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 CShow Answer
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 %
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 %
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