CAT 2025 Slot 2 DILR Question & Solution
Data Set
Question 1
What was the total number of pings made by B1, B2, and B3?
Solution:
Let us try to arrange the hoops and the spheres in increasing order of their diameter according to the information given,
In clue 1, we are given that B1 and B6 made a ping on H4, and B5 did not.
So, we can definitely say that,
(B1, B6) < H4 < B5 --(1)
In clue 2, we are given that B4 made a ping on H3, but B1 did not.
So, we can definitely say that,
B4 < H3 < B1 --(2)
In clue 3, we are given that all balls, except B3, made pings on H1.
So, we can definitely say that,
(B1, B2, B4, B5, B6) < H1 < B3 --(3)
In clue 4, we are given that none of the balls, except B2, made a ping on H2.
So, we can definitely say that,
B2 < H2 < (B1, B3, B4, B5, B6) --(4)
Combining (1) and (2), we can definitely say that
B4 < H3 < B1 < H4 < B5 --(5)
The only ball that we do not have enough information is about B6, as it can be either less than H3 or greater than H3 and less than H4.
Combining (3), (4) and (5), we get,
B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3
The sphere B6 can be placed in 2 positions, giving us two possible inequalities, which are
B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3
B2 < H2 < B4 < H3 < (B1,B6) < H4 < B5 < H1 < B3
The position of B6 is not clear, and except for that, all the positions are fixed.
Pings made by B1 = H4, H1 = 2
Pings made by B2 = H2, H3, H4, H1 = 4
Pings made by B3 = 0
Total number of pings by B1, B2 and B3 = 2 + 4 + 0 = 6.
Hence, the correct answer is 6.
Question 2
Which of the following statements about the relative sizes of the balls is NOT NECESSARILY true?
Solution:
Let us try to arrange the hoops and the spheres in increasing order of their diameter according to the information given,
In clue 1, we are given that B1 and B6 made a ping on H4, and B5 did not.
So, we can definitely say that,
(B1, B6) < H4 < B5 --(1)
In clue 2, we are given that B4 made a ping on H3, but B1 did not.
So, we can definitely say that,
B4 < H3 < B1 --(2)
In clue 3, we are given that all balls, except B3, made pings on H1.
So, we can definitely say that,
(B1, B2, B4, B5, B6) < H1 < B3 --(3)
In clue 4, we are given that none of the balls, except B2, made a ping on H2.
So, we can definitely say that,
B2 < H2 < (B1, B3, B4, B5, B6) --(4)
Combining (1) and (2), we can definitely say that
B4 < H3 < B1 < H4 < B5 --(5)
The only ball for which we do not have enough information is about B6, as it can be either less than H3 or greater than H3 and less than H4.
Combining (3), (4) and (5), we get,
B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3
The sphere B6 can be placed in 2 positions, giving us two possible inequalities, which are
B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3
B2 < H2 < B4 < H3 < (B1,B6) < H4 < B5 < H1 < B3
The position of B6 is unclear, and except for that, all the other positions are fixed.
Option A) B4 < B5 < B3. This is definitely true.
Option B) B2 < B1 < B5. This is definitely true.
Option C) B1 < B6 < B3. This need not be true, as we do not know which of B1 and B6 has the larger diameter.
Option D) B1 < B5 < B3. This is definitely true.
Hence, the correct answer is option C.
Question 3
Which of the following statements about the relative sizes of the hoops is true?
Solution:
Let us try to arrange the hoops and the spheres in increasing order of their diameter according to the information given,
In clue 1, we are given that B1 and B6 made a ping on H4, and B5 did not.
So, we can definitely say that,
(B1, B6) < H4 < B5 --(1)
In clue 2, we are given that B4 made a ping on H3, but B1 did not.
So, we can definitely say that,
B4 < H3 < B1 --(2)
In clue 3, we are given that all balls, except B3, made pings on H1.
So, we can definitely say that,
(B1, B2, B4, B5, B6) < H1 < B3 --(3)
In clue 4, we are given that none of the balls, except B2, made a ping on H2.
So, we can definitely say that,
B2 < H2 < (B1, B3, B4, B5, B6) --(4)
Combining (1) and (2), we can definitely say that
B4 < H3 < B1 < H4 < B5 --(5)
The only ball for which we do not have enough information is about B6, as it can be either less than H3 or greater than H3 and less than H4.
Combining (3), (4) and (5), we get,
B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3
The sphere B6 can be placed in 2 positions, giving us two possible inequalities, which are
B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3
B2 < H2 < B4 < H3 < (B1,B6) < H4 < B5 < H1 < B3
The position of B6 is unclear, and except for that, all the other positions are fixed.
We know the correct order of the diameters of the hoops is H2 < H3 < H4 < H1.
Hence, the correct answer is option A.
Question 4
What BEST can be said about the total number of pings from all the tests undertaken?
Solution:
Let us try to arrange the hoops and the spheres in increasing order of their diameter according to the information given,
In clue 1, we are given that B1 and B6 made a ping on H4, and B5 did not.
So, we can definitely say that,
(B1, B6) < H4 < B5 --(1)
In clue 2, we are given that B4 made a ping on H3, but B1 did not.
So, we can definitely say that,
B4 < H3 < B1 --(2)
In clue 3, we are given that all balls, except B3, made pings on H1.
So, we can definitely say that,
(B1, B2, B4, B5, B6) < H1 < B3 --(3)
In clue 4, we are given that none of the balls, except B2, made a ping on H2.
So, we can definitely say that,
B2 < H2 < (B1, B3, B4, B5, B6) --(4)
Combining (1) and (2), we can definitely say that
B4 < H3 < B1 < H4 < B5 --(5)
The only ball for which we do not have enough information is about B6, as it can be either less than H3 or greater than H3 and less than H4.
Combining (3), (4) and (5), we get,
B2 < H2 < B4 < H3 < B1 < H4 < B5 < H1 < B3
The sphere B6 can be placed in 2 positions, giving us two possible inequalities, which are
B2 < H2 < (B4, B6) < H3 < B1 < H4 < B5 < H1 < B3
B2 < H2 < B4 < H3 < (B1,B6) < H4 < B5 < H1 < B3
The position of B6 is unclear, and except for that, all the other positions are fixed.
In the first case, the number of pings possible can be calculated as 4 for B2, 3 for B4, 3 for B6, 2 for B1, 1 for B5 and 0 for B3. In total, the number of pings possible in the 1st case is 4 + 3 + 3 + 2 + 1 = 13.
In the second case, the number of pings possible can be calculated as 4 for B2, 3 for B4, 2 for B6, 2 for B1, 1 for B5 and 0 for B3. In total, the number of pings possible in the 2nd case is 4 + 3 + 2 + 2 + 1 = 12.
So, the total number of pings can be 12 or 13.
Hence, the correct answer is Option A.
