# Mixtures – removal and replacement questions made easy

## Mixtures – removal and replacement questions made easy

We often get questions involving  removal and replacement of mixtures where a part of a mixture is removed and replaced by some other solution. Say, a vessel contains different constituents in some given ratio. A fraction of the solution is removed and some quantity of some constituent is added to the solution. This process is repeated multiple times and the final ratio of constituents in the mixture has to be determined.

Though this looks very complicated, we can solve it simply, in fact, in a single line!

We just need to understand the fact that if a fraction, f of the solution is removed, the same fraction of each constituent also gets removed, and hence, the fraction (1-f) of each constituent is left, and as a result, after removal, the ratio of the constituents remains unchanged.

This is very easy to understand. Say, you have a sugar solution in a glass and you drink a part of it. The remaining part would taste just as sweet, implying the concentration of sugar in the remaining solution is the same as it was initially. The only time when the ratio of the constituents changes, is when a different solution is added to the existing mixture.

Let us use this and develop a process, as shown below, to solve such questions. We have to follow the steps shown for the same number of times as the process continues.

Say the solution has constituents A and B in some ratio a : b.

1. Note the TOTAL initial volume of the mixture, say V1
2. Determine the volume of constituent A in the mixture = V × [a ÷ (a+b)] = A1
3. Note the volume of the solution removed, say r1
4. Determine what fraction, say f, of the TOTAL volume is removed = (r1 ÷ V1) = f
5. Thus, fraction of the solution remaining = (1 − f)
6. Thus, volume of constituent A remaining = A1 × (1 − f)
7. Let volume of A added after the removal process = v1
8. Thus, final volume of A = A1 × (1 − f) + v1 = A2
9. Determine the TOTAL volume of the mixture = V1 − r1 + v1 = V2

Let us take an example:

A beaker has 80 litres of alcohol and water in the ratio 4 : 1. The following steps are carried out:

• Step 1: 10 liters of the solution is removed and filled with 10 liters water
• Step 2: 20 liters of the solution is removed and filled with 10 liters water
• Step 3: 30 liters of the solution is removed and filled with 20 liters alcohol
• Step 4: 15 liters of the solution is removed and filled with 10 liters water

How much alcohol and water are remaining finally in the beaker?

Solution: Let V be the TOTAL volume, A be alcohol and B be water. Thus, we have:

• Step 1: V1 = 80, A1 = 80 × (4/5) = 64, r1 = 10, f = 10/80 = 1/8: Volume of alcohol left = 64 × (1 – 1/8) = 56 liters. Since 10 liters of water is added, the volume of alcohol doesn’t change, also, the TOTAL volume becomes 80 liters.
• Step 2: V2 = 80, A2 = 56, r2 = 20, f = 20/80 = 1/4: Volume of alcohol left = 56 × (1 – 1/4) = 42 liters. Since 10 liters of water is added, the volume of alcohol doesn’t change, also, the TOTAL volume becomes 80 – 20 + 10 = 70 liters
• Step 3: V3 = 70, A3 = 42, r3 = 30, f = 30/70 = 3/7: Volume of alcohol left = 42 × (1 – 3/7) = 24 liters. Since 20 liters of alcohol is added, the volume of alcohol becomes 24 + 20 = 44 liters, also, the TOTAL volume becomes 70 – 30 + 20 = 60 liters
• Step 4: V4 = 60, A4 = 44, r4 = 15, f = 15/60 = 1/4: Volume of alcohol left = 44 × (1 – 1/4) = 33 liters. Since 10 liters of water is added, the volume of alcohol doesn’t change, also, the TOTAL volume becomes 60 – 15 + 10 = 55 liters. Thus, volume of water = 55 – 33 = 22 liters.

Thus, effectively, to determine the volume of alcohol, we simply did this:

Volume of alcohol = {(80 × 4/5) × (1 – 1/8) × (1 – 1/4) × (1 – 3/7) + 20} × (1 – 1/4)

= {80 × 4/5 × 7/8 × 3/4 × 4/7 + 20} × 3/4 = 33 liters – Just one line of calculation!

Thus, alcohol and water finally are 33 liters and 22 liters, respectively.

Hope the above method helps in making problems on removal and replacements simpler.

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