CAT 2023 Slot 2 · Quant · Question 14

Each replacement removes 4 litres out of 40, that is one-tenth of whatever is in the container.

Since the liquid is uniformly mixed, one-tenth of the milk goes with it, leaving nine-tenths.

So after each operation the milk is multiplied by \frac{9}{10}.

After n operations the milk fraction is {\left( \frac{9}{10}\right)}^{n} of the original.

Milk becomes less than water when it falls below half the container, so the condition is {\left( \frac{9}{10}\right)}^{n}<\frac{1}{2}.

Computing the powers: {0.9}^{2}=0.81, {0.9}^{4}=0.6561.

Continuing, {0.9}^{6}=0.6561\times 0.81=0.5314, which is still above one-half.

One more step gives {0.9}^{7}=0.5314\times 0.9=0.4783, which is below one-half.

So the milk first drops below the water after 7 replacements.

In litres, that is 40\times 0.4783\approx 19.1 litres of milk against about 20.9 of water.

Shortcut. Squaring repeatedly is far faster than multiplying seven times: {0.9}^{2}=0.81, then {0.81}^{2}=0.6561, then {0.6561}^{2}\approx 0.43 — so {0.9}^{8} is already under half while {0.9}^{6}\approx 0.53 is not, leaving only n=7 to check.