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CAT 2021 Slot 3 DILR · Set 4 of 4Logical ReasoningHard DifficultyQ17–Q20 · 2 MCQ + 2 TITA

CAT 2021 Slot 3 DILR Set 4: Math Puzzles (4 Questions) with Solutions

The Math Puzzles set from the official CAT 2021 Slot 3 DILR paper: the caselet, all 4 questions, the official answer key and a step-by-step solution for each. 100% free, no sign-up.

The Set: Read the Caselet First

CAT 2021 Slot 3 DILR · Questions 17–20
Each of the bottles mentioned in this question contains 50 ml of liquid. The liquid in any bottle can be 100% pure content (P) or can have certain amount of impurity (I). Visually it is not possible to distinguish between P and I. There is a testing device which detects impurity, as long as the percentage of impurity in the content tested is 10% or more.
For example, suppose bottle 1 contains only P, and bottle 2 contains 80% P and 20% I. If content from bottle 1 is tested, it will be found out that it contains only P. If content of bottle 2 is tested, the test will reveal that it contains some amount of I. If 10 ml of content from bottle 1 is mixed with 20 ml content from bottle 2, the test will show that the mixture has impurity, and hence we can conclude that at least one of the two bottles has I. However, if 10 ml of content from bottle 1 is mixed with 5 ml of content from bottle 2. the test will not detect any impurity in the resultant mixture.

In the exam this set competed with 3 others for 40 minutes. Budget about 10 minutes if you attempt it.

Q17. Math Puzzles

MCQ · +3 / −1
5 ml of content from bottle A is mixed with 5 ml of content from bottle B. The resultant mixture, when tested, detects the presence of I. If it is known that bottle A contains only P, what BEST can be concluded about the volume of I in bottle B?
  1. A) 10 ml or more
  2. B) Less than 1 ml
  3. C) 1 ml
  4. D) 10 ml
Correct Answer
A) 10 ml or more
Step-by-step solution
📌 Core Concept
The test flags impurity only when I is at least 10% of the tested mixture. Work backwards from the mixture to the concentration of I in bottle B, then scale that to the full 50 ml bottle.
🔢 Step-by-Step Solution
1
Bottle A contains only P, so the 5 ml taken from A adds no impurity. All the I in the mixture comes from the 5 ml taken from B.
2
The mixture is 5+5=105 + 5 = 10 ml. The test detects I, so the impurity is at least 10% of the mixture: 0.10×10=10.10 \times 10 = 1 ml.
3
That 1 ml or more of I sits in the 5 ml drawn from B. So the impurity fraction of B's content is at least 15=20%\frac{1}{5} = 20\%.
4
B's liquid is taken to be uniform, so the fraction in the 5 ml sample equals the fraction in the whole bottle. Bottle B holds 50 ml, so its I is at least 20%×50=1020\% \times 50 = 10 ml.
5
Detection can only happen if the I in B is 10 ml or more. Any larger amount also gives detection, so the best conclusion is a lower bound, not an exact value.
Why Not the Others
B) Less than 1 ml would give an impurity of at most 2% in the 5 ml sample, and the mixture would be under 1%, so the test would not detect it.
C) 1 ml is the amount needed in the 5 ml sample, not in the whole 50 ml bottle. It would make B only 2% impure, giving 1% in the mixture, which is undetectable.
D) 10 ml is only the minimum. The test result allows more than 10 ml, so stating exactly 10 ml claims more than the data supports.
Quick Check
If B had exactly 10 ml of I in 50 ml, that is 20%, so the 5 ml sample has 1 ml of I. The mixture is 110=10%\frac{1}{10} = 10\% impure, which is just detectable. Any less and the test would fail.
Correct Answer: A

Q18. Math Puzzles

TITA · +3 / 0
There are four bottles. Each bottle is known to contain only P or only I. They will be considered to be “collectively ready for despatch” if all of them contain only P. In minimum how many tests, is it possible to ascertain whether these four bottles are “collectively ready for despatch”?.
Correct Answer
1
Step-by-step solution
📌 Core Concept
Each bottle is wholly P or wholly I, so one pooled sample of equal amounts from all four bottles settles the question in a single test.
🔢 Step-by-Step Solution
1
Take the same volume from each of the four bottles and mix them. Each bottle then forms 25% of the mixture.
2
If all four bottles are P, the mixture has 0% impurity and the test shows no I. The bottles are collectively ready for despatch.
3
If even one bottle is I, then that bottle is 100% I and makes up 25% of the mixture. Since 25%≥10%25\% \ge 10\%, the test detects I.
4
So one outcome (no impurity) means ready and the other (impurity) means not ready. The answer is fully decided by this single test.
5
At least one test is needed, because no conclusion is possible without testing. The minimum is therefore 1.
Quick Check
The worst case is exactly one I bottle. Its share is 25%, well above the 10% threshold, so it is never missed.
Correct Answer: 1

Q19. Math Puzzles

TITA · +3 / 0
There are four bottles. It is known that three of these bottles contain only P, while the remaining one contains 80% P and 20% I. What is the minimum number of tests required to definitely identify the bottle containing some amount of I?
Correct Answer
2
Step-by-step solution
📌 Core Concept
A test has only two outcomes (impurity detected or not), so it splits the candidates into two groups. Equal-volume pooling can be made sensitive enough to detect the single 20%-impure bottle.
🔢 Step-by-Step Solution
1
Four bottles could each be the impure one, so there are 4 possibilities. One test gives only 2 outcomes and cannot separate 4 cases, so at least 2 tests are needed.
2
Test 1: mix equal volumes from bottles 1 and 2. If the impure bottle is among them, the mixture has 20%2=10%\frac{20\%}{2} = 10\% impurity, which is just detectable. If both are pure, the impurity is 0%.
3
If I is detected, the impure bottle is 1 or 2. If not, it is 3 or 4.
4
Test 2: test one bottle from the identified pair on its own, for example bottle 1. Pure P shows no I, so the other bottle of the pair is impure. If I shows up, that bottle is the impure one.
5
Two tests are enough and also necessary, so the minimum is 2.
Quick Check
Two tests give 2×2=42 \times 2 = 4 outcome combinations, enough to tell apart the 4 possible bottles. A single test gives only 2, which is not enough.
Correct Answer: 2

Q20. Math Puzzles

MCQ · +3 / −1
There are four bottles. It is known that either one or two of these bottles contain(s)only P, while the remaining ones contain 85% P and 15% I. What is the minimum number of tests required to ascertain the exact number of bottles containing only P?
  1. A) 2
  2. B) 1
  3. C) 3
  4. D) 4
Correct Answer
B) 1
Step-by-step solution
📌 Core Concept
The pooled impurity percentage depends on how many of the four bottles are impure. Choose equal volumes so that the counts 3 and 2 fall on opposite sides of the 10% threshold.
🔢 Step-by-Step Solution
1
The possible cases are: one pure bottle with three impure ones (85% P, 15% I), or two pure bottles with two impure ones.
2
Take equal volumes from all four bottles and mix them. Each bottle then contributes one quarter of the mixture.
3
Case 1 (1 pure, 3 impure): impurity is 3×15%4=11.25%\frac{3 \times 15\%}{4} = 11.25\%. This is at least 10%, so I is detected.
4
Case 2 (2 pure, 2 impure): impurity is 2×15%4=7.5%\frac{2 \times 15\%}{4} = 7.5\%. This is below 10%, so no I is detected.
5
One test therefore tells the two cases apart. Detection means exactly one bottle is pure, and no detection means exactly two are pure.
6
A single test is both sufficient and the least possible.
Why Not the Others
A), C), D) 2, 3 or 4 tests are more than needed, because the one pooled test already separates the cases.
Correct Answer: B
Timed Practice

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CAT 2021 Slot 3 DILR at a Glance

Sets in this section4
Questions in this section20
Section time40 minutes
Convenor IIMIIM Ahmedabad

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