CAT 2022 Slot 3 DILR Set 2 Q6–Q10

A DILR caselet from the Data Interpretation & Logical Reasoning (DILR) section of CAT 2022 Slot 3, with 5 questions (Quant-based Puzzles). Answers are CAT's official key.

The set

Pulak, Qasim, Ritesh, and Suresh participated in a tournament comprising of eight rounds. In each round, they formed two pairs, with each of them being in exactly one pair. The only restriction in the pairing was that the pairs would change in successive rounds. For example, if Pulak formed a pair with Qasim in the first round, then he would have to form a pair with Ritesh or Suresh in the second round. He would be free to pair with Qasim again in the third round. In each round, each pair decided whether to play the game in that round or not. If they decided not to play, then no money was exchanged between them. If they decided to play, they had to bet either ₹1 or ₹2 in that round. For example, if they chose to bet ₹2, then the player winning the game got ₹2 from the one losing the game.
At the beginning of the tournament, the players had ₹10 each. The following table shows partial information about the amounts that the players had at the end of each of the eight rounds. It shows every time a player had ₹10 at the end of a round, as well as every time, at the end of a round, a player had either the minimum or the maximum amount that he would have had across the eight rounds. For example, Suresh had ₹10 at the end of Rounds 1, 3, and 8 and not after any of the other rounds. The maximum amount that he had at the end of any round was ₹13 (at the end of Round 5), and the minimum amount he had at the end of any round was ₹8 (at the end of Round 2). At the end of all other rounds, he must have had either ₹9, ₹11, or ₹12.
It was also known that Pulak and Qasim had the same amount of money with them at the end of Round 4.
Pulak Qasim Ritesh Suresh
Round 1 ₹8 ₹10 ₹10
Round 2 ₹13 ₹10 ₹8
Round 3 ₹10
Round 4
Round 5 ₹10 ₹10 ₹13
Round 6
Round 7 ₹12 ₹4
Round 8 ₹13 ₹10

Question 6

What BEST can be said about the amount of money that Ritesh had with him at the end of Round 8?
  1. A
    Exactly ₹6
  2. B
    ₹4 or ₹5
  3. C
    Exactly ₹5
  4. D
    ₹5 or ₹6

Answer: (A) Exactly ₹6 Quant-based Puzzles MCQ

Solution

  1. Suresh is 10 after R1, so his pair did not play and his partner is also 10; the other pair ended 8 and 12 (12 is unshown). In R2 the unshown 12 rises to 13, Suresh's new partner gains 2 (8 to 10), and the remaining player drops 10 to 9. Matching the column order gives R1: P12, Q8, R10, S10; R2: P13, Q10, R9, S8.
  2. Ritesh's 4 in R7 is his minimum, so he cannot be 10 in R5 (a fall of 6 in two rounds is impossible). So R5 is P10, Q10, R7, S13. Since P = Q in R4 and both reach 10, they are in different pairs; this forces R4 = P11, Q11, R6, S12 (the 12/12/5/11 option fails a case check), and R3 = P11, Q11, R8, S10.
  3. Round 6 shows nothing, and R7 shows only Qasim 12 and Ritesh 4. Checking the cases forces R6: P12, Q11, R5, S12 (S-Q and P-R pairs) and R7: P12, Q12, R4, S12 (R-Q pair, S-P did not play). So 12 is Qasim's maximum.
  4. R8: Suresh drops to 10 against Ritesh (R 4 to 6), and Pulak beats Qasim by 1 to reach his maximum 13. Final R8: P13, Q11, R6, S10. Pairs in R5 were S-P (S +1) and Q-R.
  5. Ritesh ends Round 8 with exactly ₹6.

Question 7

What BEST can be said about the amount of money that Pulak had with him at the end of Round 6?
  1. A
    Exactly ₹12
  2. B
    ₹11 or ₹12
  3. C
    Exactly ₹11
  4. D
    ₹12 or ₹13

Answer: (A) Exactly ₹12 Quant-based Puzzles MCQ

Solution

  1. Suresh is 10 after R1, so his pair did not play and his partner is also 10; the other pair ended 8 and 12 (12 is unshown). In R2 the unshown 12 rises to 13, Suresh's new partner gains 2 (8 to 10), and the remaining player drops 10 to 9. Matching the column order gives R1: P12, Q8, R10, S10; R2: P13, Q10, R9, S8.
  2. Ritesh's 4 in R7 is his minimum, so he cannot be 10 in R5 (a fall of 6 in two rounds is impossible). So R5 is P10, Q10, R7, S13. Since P = Q in R4 and both reach 10, they are in different pairs; this forces R4 = P11, Q11, R6, S12 (the 12/12/5/11 option fails a case check), and R3 = P11, Q11, R8, S10.
  3. Round 6 shows nothing, and R7 shows only Qasim 12 and Ritesh 4. Checking the cases forces R6: P12, Q11, R5, S12 (S-Q and P-R pairs) and R7: P12, Q12, R4, S12 (R-Q pair, S-P did not play). So 12 is Qasim's maximum.
  4. Pulak has exactly ₹12 at the end of Round 6.

Question 8

How much money (in ₹) did Ritesh have at the end of Round 4?

TITA: type in the answer. There are no options and no negative marking.

Answer: 6 Quant-based Puzzles TITA

Solution

  1. Suresh is 10 after R1, so his pair did not play and his partner is also 10; the other pair ended 8 and 12 (12 is unshown). In R2 the unshown 12 rises to 13, Suresh's new partner gains 2 (8 to 10), and the remaining player drops 10 to 9. Matching the column order gives R1: P12, Q8, R10, S10; R2: P13, Q10, R9, S8.
  2. Ritesh's 4 in R7 is his minimum, so he cannot be 10 in R5 (a fall of 6 in two rounds is impossible). So R5 is P10, Q10, R7, S13. Since P = Q in R4 and both reach 10, they are in different pairs; this forces R4 = P11, Q11, R6, S12 (the 12/12/5/11 option fails a case check), and R3 = P11, Q11, R8, S10.
  3. Ritesh has ₹6 at the end of Round 4.

Question 9

How many games were played with a bet of ₹2?

TITA: type in the answer. There are no options and no negative marking.

Answer: 6 Quant-based Puzzles TITA

Solution

  1. Suresh is 10 after R1, so his pair did not play and his partner is also 10; the other pair ended 8 and 12 (12 is unshown). In R2 the unshown 12 rises to 13, Suresh's new partner gains 2 (8 to 10), and the remaining player drops 10 to 9. Matching the column order gives R1: P12, Q8, R10, S10; R2: P13, Q10, R9, S8.
  2. Ritesh's 4 in R7 is his minimum, so he cannot be 10 in R5 (a fall of 6 in two rounds is impossible). So R5 is P10, Q10, R7, S13. Since P = Q in R4 and both reach 10, they are in different pairs; this forces R4 = P11, Q11, R6, S12 (the 12/12/5/11 option fails a case check), and R3 = P11, Q11, R8, S10.
  3. Round 6 shows nothing, and R7 shows only Qasim 12 and Ritesh 4. Checking the cases forces R6: P12, Q11, R5, S12 (S-Q and P-R pairs) and R7: P12, Q12, R4, S12 (R-Q pair, S-P did not play). So 12 is Qasim's maximum.
  4. R8: Suresh drops to 10 against Ritesh (R 4 to 6), and Pulak beats Qasim by 1 to reach his maximum 13. Final R8: P13, Q11, R6, S10. Pairs in R5 were S-P (S +1) and Q-R.
  5. ₹2 games (a change of 2): R1 P-Q, R2 S-Q, R3 S-P, R4 S-R, R6 P-R, R8 S-R. That makes 6.

Question 10

Which of the following pairings was made in Round 5?
  1. A
    Pulak and Ritesh
  2. B
    Pulak and Qasim
  3. C
    Pulak and Suresh
  4. D
    Qasim and Suresh

Answer: (C) Pulak and Suresh Quant-based Puzzles MCQ

Solution

  1. Suresh is 10 after R1, so his pair did not play and his partner is also 10; the other pair ended 8 and 12 (12 is unshown). In R2 the unshown 12 rises to 13, Suresh's new partner gains 2 (8 to 10), and the remaining player drops 10 to 9. Matching the column order gives R1: P12, Q8, R10, S10; R2: P13, Q10, R9, S8.
  2. Ritesh's 4 in R7 is his minimum, so he cannot be 10 in R5 (a fall of 6 in two rounds is impossible). So R5 is P10, Q10, R7, S13. Since P = Q in R4 and both reach 10, they are in different pairs; this forces R4 = P11, Q11, R6, S12 (the 12/12/5/11 option fails a case check), and R3 = P11, Q11, R8, S10.
  3. In R5 Suresh gains 1 (12 to 13). The R4 pairs were S-R and P-Q, so R5 is S-P (Pulak 11 to 10) and Q-R (Qasim 11 to 10, Ritesh 6 to 7). Confirmed by R6, where S must pair with Q (they must change).
  4. So Pulak and Suresh were paired in Round 5.

Same concept in other years

See every question on Four-player game over several rounds with points/money exchanged per round.

More Quant-based Puzzles questions from CAT

All 127 Quant-based Puzzles questions asked since 2017.