A milkman has 100 liters solution of milk and water such that the conc

Question

A milkman has 100 liters solution of milk and water such that the concentration of milk is 20%. He mixes another 200 liters of milk and water solution into it such that final concentration of milk is X%. Which one of the following options can be the value of X?

A. 5%
B. 6%
C. 45%
D. 75%
E. 80%

A. 5%
B. 6%
C. 45%
D. 75%
E. 80%

CareerGeek · Accepted Answer

In 100 liters
Milk = 20% = 20 liters and water = 80 liters

When 200 liters are mixed, milk = a% (assume)
Case1: Min value of a possible = 0
Case2: Maximum value of a possible = 200

Final concentration of milk for case1
= (20 + 0)/300*100 = 6.66%

Final concentration of milk for case2
= (20 + 200)/300*100 = 220/3 = 73.3%

So, concentration should be within 6.66% and 73.3%

Only option C 45% is possible

IMO Option C

Pls Hit kudos if you like the solution

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Fdambro294 · Answer

Great question.

We’ll call the 20% Solution = Solution A

And the unknown % Solution = Solution B

Since 100 ml of Solution A and 200 ml of Solution B are mixed, the final mixture will be more heavily weighted towards the Solution B’s Unknown Concentration%

(A) and (B)

We’ll try Mixture - Weighted Average Percentage = 6%

A=20%  B =?

Ratio of Relative Weighting (I.e., the amount of each solution used) —— 1 : 2

2 parts of B are used for every 1 part of A

This means the distance on the number line from “A” to “Mix” must = 2x parts

The actual distance = 20% - 6% = 14%

2x = 14 ————> X = 7

The distance from: “Mix=6%” to “B =?” must be 1x Ratio parts

1x ——-when X = 7 ———- (-7%)

If the Mix = 6%, the concentration of Solution B would have to be:

6% - 7% = (-1)%

Which is not possible

Similarly, (A) will not be possible

For (D) and (E) following the number line approach, because the Weighting of the Solutions used in the Mix/Weighted Average is 2 : 1 in favor of the unknown Solution, the only way the Mix could equal either percentage is if Solution B’s Unknown concentration exceeded > 100%, which is not possible

(c) 45%

A=20% ———]B = ?

because of the relative weightings, the Distance on the number line between “A” and “Mix” = 2x

2x = (45)% - (20)%

x = 12.5%

Further, the distance from “Mix” to “B = ?” must equal 1x Ratio parts where X = 12.5

Mix of 45% + 12.5% ——-> Solution B would have a Concentration of = 57.5%

Given:

A -20% ——100 liters 
B-57.5%———200 liters
________________
Mix-45%———— 300 liters

(C) 45% is possible

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ScottTargetTestPrep · Answer

Solution:

The 100 liters of solution are composed of 20% milk, or 20 liters. Similarly, the amount of water in the solution is 80%, or 80 liters.

If we add 200 liters of a second milk/water solution, let’s calculate the minimum and maximum percentages of milk in the resulting 300-liter solution.

Case 1. The second 200-liter solution is all milk.  In this case, we will have 20 liters of milk from the original solution and 200 liters of milk from the second solution. Thus, we will have 220 liters of milk in a 300-liter solution, which makes the 300-liter solution 220/300 = 73.3% milk.

Case 2. The second 200-liter solution has no milk.   In this case, we will have 20 liters of milk from the original solution and 0 liters of milk from the second solution. Thus, we will have 20 liters of milk in a 300-liter solution, which makes the 300-liter solution 20/300 = 6.7% milk.

Thus, the 300-liter solution can contain any percentage of milk between 6.7% and 73.3%. Of the answer choices, only choice C (45%) is in this interval.

Answer: C

GmatPoint · Answer

To find the solution to such questions, you need to find out the range of the possible values.
Initial volume of solution = 100l
Initial concentration = 20%
Thus, initial milk = (20/100)*100 = 20l

Let us consider that the milkman added 200l of water(or 0% milk concentration)
Total volume of solution = 100 + 200 = 300l
The volume of milk = 20l
Thus, final concentration = (20/300)*100 = 6.66%

Let us consider that the milkman added 200l of milk(or 100% milk concentration)
Total volume of solution = 100 + 200 = 300l
Volume of milk = 20 + 200 = 220l
Thus, final concentration = (220/300)*100 = 73.33%

Now, the value of X will be between 6.66% and 73.33%.

Since only option C is in the given range, the correct option is C.

bumpbot · Answer

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