To 100 litres of milk, 10 litres of water is added and then 20 litres

Water is added so we work with the amount of milk. Our iteration begins when we "REMOVE".

We have 100 lts of milk and 10 lts of water so concentration of milk = 10/11 and volume of mix is 110 lts.
We need the final concentration of milk Cf

Cf = Ci *(Vi/Vf)^n

Cf = (10/11) * (90/120) = 30/44 (20 lts is removed and 30 lts added so Vi = 110 - 20 = 90 and Vf = 90 + 30 = 120)

This is the concentration of milk in 120 lts of solution. Now 20 lts is removed with doesn't change the concentration of milk.

Volume of solution after removing 20 lts = 120 - 20 = 100 lts
Amount of milk in it = 100*(30/44) = 750/11 litres

Answer (D)
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After 10 liters of water are added to 100 liters of milk, we have a 110-liter solution with the milk-to-water ratio being 10 : 1. After 20 liters of solution are removed, there are 90 liters left. Since the milk to water ratio at this point is still 10 : 1, we can set up the equation:

10x + x = 90

11x = 90

x = 90/11

Therefore, there are 900/11 liters of milk and 90/11 liters of water in the 90-liter solution.

Now, after 30 liters of water are added to the 90-liter solution, we have a 120-liter solution, of which 900/11 liters is milk and 90/11 + 30 = 90/11 + 330/11 = 420/11 liters is water. So the milk-to-water ratio is 900/11 : 420/11 = 900 : 420 = 15 : 7.

Finally, after 20 liters of solution are removed, we have a 100-liter solution left. Since the milk to water ratio at this point is still 15 : 7, we can set up the equation:

15y + 7y = 100

22y = 100

y = 100/22 = 50/11

Therefore, there are 15(50/11) = 750/11 liters of milk (and 7(50/11) = 350/11 liters of water) in the 100-liter solution.

Answer: D
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100l ( milk) + 10l (water) = 110liters solution
—> \(\frac{100}{110}= \frac{10}{11}\) of 1 liter solution is milk

20 liters solution is removed.
—> \((\frac{10}{11})*20 = (\frac{200}{11})\) liters milk is removed

—> \(100–\frac{200}{11}=( \frac{900}{11})\) liters milk is left.

Then, 90l(solution) + 30l(water) = 120 liters solution
—> from 120 liters solution, 20 liters solution is removed.

In the 120 liters solution:

\(\frac{900}{(11*120)} = \frac{90}{132}\)
—> \(\frac{90}{132}\) of 1 liter solution is milk

\(20*\frac{90}{132}= \frac{1800}{132}= (\frac{150}{11})\)
—> (\(\frac{150}{11}\)) liters of 20 liters solution is milk —> \(\frac{900}{11}\) —\(\frac{150}{11}\)= \(\frac{750}{11}\) liters

—> What is the amount milk in the solution( in liters) ???
\(\frac{750}{11}\) liters milk is left in the solution now

The answer is D

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lacktutor can you or anyone explain how to use the short winded method (cf= ci*vi/vf) to solve this question?

VeritasKarishma
Could you help me with the solution using Cf = Ci * (Vi/Vf).

VeritasKarishma Thanks a lot for the reply! That explanation clears the doubt

VeritasKarishma In the formula you used is Ci (Initial concentration) 100 l of milk ( from the first sentence of the question. just wanted to clarify because you also get 100 when you add and subtract from the solution eventually?

Ci = Initial concentration of milk in the solution (so it cannot be 100 lt because that is initial volume of milk)

To 100 lt of milk, 10 lt of water is added. So you get 110 lt of solution.
The concentration of milk in this is 100/110 = 10/11
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Scenario Milk Water Total Calculation using concentration of milk

1 100 0 100
2 100 10 110
100/110 of 1 L thus (100/110)*20 = 200/11 L is removed from milk
, thus 100-(200/11)=900/11 L milk remains when 20Lsolution is removed
3 900/11 90
4 900/11 120 30L water is added
(900/11)/120 of 1 L thus (900/11)/120*20 = 150/11 L is removed from milk
, thus (900/11)-(150/11)=750/11 L milk remains when 20L solution is further removed

Couldn't understand the power i.e. ^n part, can anyone explain?

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