 ### From a container, full of pure milk, 20% is replaced by water and this process is repeated three times. At the end of third operation, the quantity of pure milk reduces to:

A. 40.0% B. 50.0% C. 51.2% D. 58.8% Answer: Option C

### Solution(By Apex Team)

Let pure milk was 100L. So, Water is replaced 20% in per process = 20% of 100 = 20L. Now, we use short-cut formula for it. Quantity of Milk reduced to, \begin{aligned}&=\mathrm{X}\times\left[1-\frac{\mathrm{Y}}{\mathrm{X}}\right]^{\mathrm{n}}\\ &=100\times\left[1-\frac{20}{100}\right]^3\\ &=\frac{100\times64}{125}\\ &=51.2\%\end{aligned} Here, X = Initial quantity of milk. Y = Replaced water in per process. n = No. of process repeated.
Alternatively, Let pure milk be 100 litres initially. After third operation, milk will be 100 = 20%↓(- 20L) ⇒ 80 = 20%↓(- 16L) ⇒ 64 = 20%↓(-12.8L) ⇒ 51.2 %

## Related Questions On Alligation

A. 2 : 5
B. 3 : 5
C. 5 : 3
D. 5 : 2

### An alloy contains zinc, copper and tin in the ratio 2 : 3 : 1 and another contains copper, tin and lead in the ratio 5 : 4 : 3. If equal weights of both alloys are melted together to form a third alloy, then the weight of lead per kg in new alloy will be:

A. $\large\frac{1}{2} \mathrm{~kg}$
B. $\large\frac{1}{8} \mathrm{~kg}$
C. $\large\frac{3}{14} \mathrm{~kg}$
D. $\large\frac{7}{9} \mathrm{~kg}$

A. 81 litres
B. 71 litres
C. 56 litres
D. 50 litres