CAT 2024 Slot 1 DILR Set 1 Q1–Q4

A DILR caselet from the Data Interpretation & Logical Reasoning (DILR) section of CAT 2024 Slot 1, with 4 questions (Selection & Distribution). Answers are CAT's official key.

The set

This set relies on a figure in the original paper, which is not reproduced here. The questions and answers are below.

Six web surfers M, N, O, P, X, and Y each had 30 stars which they distributed among four bloggers A, B, C, and D. The number of stars received by A and B from the six web surfers is shown in the figure below.

[Figure in the original paper]

The following additional facts are known regarding the number of stars received by the bloggers from the surfers.

1. The numbers of stars received by the bloggers from the surfers were all multiples of 5 (including 0).
2. The total numbers of stars received by the bloggers were the same.
3. Each blogger received a different number of stars from M.
4. Two surfers gave all their stars to a single blogger.
5. D received more stars than C from Y.

Question 1

What was the total number of stars received by D?

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

Answer: 45 Selection & Distribution TITA

Solution

  1. Total stars \(= 6 \times 30 = 180\), so each blogger got \(45\). From the chart, A gets \(10+25+0+5+0+5 = 45\) and B gets \(25+20 = 45\), so C and D share the rest: M 20, N 5, O 30, P 0, X 30, Y 5.
  2. From M, all four bloggers got different amounts. A got 10 and B got 0, so C and D must take 5 and 15 in some order. From Y, \(D > C\) with a total of 5, so D gets 5 and C gets 0.
  3. Only O and X can give all 30 stars to one blogger (N and P already gave some to A or B). If both went to the same blogger, that blogger would have at least 60, so one goes to C and the other to D.
  4. C's total: \(30 + M_C + N_C = 45\), so \(M_C + N_C = 15\). This forces \(M_C = 15\) and \(N_C = 0\). Then D gets M 5, N 5, Y 5 and one of O or X (30), which also totals 45.
  5. All bloggers received equal totals, so D received \(180 / 4 = 45\).

Shortcut. Equal totals: \(6 \times 30 / 4 = 45\). No grid is needed.

Question 2

What was the number of stars received by D from Y?
  1. A
    5
  2. B
    cannot be determined
  3. C
    0
  4. D
    10

Answer: (A) 5 Selection & Distribution MCQ

Solution

  1. Total stars \(= 6 \times 30 = 180\), so each blogger got \(45\). From the chart, A gets \(10+25+0+5+0+5 = 45\) and B gets \(25+20 = 45\), so C and D share the rest: M 20, N 5, O 30, P 0, X 30, Y 5.
  2. From M, all four bloggers got different amounts. A got 10 and B got 0, so C and D must take 5 and 15 in some order. From Y, \(D > C\) with a total of 5, so D gets 5 and C gets 0.
  3. Y has only 5 stars left for C and D, and D got more than C, so D got 5 from Y.

Question 3

How many surfers distributed their stars among exactly 2 bloggers?

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

Answer: 2 Selection & Distribution TITA

Solution

  1. Total stars \(= 6 \times 30 = 180\), so each blogger got \(45\). From the chart, A gets \(10+25+0+5+0+5 = 45\) and B gets \(25+20 = 45\), so C and D share the rest: M 20, N 5, O 30, P 0, X 30, Y 5.
  2. From M, all four bloggers got different amounts. A got 10 and B got 0, so C and D must take 5 and 15 in some order. From Y, \(D > C\) with a total of 5, so D gets 5 and C gets 0.
  3. Only O and X can give all 30 stars to one blogger (N and P already gave some to A or B). If both went to the same blogger, that blogger would have at least 60, so one goes to C and the other to D.
  4. C's total: \(30 + M_C + N_C = 45\), so \(M_C + N_C = 15\). This forces \(M_C = 15\) and \(N_C = 0\). Then D gets M 5, N 5, Y 5 and one of O or X (30), which also totals 45.
  5. Bloggers per surfer: M (A, C, D) = 3, N (A, D) = 2, O = 1, P (A, B) = 2, X = 1, Y (A, B, D) = 3. So exactly 2 surfers (N and P) split between two bloggers.

Question 4

Which of the following can be determined with certainty?

1)The number of stars received by C from M
2) The number of stars received by D from O
  1. A
    Neither I nor II
  2. B
    Only I
  3. C
    Only II
  4. D
    Both I and II

Answer: (B) Only I Selection & Distribution MCQ

Solution

  1. Total stars \(= 6 \times 30 = 180\), so each blogger got \(45\). From the chart, A gets \(10+25+0+5+0+5 = 45\) and B gets \(25+20 = 45\), so C and D share the rest: M 20, N 5, O 30, P 0, X 30, Y 5.
  2. From M, all four bloggers got different amounts. A got 10 and B got 0, so C and D must take 5 and 15 in some order. From Y, \(D > C\) with a total of 5, so D gets 5 and C gets 0.
  3. Only O and X can give all 30 stars to one blogger (N and P already gave some to A or B). If both went to the same blogger, that blogger would have at least 60, so one goes to C and the other to D.
  4. C's total: \(30 + M_C + N_C = 45\), so \(M_C + N_C = 15\). This forces \(M_C = 15\) and \(N_C = 0\). Then D gets M 5, N 5, Y 5 and one of O or X (30), which also totals 45.
  5. C got 15 from M in every case, so I is fixed. O's 30 stars could go to C or to D (X takes the other), so D's stars from O are 0 or 30 and II is not fixed. The answer is Only I.

Same concept in other years

See every question on Distribute items among people under value/total constraints.

More Selection & Distribution questions from CAT

All 41 Selection & Distribution questions asked since 2017.