11.

Total runs scored by the team before adding E = 27x = 27 * 11 = 297

Now, let's find the individual runs scored by each player.
Since no player scored less than E or more than 65 runs, we can conclude that A scored 65 runs and B scored 65 - 13 = 52 runs.

Let's assume that C scored y runs. Since C scored more than A, y > 65. Also, since y is a natural number, the minimum value of y is 66.

Total runs scored by the team = 65 + 52 + y + 2x = 297
117 + y + 22 = 297
y = 158

So, C scored 158 runs and D scored 297 - 65 - 52 - 158 = 22 runs.

The ratio of the runs scored by B to the average runs scored by C & D = 52 : ((158 + 22)/2) = 52 : 90 = 4 : 5

Therefore, the correct answer is A: 4:5.

Step 1: 10 liters of water is added to the 100 liters of milk, resulting in a total volume of 100 + 10 = 110 liters. The amount of milk remains unchanged.

Step 2: 20 liters of solution is removed. Since the solution consists of both milk and water, we can consider this operation as removing a proportional amount of milk and water. The ratio of milk to the total volume is 100/110, so the amount of milk removed is (20 * 100) / 110 = 2000 / 11 liters.

After step 2, the remaining amount of milk is 100 - (2000 / 11) = (11000 - 2000) / 11 = 9000 / 11 liters.

Step 3: 30 liters of water is added to the remaining solution. This means the total volume becomes 110 + 30 = 140 liters. The amount of milk remains unchanged.

Step 4: 20 liters of solution is removed. Similar to step 2, we remove a proportional amount of milk and water. The ratio of milk to the total volume is (9000 / 11) / 140, so the amount of milk removed is (20 * 9000) / (11 * 140) = 180000 / (1540) liters.

After step 4, the remaining amount of milk is (9000 / 11) - (180000 / 1540) = (693000 - 180000) / 1540 = 513000 / 1540 liters.

Simplifying the fraction, we have 513000 / 1540 = (3 * 171000) / (2 * 770) = 513000 / 1540 = 750 / 11 liters.

Therefore, the amount of milk in the solution now is 750 / 11 liters, which corresponds to option D: 750/11 liters.

Given that a jar contains a mixture of 175 ml water and 700 ml alcohol.
It is given that 10% of the mixture is removed and it is substituted by water of the same amount and the process is repeated once again
Now we have to find the percentage of water in the mixture.
Since the mixture is removed and substituted with water, we can deal with alcohol and the second step we can find how much amount of alcohol is retained and not about how much amount of alcohol is removed
As 10% of alcohol is removed, 90% of alcohol is retained
So alcohol remaining = 700 × 90% × 90%
⟹ 700 × 0.9 × 0.9 = 567
We totally have 875 ml overall mixture and of this 567 ml is alcohol.
Remaining 875 – 567 = 308 is the amount of water.
We have to find the percentage of water in the mixture i.e. 308/875

Approximately 308 is 30% of 1000 so by this we know that 308 is more than 30%
Hence 35.2% is the percentage of water in the given mixture.

The question is "A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now"

Hence, the answer is 35.2%
Choice D is the correct answer.

After the replacement, the concentration of the combined solution becomes 35%. We can use the concept of weighted averages to solve this problem.

The initial solution of 40% concentration is being replaced with a 25% concentration solution. The resulting concentration of the combined solution is 35%. This means that the 40% solution is being diluted.

To find the ratio of the amounts, we can set up the following equation based on the concentrations and quantities:

(40% * x) + (25% * y) = 35% * (x + y)

where "y" represents the amount of the 25% concentration solution that was added.

Simplifying the equation:

(0.4x) + (0.25y) = 0.35x + 0.35y

0.25y - 0.35y = 0.35x - 0.4x

-0.1y = -0.05x

y/x = 0.05/0.1

y/x = 1/2

From the equation, we can see that the ratio of the amount of solution that was replaced (y) to the amount of solution that was not replaced (x) is 1:2.

Therefore, the correct answer is B: 1:2.

We can establish the following ratios for two vessels:

Vessel 1: water to milk ratio is x to 2x.

Vessel 2: water to milk ratio is 2y to 5y.

We can set up the equation:

(x + 2x)/(2y + 5y) = 1/4

Simplifying, we get:

3x/7y = 1/4

Multiplying both sides by 12, we have:

12x = 7y

By choosing x = 7 and y = 12, we find:

Vessel 1: water = 7 and milk = 14

Vessel 2: water = 24 and milk = 60

Consequently, in the final mixture, the ratio of water to milk is:

(7 + 24)/(14 + 60) = 31/74

Given that the answer choices are expressed as percentages, we can manipulate the numbers to obtain our solution. Let's assume that the milkman purchased 100 liters of milk for $100 (his cost price). Now, we are aware that the quantity of water in the mixture is 74% of the quantity of milk. Therefore, out of the 100 liters of milk, the milkman added 74 liters of water. Consequently, the total mixture amounts to 100 + 74 = 174 liters.

Ideally, the milkman should sell this 174-liter mixture for $174. However, he provides a discount of 16.66%. Thus, the selling price (SP) is calculated as follows: SP = 174 - (16.66/100 × 174) = 174 - (1/6 × 174) = 174 - 29 = 145.

Consequently, the milkman purchased the milk for $100 and sold it for $145, resulting in a profit of 145 - 100 = 45. The profit percentage is determined by 45/100 × 100 = 45%.