If a container contains ‘\( t \)’ units of a particular liquid. Out of which ‘\( s \)’ units are taken out and replaced by water. Then after \( p-1 \) operations, the quantity of pure liquid left in the container is given by:-
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \left[t\left(1-\displaystyle\frac{s}{t}\right)^{p-1}\right] \) units
At the start the container has t units of pure liquid. In each operation s of the t units of the mixture are taken out, so the pure liquid is reduced in the ratio \( \frac{t-s}{t}=1-\frac{s}{t} \):
\( \displaystyle \text{After 1 operation: } t\left(1-\frac{s}{t}\right) \)
\( \displaystyle \text{After 2 operations: } t\left(1-\frac{s}{t}\right)^{2} \)
After n operations the pure liquid left is \( t\left(1-\frac{s}{t}\right)^{n} \). With \( n=p-1 \) operations:
\( \displaystyle \text{Pure liquid left}=\left[t\left(1-\frac{s}{t}\right)^{p-1}\right]\text{ units} \)
CBSE Applied Mathematics (Class XII):
Alligation and Mixture
