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CAT 2024 Slot 2 — DILR questions with answers

All 22 questions of the Data Interpretation & Logical Reasoning section (12 MCQs, 10 TITA, 5 sets). Try each one, then open its answer and solution.

CAT 2024 Slot 2, timed like the real exam (120 minutes, 40 a section) and scored the CAT way: +3 right, −1 for a wrong MCQ, 0 for a wrong TITA, with the full post-mock analysis. Free — sign in and you come straight back to this paper.

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Data Interpretation & Logical Reasoning

CAT 2024 Slot 2 · 40 minutes · +3 for a right answer, −1 for a wrong MCQ, 0 for a wrong TITA

Data set

Set for questions 25–29

DIRECTIONS for questions 25–29: Read the information given below and answer the question(s) that follow(s).

The numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 are placed in ten slots of the following grid based on the conditions below.

Data for questions 25–29, CAT 2024 Slot 2 DILR
  1. Numbers in any row appear in an increasing order from left to right.
  2. Numbers in any column appear in a decreasing order from top to bottom.
  3. 1 is placed either in the same row or in the same column as 10.
  4. Neither 2 nor 3 is placed in the same row or in the same column as 10.
  5. Neither 7 nor 8 is placed in the same row or in the same column as 9.
  6. 4 and 6 are placed in the same row.

Q25TITALogical Puzzles

What is the row number which has the least sum of numbers placed in that row? 

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 4

Rows increase left to right and columns decrease downwards, so 10 is in the top-right slot, R1C4. 1 needs a slot with nothing to its left or below (R1C1, R2C2, R3C3 or R4C4) and must share a row or column with 10, so it is in R1C1 or R4C4. 2 and 3 avoid Row 1 and Column 4, leaving R2C2, R2C3 and R3C3. R2C3 exceeds both R2C2 and R3C3, which cannot hold 1, so it cannot be 2 or 3. Hence 2 and 3 fill R2C2 and R3C3, in either order. 9 is in R1C3 or R2C4. In R2C4 it would push 7 and 8 into Row 1, leaving no row with two free slots for 4 and 6. So 9 is in R1C3, and 7 and 8, kept out of Row 1 and Column 3, go in Column 4. 4 and 6 then need R1C1–R1C2 or R2C3–R2C4. Row 2 would put 6 above 7 and 8 in Column 4, which is impossible. So R1C1 = 4, R1C2 = 6, 1 is in R4C4, R2C4 = 8, R3C4 = 7 and R2C3 = 5. Solution figure for question 25, CAT 2024 Slot 2 Row sums: Row 1 is 4+6+9+10=294+6+9+10 = 29, Row 2 is 15 or 16, Row 3 is 9 or 10, and Row 4 is just 1. The answer is 4.

Q26MCQLogical Puzzles

Which of the following statements MUST be true? I. 10 is placed in a slot in Row 1. II. 1 is placed in a slot in Row 4.
  1. Both I and II
  2. Neither I nor II
  3. Only II
  4. Only I
Answer and solution

Answer: (A) Both I and II

Rows increase rightwards and columns decrease downwards. RrCc is the slot in row r, column c. 10 needs nothing to its right or above it, so it is in R1C4. Statement I is true. 2 and 3 avoid Row 1 and Column 4, so they lie among R2C2, R2C3 and R3C3. R2C3 exceeds both others, so if 2 or 3 sat there, the third slot would hold 1, which must share a row or column with 10. So 2 and 3 take R2C2 and R3C3. 9 is in R1C3 or R2C4. In R2C4, it would push 7 and 8 into Row 1, leaving no row with two free slots for 4 and 6. So 9 is in R1C3, and 7 and 8 (avoiding Row 1 and Column 3) go in Column 4. In R4C4, 7 or 8 would need two larger numbers below 10 above it, but with 9 used at most one is left. So R2C4 = 8 and R3C4 = 7. Only Row 1 now has two free slots, so 4 and 6 take R1C1 and R1C2. The only free slot in Row 1 or Column 4 is R4C4, so 1 goes there, and 5 in R2C3. Statement II is true. Solution figure for question 26, CAT 2024 Slot 2 Both statements hold, so options B, C and D fail. Hence, option A (Both I and II).

Q27MCQLogical Puzzles

Which of the following statements MUST be true? I. 2 is placed in a slot in Column 2. II. 3 is placed in a slot in Column 3.
  1. Only I
  2. Both I and II
  3. Neither I nor II
  4. Only II
Answer and solution

Answer: (C) Neither I nor II

Rows increase left to right and columns decrease downwards, so 10 is in the top-right slot, R1C4. 1 needs a slot with nothing to its left or below (R1C1, R2C2, R3C3 or R4C4) and must share a row or column with 10, so it is in R1C1 or R4C4. 2 and 3 avoid Row 1 and Column 4, leaving R2C2, R2C3 and R3C3. R2C3 exceeds both R2C2 and R3C3, which cannot hold 1, so it cannot be 2 or 3. Hence 2 and 3 fill R2C2 and R3C3, in either order. 9 is in R1C3 or R2C4. In R2C4 it would push 7 and 8 into Row 1, leaving no row with two free slots for 4 and 6. So 9 is in R1C3, and 7 and 8, kept out of Row 1 and Column 3, go in Column 4. 4 and 6 then need R1C1–R1C2 or R2C3–R2C4. Row 2 would put 6 above 7 and 8 in Column 4, which is impossible. So R1C1 = 4, R1C2 = 6, 1 is in R4C4, R2C4 = 8, R3C4 = 7 and R2C3 = 5. Solution figure for question 27, CAT 2024 Slot 2 Only 2 and 3 can swap, between Column 2 (R2C2) and Column 3 (R3C3), so neither statement must hold, and options A, B and D all fail. Hence, option C (Neither I nor II).

Q28TITALogical Puzzles

For how many slots in the grid, placement of numbers CANNOT be determined with certainty? 

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 2

Rows increase left to right and columns decrease downwards, so 10 is in the top-right slot, R1C4. 1 needs a slot with nothing to its left or below (R1C1, R2C2, R3C3 or R4C4) and must share a row or column with 10, so it is in R1C1 or R4C4. 2 and 3 avoid Row 1 and Column 4, leaving R2C2, R2C3 and R3C3. R2C3 exceeds both R2C2 and R3C3, which cannot hold 1, so it cannot be 2 or 3. Hence 2 and 3 fill R2C2 and R3C3, in either order. 9 is in R1C3 or R2C4. In R2C4 it would push 7 and 8 into Row 1, leaving no row with two free slots for 4 and 6. So 9 is in R1C3, and 7 and 8, kept out of Row 1 and Column 3, go in Column 4. 4 and 6 then need R1C1–R1C2 or R2C3–R2C4. Row 2 would put 6 above 7 and 8 in Column 4, which is impossible. So R1C1 = 4, R1C2 = 6, 1 is in R4C4, R2C4 = 8, R3C4 = 7 and R2C3 = 5. Solution figure for question 28, CAT 2024 Slot 2 Every slot is fixed except R2C2 and R3C3, where 2 and 3 can be swapped. The answer is 2.

Q29TITALogical Puzzles

What is the sum of the numbers placed in Column 4? 

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 26

Rows increase left to right and columns decrease downwards, so 10 is in the top-right slot, R1C4. 1 needs a slot with nothing to its left or below (R1C1, R2C2, R3C3 or R4C4) and must share a row or column with 10, so it is in R1C1 or R4C4. 2 and 3 avoid Row 1 and Column 4, leaving R2C2, R2C3 and R3C3. R2C3 exceeds both R2C2 and R3C3, which cannot hold 1, so it cannot be 2 or 3. Hence 2 and 3 fill R2C2 and R3C3, in either order. 9 is in R1C3 or R2C4. In R2C4 it would push 7 and 8 into Row 1, leaving no row with two free slots for 4 and 6. So 9 is in R1C3, and 7 and 8, kept out of Row 1 and Column 3, go in Column 4. 4 and 6 then need R1C1–R1C2 or R2C3–R2C4. Row 2 would put 6 above 7 and 8 in Column 4, which is impossible. So R1C1 = 4, R1C2 = 6, 1 is in R4C4, R2C4 = 8, R3C4 = 7 and R2C3 = 5. Solution figure for question 29, CAT 2024 Slot 2 Column 4 holds 10, 8, 7 and 1, so its sum is 10+8+7+1=2610+8+7+1 = 26. The answer is 26.

Data set

Set for questions 30–33

DIRECTIONS for questions 30–33: Read the information given below and answer the question(s) that follow(s).

The above is a schematic diagram of walkways (indicated by all the straight-lines) and lakes (3 of them, each in the shape of rectangles — shaded in the diagram) of a gated area. Different points on the walkway are indicated by letters (A through P) with distances being OP = 150 m, ON = MN = 300 m, ML = 400 m, EL = 200 m, DE = 400 m.

Data for questions 30–33, CAT 2024 Slot 2 DILR

The following additional information about the facilities in the area is known.

  1. The only entry/exit point is at C.
  2. There are many residences within the gated area; all of them are located on the path AH and ML with four of them being at A, H, M, and L.
  3. The post office is located at P and the bank is located at B.

Q30MCQNetworks & Routes

One resident whose house is located at L, needs to visit the post office as well as the bank. What is the minimum distance (in m) he has to walk starting from his residence and returning to his residence after visiting both the post office and the bank?
  1. 2700
  2. 3200
  3. 3000
  4. 3400
Answer and solution

Answer: (B) 3200

Along each row the gaps are 150 m, 300 m and 300 m (like OPOP, ONON, NMNM); down each column they are 400 m, 200 m and 400 m (like DEDE, ELEL, LMLM). The diagonals are IG=1502+2002=250IG = \sqrt{150^2 + 200^2} = 250 m and OK=3002+4002=500OK = \sqrt{300^2 + 400^2} = 500 m. The shortest round trip is the sum of the shortest paths L to B, B to P and P to L, in either order. L to B: B is 600 m left of and 600 m above L, and neither diagonal lies on the way, so the best is 1200 m, e.g. L–E–D–C–B. B to P: B–G–I–P =400+250+400=1050= 400 + 250 + 400 = 1050 m; the diagonal GI saves 100 m over G–J–I. P to L: P–O–K–L =150+500+300=950= 150 + 500 + 300 = 950 m, using the diagonal OK. Solution figure for question 30, CAT 2024 Slot 2 Total =1200+1050+950=3200= 1200 + 1050 + 950 = 3200 m. Each leg is already at its minimum, so a smaller total such as 3000 (option C) is impossible. Hence, option B (3200).

Q31MCQNetworks & Routes

One person enters the gated area and decides to walk as much as possible before leaving the area without walking along any path more than once and always walking next to one of the lakes. Note that he may cross a point multiple times. How much distance (in m) will he walk within the gated area?
  1. 2800
  2. 3000
  3. 3800
  4. 3200
Answer and solution

Answer: (C) 3800

He enters and leaves at C and may use only walkways that border a lake, each at most once. These are the sides of the three lakes: Lake CDEF: CD+DE+EF+FC=300+400+300+400=1400CD + DE + EF + FC = 300 + 400 + 300 + 400 = 1400 m Lake GFKJ: GF+FK+KJ+JG=300+200+300+200=1000GF + FK + KJ + JG = 300 + 200 + 300 + 200 = 1000 m Lake KLMN: KL+LM+MN+NK=300+400+300+400=1400KL + LM + MN + NK = 300 + 400 + 300 + 400 = 1400 m So he can walk at most 1400+1000+1400=38001400 + 1000 + 1400 = 3800 m. He can in fact walk all twelve once and return to C: C–D–E–F–K–L–M–N–K–J–G–F–C. Points F and K are crossed twice, which is allowed. Solution figure for question 31, CAT 2024 Slot 2 Any smaller total, such as 3200 (option D), leaves some lakeside walkway unused. Hence, option C (3800).

Q32TITANetworks & Routes

One resident takes a walk within the gated area starting from A and returning to A without going through any point (other than A) more than once. What is the maximum distance (in m) she can walk in this way? 

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 5100

She may not pass any point except A twice, so she walks a single loop from A. To make it long she should visit as many of the 16 points as possible and avoid the diagonals IG and OK, each of which is shorter than the two walkways it cuts across. A loop through all 16 points without diagonals is A–H–I–P–O–J–K–N–M–L–E–D–C–F–G–B–A. Solution figure for question 32, CAT 2024 Slot 2 Its length, in metres: Left edge A–H–I–P: 400+200+400=1000400 + 200 + 400 = 1000 Right edge M–L–E–D: 400+200+400=1000400 + 200 + 400 = 1000 Column walkways O–J, K–N, C–F, G–B: 4×400=16004 \times 400 = 1600 Row walkways P–O, J–K, N–M, D–C, F–G, B–A: 150+300+300+300+300+150=1500150 + 300 + 300 + 300 + 300 + 150 = 1500 Total =1000+1000+1600+1500=5100= 1000 + 1000 + 1600 + 1500 = 5100 m. The answer is 5100.

Q33TITANetworks & Routes

Visitors coming for morning walks are allowed to enter as long as they do not pass by any of the residences and do not cross any point (except C) more than once. What is the maximum distance (in m) that such a visitor can walk within the gated area? 

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 3500

The visitor enters and leaves at C. Residences lie along AH and ML, including A, H, M and L, so he cannot use those walkways or any walkway touching these points: AB, HG, HI, EL, KL and MN are also barred. He may cross each point except C only once, so he wants the longest loop from C through the remaining points B, D, E, F, G, I, J, K, N, O and P. The diagonals IG and OK are shorter than the walkways they cut across, so he avoids them. The loop C–D–E–F–K–N–O–P–I–J–G–B–C visits every allowed point once: Solution figure for question 33, CAT 2024 Slot 2 CD+DE+EF+FK+KN+NO=300+400+300+200+400+300=1900CD + DE + EF + FK + KN + NO = 300 + 400 + 300 + 200 + 400 + 300 = 1900 OP+PI+IJ+JG+GB+BC=150+400+150+200+400+300=1600OP + PI + IJ + JG + GB + BC = 150 + 400 + 150 + 200 + 400 + 300 = 1600 Total =1900+1600=3500= 1900 + 1600 = 3500 m. The answer is 3500.

Data set

Set for questions 34–37

DIRECTIONS for questions 34–37: Read the information given below and answer the question(s) that follow(s).

An online e-commerce firm receives daily integer product ratings from 1 through 5 given by buyers. The daily average is the average of the ratings given on that day. The cumulative average is the average of all ratings given on or before that day. The rating system began on Day 1, and the cumulative averages were 3 and 3.1 at the end of Day 1 and Day 2, respectively.

The distribution of ratings on Day 2 is given in the figure below.

Data for questions 34–37, CAT 2024 Slot 2 DILR

The following information is known about ratings on Day 3.

  1. 100 buyers gave product ratings on Day 3.
  2. The modes of the product ratings were 4 and 5.
  3. The numbers of buyers giving each product rating are non-zero multiples of 10.
  4. The same number of buyers gave product ratings of 1 and 2, and that number is half the number of buyers who gave a rating of 3.

Q34TITABar & Line Charts

How many buyers gave ratings on Day 1? 

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 150

Day 2 ratings from the chart: 5 buyers gave 1, 10 gave 2, 5 gave 3, 20 gave 4 and 10 gave 5. Day 2 total =5(1)+10(2)+5(3)+20(4)+10(5)=170= 5(1) + 10(2) + 5(3) + 20(4) + 10(5) = 170, from 5050 buyers. Let xx buyers rate on Day 1. Their average is 3, so their total is 3x3x. The cumulative average after Day 2 is 3.1: 3x+170x+50=3.1\dfrac{3x + 170}{x + 50} = 3.1 3x+170=3.1x+1553x + 170 = 3.1x + 155 0.1x=150.1x = 15, so x=150x = 150. So 150 buyers gave ratings on Day 1. The answer is 150.

Q35MCQBar & Line Charts

What is the daily average rating of Day 3?
  1. 3.6
  2. 3.0
  3. 3.2
  4. 3.5
Answer and solution

Answer: (A) 3.6

On Day 3, 100 buyers rated. Let 10a10a buyers give 1 and 10a10a give 2; then 20a20a give 3. The modes are 4 and 5, so they have equal, largest counts; call each 10b10b. Every count is a non-zero multiple of 10, so aa and bb are positive integers. 10a+10a+20a+10b+10b=10010a + 10a + 20a + 10b + 10b = 100 gives 2a+b=52a + b = 5, so (a,b)=(1,3)(a, b) = (1, 3) or (2,1)(2, 1). If (a,b)=(2,1)(a, b) = (2, 1), rating 3 has 40 buyers and ratings 4 and 5 only 10 each, so 3 would be the mode. Hence a=1a = 1, b=3b = 3: ratings 1 to 5 have 10, 10, 20, 30 and 30 buyers. Total =10+20+60+120+150=360= 10 + 20 + 60 + 120 + 150 = 360, so the daily average is 360100=3.6\dfrac{360}{100} = 3.6. This is the only possible distribution, so 3.5 (option D) and the other values cannot occur. Hence, option A (3.6).

Q36TITABar & Line Charts

What is the median of all ratings given on Day 3? 

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 4

On Day 3, 100 buyers rated. If 10a10a buyers gave 1 and 10a10a gave 2, then 20a20a gave 3, and the modes 4 and 5 each have the same larger count 10b10b. So 40a+20b=10040a + 20b = 100, or 2a+b=52a + b = 5. The case a=2a = 2, b=1b = 1 makes 3 the mode, so a=1a = 1, b=3b = 3. Counts: rating 1: 10, rating 2: 10, rating 3: 20, rating 4: 30, rating 5: 30. With 100 ratings in order, the median is the average of the 50th and 51st. Ratings 1 to 3 fill positions 1 to 40, and rating 4 fills positions 41 to 70, so both are 4. Solution figure for question 36, CAT 2024 Slot 2 The median is 4+42=4\dfrac{4 + 4}{2} = 4. The answer is 4.

Q37MCQBar & Line Charts

Which of the following is true about the cumulative average ratings of Day 2 and Day 3?
  1. The cumulative average of Day 3 increased by less than 5% from Day 2.
  2. The cumulative average of Day 3 decreased from Day 2.
  3. The cumulative average of Day 3 increased by a percentage between 5% and 8% from Day 2.
  4. The cumulative average of Day 3 increased by more than 8% from Day 2.
Answer and solution

Answer: (C) The cumulative average of Day 3 increased by a percentage between 5% and 8% from Day 2.

Day 2 has 50 ratings with total 5(1)+10(2)+5(3)+20(4)+10(5)=1705(1) + 10(2) + 5(3) + 20(4) + 10(5) = 170. If Day 1 has xx ratings (total 3x3x), then 3x+170x+50=3.1\dfrac{3x + 170}{x + 50} = 3.1 gives x=150x = 150. So by Day 2 there are 200 ratings with total 200×3.1=620200 \times 3.1 = 620. Day 3 has 100 ratings: 10a10a each of 1 and 2, 20a20a of 3 and 10b10b each of 4 and 5, so 2a+b=52a + b = 5. Taking a=2a = 2 would make 3 the mode, so a=1a = 1, b=3b = 3 and the Day 3 total is 10+20+60+120+150=36010 + 20 + 60 + 120 + 150 = 360. Cumulative average after Day 3 =620+360300=980300≈3.267= \dfrac{620 + 360}{300} = \dfrac{980}{300} \approx 3.267. Change =3.267−3.13.1×100%≈5.4%= \dfrac{3.267 - 3.1}{3.1} \times 100\% \approx 5.4\%, an increase. This is above 5%, so option A fails, and below 8%, so option D fails. Hence, option C (The cumulative average of Day 3 increased by a percentage between 5% and 8% from Day 2.).

Data set

Set for questions 38–41

DIRECTIONS for questions 38–41: Read the information given below and answer the question(s) that follow(s).

The two plots below give the following information about six firms A, B, C, D, E, and F for 2019 and 2023.

  • PAT: The firm's profits after taxes in Rs. crores
  • ES: The firm's employee strength, that is the number of employees in the firm
  • PRD: The percentage of the firm's PAT that they spend on Research and Development (R&D)

In the plots, the horizontal and vertical coordinates of point representing each firm gives their ES and PAT values respectively. The PRD values of each firm are proportional to the areas around the points representing each firm. The areas are comparable between the two plots, i.e., equal areas in the two plots represent the same PRD values for the two years.

Data for questions 38–41, CAT 2024 Slot 2 DILR

Q38MCQPie & Scatter Charts

Assume that the annual rate of growth in PAT over the previous year (ARG) remained constant over the years for each of the six firms. Which among the firms A, B, C, and E had the highest ARG?
  1. Firm A
  2. Firm C
  3. Firm E
  4. Firm B
Answer and solution

Answer: (C) Firm E

With a constant annual growth rate, the firm whose PAT grew most over 2019–2023 has the highest ARG. So compare PAT in 2023PAT in 2019\dfrac{\text{PAT in 2023}}{\text{PAT in 2019}}, read from the plots (Rs. crores): A: 39003000=1.30\dfrac{3900}{3000} = 1.30 B: 38002800≈1.36\dfrac{3800}{2800} \approx 1.36 C: 30002400=1.25\dfrac{3000}{2400} = 1.25 E: 35002400≈1.46\dfrac{3500}{2400} \approx 1.46 Firm E has the largest ratio, so it has the highest ARG. Firm B, the closest rival at about 1.36, falls well short. Hence, option C (Firm E).

Q39MCQPie & Scatter Charts

The ratio of the amount of money spent by Firm C on R&D in 2019 to that in 2023 is closest to
  1. 9 : 4
  2. 5 : 6
  3. 5 : 9
  4. 9 : 5
Answer and solution

Answer: (D) 9 : 5

R&D spending = PAT × PRD, and PRD is proportional to the area of the circle around each firm's point. Firm C's PAT is 2400 in 2019 and 3000 in 2023. Measured against the PAT gridlines (200 apart), C's circle is about 3 gaps across in 2019 and about 2 gaps across in 2023. So the radii are in the ratio 3:23 : 2, and the areas, and hence the PRDs, are in the ratio 9:49 : 4. R&D spending, 2019 to 2023: 2400×93000×4=2160012000=95\dfrac{2400 \times 9}{3000 \times 4} = \dfrac{21600}{12000} = \dfrac{9}{5} So the ratio is 9:59 : 5. Option A (9:49 : 4) is only the ratio of the PRDs; it ignores the rise in PAT. Hence, option D (9 : 5).

Q40MCQPie & Scatter Charts

Which among the firms A, C, E, and F had the maximum PAT per employee in 2023?
  1. Firm E
  2. Firm A
  3. Firm F
  4. Firm C
Answer and solution

Answer: (D) Firm C

PAT per employee is PAT divided by ES. Reading both from the 2023 plot (PAT in Rs. crores, ES in employees): E: 35001400=2.5\dfrac{3500}{1400} = 2.5 A: 39001300=3\dfrac{3900}{1300} = 3 F: 32001000=3.2\dfrac{3200}{1000} = 3.2 C: 3000800=3.75\dfrac{3000}{800} = 3.75 Firm C has the highest PAT per employee. Firm F, the nearest rival at 3.2, is well below 3.75, and Firms A and E are lower still. Hence, option D (Firm C).

Q41MCQPie & Scatter Charts

Which among the firms C, D, E, and F had the least amount of R&D spending per employee in 2023?
  1. Firm F
  2. Firm D
  3. Firm C
  4. Firm E
Answer and solution

Answer: (B) Firm D

R&D spending per employee =PAT×PRDES= \dfrac{\text{PAT} \times \text{PRD}}{\text{ES}}, that is, PAT per employee times PRD, and PRD is proportional to the circle's area. 2023 PAT per employee: C =3000800=3.75= \dfrac{3000}{800} = 3.75, D =2400800=3= \dfrac{2400}{800} = 3, E =35001400=2.5= \dfrac{3500}{1400} = 2.5, F =32001000=3.2= \dfrac{3200}{1000} = 3.2. In 2023 the circles of C, D and F are the same size (about 2 PAT gaps across), so their PRDs are equal, and D, with the lowest PAT per employee, spends least among these three. E has the lowest PAT per employee, but its circle is about 3 gaps across, so its PRD is (32)2=2.25\left(\tfrac{3}{2}\right)^2 = 2.25 times D's. Taking D's PRD as 1, D scores 3×1=33 \times 1 = 3 while E scores 2.5×2.25=5.6252.5 \times 2.25 = 5.625. So E spends more per employee than D. Hence, option B (Firm D).

Data set

Set for questions 42–46

DIRECTIONS for questions 42–46: Read the information given below and answer the question(s) that follow(s).

Eight gymnastics players numbered 1 through 8 underwent a training camp where they were coached by three coaches — Xena, Yuki, and Zara. Each coach trained at least two players. Yuki trained only even numbered players, while Zara trained only odd numbered players. After the camp, the coaches evaluated the players and gave integer ratings to the respective players trained by them on a scale of 1 to 7, with 1 being the lowest rating and 7 the highest.

The following additional information is known.

  1. Xena trained more players than Yuki.
  2. Player-1 and Player-4 were trained by the same coach, while the coaches who trained Player-2, Player-3 and Player-5 were all different.
  3. Player-5 and Player-7 were trained by the same coach and got the same rating. All other players got a unique rating.
  4. The average of the ratings of all the players was 4.
  5. Player-2 got the highest rating.
  6. The average of the ratings of the players trained by Yuki was twice that of the players trained by Xena and two more than that of the players trained by Zara.
  7. Player-4's rating was double of Player-8's and less than Player-5's.

Q42MCQLogical Puzzles

What best can be concluded about the number of players coached by Zara?
  1. Either 2 or 3 or 4
  2. Exactly 2
  3. Either 2 or 3
  4. Either 3
Answer and solution

Answer: (B) Exactly 2

Player 4 is even and Player 1 is odd, so the coach they share is Xena. Players 2, 3 and 5 have different coaches; 3 and 5 are odd, so they take Xena and Zara, and Player 2 is Yuki's. If Yuki had 3 players, Xena would have at least 4, leaving Zara 1; so Yuki has 2. Only Players 5 and 7 share a rating, so the ratings are 1 to 7 with one repeated. The total is 8×4=328 \times 4 = 32 and 1+2+⋯+7=281 + 2 + \dots + 7 = 28, so Players 5 and 7 got 4. Player 2 got 7. Player 4's rating is below 4 and double Player 8's, so Player 4 got 2 and Player 8 got 1. If Player 3 were Zara's, Players 5 and 7 would be Xena's and Zara would have one player. So Zara has only Players 5 and 7 (average 4), and Player 3 is Xena's. Yuki's average is then 6, so Yuki's other player scored 5: Player 6. Xena has Players 1, 3, 4 and 8 (average 3), and Players 1 and 3 share 3 and 6 in either order. Solution figure for question 42, CAT 2024 Slot 2 Zara cannot have 3 or more players, since the only other odd players, 1 and 3, are Xena's. Hence, option B (Exactly 2).

Q43TITALogical Puzzles

What was the rating of Player-7? 

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 4

Player 4 is even and Player 1 is odd, so the coach they share is Xena. Players 2, 3 and 5 have different coaches; 3 and 5 are odd, so they take Xena and Zara, and Player 2 is Yuki's. If Yuki had 3 players, Xena would have at least 4, leaving Zara 1; so Yuki has 2. Only Players 5 and 7 share a rating, so the ratings are 1 to 7 with one repeated. The total is 8×4=328 \times 4 = 32 and 1+2+⋯+7=281 + 2 + \dots + 7 = 28, so Players 5 and 7 got 4. Player 2 got 7. Player 4's rating is below 4 and double Player 8's, so Player 4 got 2 and Player 8 got 1. If Player 3 were Zara's, Players 5 and 7 would be Xena's and Zara would have one player. So Zara has only Players 5 and 7 (average 4), and Player 3 is Xena's. Yuki's average is then 6, so Yuki's other player scored 5: Player 6. Xena has Players 1, 3, 4 and 8 (average 3), and Players 1 and 3 share 3 and 6 in either order. Solution figure for question 43, CAT 2024 Slot 2 Player 7 has the same rating as Player 5. The answer is 4.

Q44TITALogical Puzzles

What was the rating of Player-6? 

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 5

Player 4 is even and Player 1 is odd, so the coach they share is Xena. Players 2, 3 and 5 have different coaches; 3 and 5 are odd, so they take Xena and Zara, and Player 2 is Yuki's. If Yuki had 3 players, Xena would have at least 4, leaving Zara 1; so Yuki has 2. Only Players 5 and 7 share a rating, so the ratings are 1 to 7 with one repeated. The total is 8×4=328 \times 4 = 32 and 1+2+⋯+7=281 + 2 + \dots + 7 = 28, so Players 5 and 7 got 4. Player 2 got 7. Player 4's rating is below 4 and double Player 8's, so Player 4 got 2 and Player 8 got 1. If Player 3 were Zara's, Players 5 and 7 would be Xena's and Zara would have one player. So Zara has only Players 5 and 7 (average 4), and Player 3 is Xena's. Yuki's average is then 6, so Yuki's other player scored 12−7=512 - 7 = 5; Player 8 scored 1, so this is Player 6. Solution figure for question 44, CAT 2024 Slot 2 The answer is 5.

Q45TITALogical Puzzles

For how many players the ratings can be determined with certainty? 

Type in your answer (TITA). No negative mark for a wrong answer.

Answer and solution

Answer: 6

Player 4 is even and Player 1 is odd, so the coach they share is Xena. Players 2, 3 and 5 have different coaches; 3 and 5 are odd, so they take Xena and Zara, and Player 2 is Yuki's. If Yuki had 3 players, Xena would have at least 4, leaving Zara 1; so Yuki has 2. Only Players 5 and 7 share a rating, so the ratings are 1 to 7 with one repeated. The total is 8×4=328 \times 4 = 32 and 1+2+⋯+7=281 + 2 + \dots + 7 = 28, so Players 5 and 7 got 4. Player 2 got 7. Player 4's rating is below 4 and double Player 8's, so Player 4 got 2 and Player 8 got 1. If Player 3 were Zara's, Players 5 and 7 would be Xena's and Zara would have one player. So Zara has only Players 5 and 7 (average 4), and Player 3 is Xena's. Yuki's average is then 6, so Yuki's other player scored 5: Player 6. Xena has Players 1, 3, 4 and 8 (average 3), and Players 1 and 3 share 3 and 6 in either order. Solution figure for question 45, CAT 2024 Slot 2 Ratings are fixed for Players 2, 4, 5, 6, 7 and 8; only Players 1 and 3 are undecided. The answer is 6.

Q46MCQLogical Puzzles

Who all were the players trained by Xena?
  1. Player-1, Player-4, Player-6, Player-8
  2. Player-1, Player-3, Player-4, Player-8
  3. Player-1, Player-3, Player-4, Player-6
  4. Player-1, Player-3, Player-4
Answer and solution

Answer: (B) Player-1, Player-3, Player-4, Player-8

Player 4 is even and Player 1 is odd, so the coach they share is Xena. Players 2, 3 and 5 have different coaches; 3 and 5 are odd, so they take Xena and Zara, and Player 2 is Yuki's. If Yuki had 3 players, Xena would have at least 4, leaving Zara 1; so Yuki has 2. Only Players 5 and 7 share a rating, so the ratings are 1 to 7 with one repeated. The total is 8×4=328 \times 4 = 32 and 1+2+⋯+7=281 + 2 + \dots + 7 = 28, so Players 5 and 7 got 4. Player 2 got 7. Player 4's rating is below 4 and double Player 8's, so Player 4 got 2 and Player 8 got 1. If Player 3 were Zara's, Players 5 and 7 would be Xena's and Zara would have one player. So Zara has only Players 5 and 7 (average 4), and Player 3 is Xena's. Yuki's average is then 6, so Yuki's other player scored 5: Player 6. Xena has Players 1, 3, 4 and 8 (average 3). Solution figure for question 46, CAT 2024 Slot 2 Options A and C include Player 6, who is Yuki's; option D leaves out Player 8. Hence, option B (Player-1, Player-3, Player-4, Player-8).