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CAT 2023 Slot 2 — QA Question 14

Removal & ReplacementEasy

A container has 40 liters of milk. Then, 4 liters are removed from the container and replaced with 4 liters of water. This process of replacing 4 liters of the liquid in the container with an equal volume of water is continued repeatedly. The smallest number of times of doing this process, after which the volume of milk in the container becomes less than that of water, is

Answer & solution

Answer: 7

Solution

Easy

This is the classic "replacement" mixture. Each step removes 440=110\tfrac{4}{40}=\tfrac{1}{10} of the liquid, so a fraction 910\tfrac{9}{10} of the milk survives every step. Milk falls below water exactly when milk drops under half of 4040, i.e. below 2020 L. Find the smallest nn with (0.9)n<0.5(0.9)^n<0.5.

1

Milk remaining after nn replacements:

fraction kept each step=1−440=910Milk=40×(910)n\begin{aligned} &\text{fraction kept each step}=1-\tfrac{4}{40}=\tfrac{9}{10}\\ &\text{Milk}=40\times\left(\tfrac{9}{10}\right)^n \end{aligned}
2

Set the "less than water" condition (milk <20<20 L):

40×(0.9)n<20⇒ (0.9)n<0.5\begin{aligned} &40\times(0.9)^n<20\\ &\Rightarrow\ (0.9)^n<0.5 \end{aligned}
3

Test successive powers of 0.90.9:

(0.9)6=0.5314…>0.5(0.9)7=0.4783…<0.5 ✓\begin{aligned} &(0.9)^6=0.5314\ldots>0.5\\ &(0.9)^7=0.4783\ldots<0.5\ \checkmark \end{aligned}

The smallest number of times is 7\mathbf{7}.

Use logs to skip the trials: n>log⁡0.5log⁡0.9=−0.3010−0.0458≈6.58n>\dfrac{\log 0.5}{\log 0.9}=\dfrac{-0.3010}{-0.0458}\approx6.58, so the least integer is n=7n=7.

CAT 2023 Slot 2 QA Q14: A container has 40 liters of milk. Then, 4 liters are removed from the container and replaced with 4 liters of — Solution | TheCATExam