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CAT 2023 Slot 2 QA Question 16

Type-in-the-answer (no negative marking) · Arithmetic · Mixture & Alligation · Try it, then check the answer and solution below.

CAT 2023 Slot 2QAArithmetic • Mixture & AlligationEasy
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
TITA Answer:
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Answer and solution

Answer: 7

📌 Core Concept
The problem involves repeated replacement of a portion of a solution, which can be modeled using the formula for concentration after multiple replacements:
Remaining Amount=Initial Amount×(1−Volume ReplacedTotal Volume)n\text{Remaining Amount} = \text{Initial Amount} \times \left(1 - \frac{\text{Volume Replaced}}{\text{Total Volume}}\right)^n
🔢 Step-by-Step Solution
1
Initial Setup:
Total volume = 40 liters.
Volume replaced each time = 4 liters.
Fraction replaced = 440\frac{4}{40} = 110\frac{1}{10}$.
2
Milk Remaining After n Replacements:
Milk=40×(1−110)n=40×(910)n\text{Milk} = 40 \times \left(1 - \frac{1}{10}\right)^n = 40 \times \left(\frac{9}{10}\right)^n
3
Volume of Water:
Water=40−Milk=40−40×(910)n\text{Water} = 40 - \text{Milk} = 40 - 40 \times \left(\frac{9}{10}\right)^n
4
Inequality for Milk < Water:
40×(910)n<40−40×(910)n40 \times \left(\frac{9}{10}\right)^n < 40 - 40 \times \left(\frac{9}{10}\right)^n
5
Simplify the Inequality:
Divide both sides by 40:
(910)n<1−(910)n−Combineterms:\left(\frac{9}{10}\right)^n < 1 - \left(\frac{9}{10}\right)^n - Combine terms: 2 × \left(910\frac{9}{10}\right)^n < 1 \left(910\frac{9}{10}\right)^n < 12\frac{1}{2} $
6
Solve for n Using Logarithms:
Take natural logarithm:
ln⁡((910)n)<ln⁡(12)n×ln⁡(910)<ln⁡(12)−Sinceln⁡(910)\ln\left(\left(\frac{9}{10}\right)^n\right) < \ln\left(\frac{1}{2}\right) n \times \ln\left(\frac{9}{10}\right) < \ln\left(\frac{1}{2}\right) - Since \ln\left(\frac{9}{10}\right) is negative, divide both sides and reverse inequality: n>ln⁡(12)ln⁡(910)−Calculate:n > \frac{\ln\left(\frac{1}{2}\right)}{\ln\left(\frac{9}{10}\right)} - Calculate: n > −0.6931−0.1054\frac{-0.6931}{-0.1054} ≈ 6.579 $
7
Determine the Smallest Integer n:
Since n must be an integer, round up to the next whole number:
n=7n = 7
⚡ 30-Second Shortcut
Use the formula for concentration after n replacements. Calculate when the remaining milk is less than half the total volume:
(910)n<12\left(\frac{9}{10}\right)^n < \frac{1}{2}
Solve for n using logarithms or estimation. The smallest integer n satisfying this is 7.
🎯 Final Answer
The smallest number of replacements needed is 7.
Correct Answer: 7

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