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CAT 2025 Slot 2 DILR · Set 2 of 5Logical ReasoningModerate DifficultyQ6–Q9 · 3 MCQ + 1 TITA

CAT 2025 Slot 2 DILR Set 2: Grids (4 Questions) with Solutions

The Grids set from the official CAT 2025 Slot 2 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 2025 Slot 2 DILR · Questions 6–9
There are six spherical balls, B1, B2, B3, B4, B5, and B6, and four circular hoops H1, H2, H3, and H4. Each ball was tested on each hoop once, by attempting to pass the ball through the hoop. If the diameter of a ball is not larger than the diameter of the hoop, the ball passes through the hoop and makes a "ping". Any ball having a diameter larger than that of the hoop gets stuck on that hoop and does not make a ping. The following additional information is known:
1
B1 and B6 each made a ping on H4, but B5 did not.
2
B4 made a ping on H3, but B1 did not.
3
All balls, except B3, made pings on H1.
4
None of the balls, except B2, made a ping on H2.

In the exam this set competed with 4 others for 40 minutes. Budget about 8 minutes if you attempt it.

Q6. Grids

TITA · +3 / 0
What was the total number of pings made by B1, B2, and B3?
Correct Answer
6
Step-by-step solution
📌 Core Concept
A ball pings on a hoop exactly when its diameter is at most the hoop's diameter. Each clue therefore gives an inequality, and the inequalities fix the full ping grid for B1, B2 and B3.
🔢 Step-by-Step Solution
1
Clue 2 gives B4≤H3<B1B4 \le H3 < B1. Clue 1 gives B1≤H4B1 \le H4. So H3<H4H3 < H4.
2
Clue 4: only B2 pings on H2, so B2≤H2B2 \le H2 and every other ball is larger than H2H2. In particular H2<B4≤H3H2 < B4 \le H3, so H2<H3H2 < H3.
3
Clue 3: B3 does not ping on H1, so B3>H1B3 > H1. B5 pings on H1 but not on H4 (clue 1), so H4<B5≤H1H4 < B5 \le H1. Hence H4<H1H4 < H1, and the hoop order is H2<H3<H4<H1H2 < H3 < H4 < H1.
4
B1: pings on H1 (clue 3) and H4 (clue 1). It misses H2 (clue 4) and H3 (clue 2). That is 2 pings.
5
B2: B2≤H2B2 \le H2, the smallest hoop, so it passes through all four hoops. That is 4 pings.
6
B3: B3>H1B3 > H1, the largest hoop, so it passes through none. That is 0 pings.
7
Total =2+4+0=6= 2 + 4 + 0 = 6.
Quick Check
B3 is larger than H1, which is larger than every other hoop, so 0 pings for B3 is consistent.
Correct Answer: 6

Q7. Grids

MCQ · +3 / −1
Which of the following statements about the relative sizes of the balls is NOT NECESSARILY true?
  1. A) B1 < B5 < B3
  2. B) B4 < B5 < B3
  3. C) B1 < B6 < B3
  4. D) B2 < B1 < B5
Correct Answer
C) B1 < B6 < B3
Step-by-step solution
📌 Core Concept
Turn the clues into a chain of inequalities between hoops and balls, then check which ball orderings are forced and which are not.
🔢 Step-by-Step Solution
1
Clue 4 gives B2≤H2B2 \le H2 and H2<H2 < every other ball.
2
Clue 2 gives B4≤H3<B1B4 \le H3 < B1. Clue 1 gives B1≤H4B1 \le H4.
3
B5 pings on H1 but not on H4, so H4<B5≤H1H4 < B5 \le H1. B3 does not ping on H1, so H1<B3H1 < B3.
4
Chaining these gives B2<B4<B1≤H4<B5<B3B2 < B4 < B1 \le H4 < B5 < B3. This order is forced.
5
B6 pings on H4 (so B6≤H4B6 \le H4) and misses H2 (so B6>H2B6 > H2). Nothing compares B6 with B1 or B4. All that is forced is B6<B5<B3B6 < B5 < B3.
6
Check the options: A, B and D all follow from the forced chain in step 4. C needs B1<B6B1 < B6, which is not forced.
Why Not the Others
A: B1<B5<B3B1 < B5 < B3 is part of the forced chain.
B: B4<B5<B3B4 < B5 < B3 is also part of the chain.
D: B2<B1<B5B2 < B1 < B5 is also part of the chain.
Correct Answer: C

Q8. Grids

MCQ · +3 / −1
Which of the following statements about the relative sizes of the hoops is true?
  1. A) H2 < H4 < H3 < H1
  2. B) H2 < H3 < H4 < H1
  3. C) H1 < H4 < H3 < H2
  4. D) H1 < H3 < H4 < H2
Correct Answer
B) H2 < H3 < H4 < H1
Step-by-step solution
📌 Core Concept
The ping and no-ping results compare hoops with the same ball. A ball that pings on one hoop but not on another shows which hoop is larger.
🔢 Step-by-Step Solution
1
B4 pings on H3 but is larger than H2, because only B2 pings on H2 (clue 4). So H2<B4≤H3H2 < B4 \le H3, which gives H2<H3H2 < H3.
2
B1 pings on H4 but not on H3. So H3<B1≤H4H3 < B1 \le H4, which gives H3<H4H3 < H4.
3
B5 pings on H1 but not on H4. So H4<B5≤H1H4 < B5 \le H1, which gives H4<H1H4 < H1.
4
Combining the three results gives H2<H3<H4<H1H2 < H3 < H4 < H1.
Why Not the Others
A puts H3 above H4, which contradicts step 2.
C and D put H1 as the smallest hoop, but H1 is the largest (step 3).
Correct Answer: B

Q9. Grids

MCQ · +3 / −1
What BEST can be said about the total number of pings from all the tests undertaken?
  1. A) 12 or 13
  2. B) At least 9
  3. C) 12 or 13 or 14
  4. D) 13 or 14
Correct Answer
A) 12 or 13
Step-by-step solution
📌 Core Concept
Using the ball and hoop orderings, fill in every test result that is forced. Count the pings in each hoop's column and see which cells stay open.
🔢 Step-by-Step Solution
1
Hoop order: H2<H3<H4<H1H2 < H3 < H4 < H1. Ball order: B2<B4<B1<B5<B3B2 < B4 < B1 < B5 < B3, with H2<B6≤H4H2 < B6 \le H4.
2
H1: all balls except B3 ping (clue 3). That is 5 pings.
3
H2: only B2 pings (clue 4). That is 1 ping.
4
H4: B1 and B6 ping (clue 1). B2 and B4 are smaller than B1, so they ping. B5 and B3 are larger than H4, so they do not. That is 4 pings.
5
H3: B2 and B4 ping, because B2<B4≤H3B2 < B4 \le H3. B1, B5 and B3 do not, because each is larger than B1, which is larger than H3. B6 is the only open cell. It satisfies H2<B6≤H4H2 < B6 \le H4, so it may be at most H3 or larger than H3. That is 2 or 3 pings.
6
Total =5+1+4+(2 or 3)=12= 5 + 1 + 4 + (2 \text{ or } 3) = 12 or 1313.
Why Not the Others
B is true but weaker than A, so it is not the best statement.
C and D include 14, which cannot occur because the only open cell adds at most 1.
Correct Answer: A
Timed Practice

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CAT 2025 Slot 2 DILR at a Glance

Sets in this section5
Questions in this section22
Section time40 minutes
Convenor IIMIIM Kozhikode

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