CATin
  1. CATin
  2. CAT past papers
  3. CAT 2020 Slot 1
  4. DILR

CAT 2020 Slot 1 — DILR questions with answers

All 24 questions of the Data Interpretation & Logical Reasoning section (18 MCQs, 6 TITA, 5 sets). Try each one, then open its answer and solution.

CAT 2020 Slot 1, 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. Older official papers are in CATin Pro, with 80+ mocks and the full planner. The last three years of papers are free.

Sit this paper as a timed mock with CATin Pro

Data Interpretation & Logical Reasoning

CAT 2020 Slot 1 · 40 minutes · +3 for a right answer, −1 for a wrong MCQ, 0 for a wrong TITA

Data set

Set for questions 27–32

In a certain board examination, students were to appear for examination in five subjects: English, Hindi, Mathematics, Science and Social Science. Due to a certain emergency situation, a few of the examinations could not be conducted for some students. Hence, some students missed one examination and some others missed two examinations. Nobody missed more than two examinations. The board adopted the following policy for awarding marks to students. If a student appeared in all five examinations, then the marks awarded in each of the examinations were on the basis of the scores obtained by them in those examinations. If a student missed only one examination, then the marks awarded in that examination was the average of the best three among the four scores in the examinations they appeared for. If a student missed two examinations, then the marks awarded in each of these examinations was the average of the best two among the three scores in the examinations they appeared for. The marks obtained by six students in the examination are given in the table below. Each of them missed either one or two examinations. Data for questions 27–32, CAT 2020 Slot 1 DILR The following facts are also known. I. Four of these students appeared in each of the English, Hindi, Science, and Social Science examinations. II. The student who missed the Mathematics examination did not miss any other examination. III. One of the students who missed the Hindi examination did not miss any other examination. The other student who missed the Hindi examination also missed the Science examination.

Q27MCQTables & Caselets

Who among the following did not appear for the Mathematics examination?
  1. Alva
  2. Carl
  3. Foni
  4. Esha
Answer and solution

Answer: (B) Carl

By fact II, one student missed Mathematics and nothing else. That student's Mathematics mark must therefore equal the average of the best three of his or her other four marks. Such an average can never exceed all the marks it is taken from. Deep (100) and Esha (95) have Mathematics marks above every other mark of theirs, so neither missed it. Check the rest: Alva: the best three of 80, 75, 75, 60 give 2303≈76.67≠70\frac{230}{3} \approx 76.67 \ne 70. Bithi: the best three of 90, 80, 85, 85 give 2603≈86.67≠55\frac{260}{3} \approx 86.67 \ne 55. Foni: the best three of 83, 72, 88, 83 give 2543≈84.67≠78\frac{254}{3} \approx 84.67 \ne 78. Carl: the best three of 75, 80, 100, 90 are 100, 90 and 80, giving 2703=90\frac{270}{3} = 90, which equals his Mathematics mark. Only Carl fits, so Alva, Foni and Esha are ruled out. Hence, option B (Carl).

Q28MCQTables & Caselets

Which students did not appear for the English examination?
  1. Carl and Deep
  2. Cannot be determined
  3. Alva and Bithi
  4. Esha and Foni
Answer and solution

Answer: (D) Esha and Foni

By fact I, four of the six sat English, so exactly two missed it. A missed mark is an average of the student's other marks: the best three of the other four if one exam was missed, or the best two of the other three if two were missed (and then both missed exams show the same mark). Test each English mark: Alva (80): the best three of 75, 70, 75, 60 give 2203≈73.3\frac{220}{3} \approx 73.3, and no other mark of hers is 80. Bithi (90): higher than all her other marks, so it cannot be an average of them. Carl (75) and Deep (70): each is far below the best-three average of the other marks (2803≈93.3\frac{280}{3} \approx 93.3 in both cases) and matches no other mark. Esha (80): the best three of 85, 95, 60, 55 give 2403=80\frac{240}{3} = 80. Fits. Foni (83): the best three of 72, 78, 88, 83 give 2493=83\frac{249}{3} = 83. Fits. Only Esha and Foni can have missed English, and two students did, so it is both of them. The answer is fixed, so option B fails. Hence, option D (Esha and Foni).

Q29MCQTables & Caselets

What BEST can be concluded about the students who did not appear for the Hindi examination?
  1. Deep and Esha
  2. Alva and Deep
  3. Alva and Esha
  4. Two among Alva, Deep and Esha
Answer and solution

Answer: (B) Alva and Deep

By fact I, exactly two of the six missed Hindi. A missed mark equals the average of the best three of the student's other four marks (one exam missed), or of the best two of the other three (two missed, and then both missed exams carry the same mark). Test each Hindi mark: Alva (75): the best three of 80, 70, 75, 60 give 2253=75\frac{225}{3} = 75. Fits. Deep (90): the best three of 70, 100, 90, 80 give 2703=90\frac{270}{3} = 90. Fits. Esha (85): the best three of 80, 95, 60, 55 give 2353≈78.3\frac{235}{3} \approx 78.3, and no other mark of hers is 85. Does not fit. Bithi (80), Carl (80) and Foni (72): each is below the best-three average of the other marks (about 86.7, 93.3 and 84.7) and matches none of them. Only Alva and Deep fit, and two students missed Hindi, so it is exactly these two. Option D is the trap: Esha cannot be one of them, so the answer is definite, not 'two among three'. Hence, option B (Alva and Deep).

Q30MCQTables & Caselets

What BEST can be concluded about the students who missed the Science examination?
  1. Bithi and one out of Alva and Deep
  2. Alva and Bithi
  3. Deep and Bithi
  4. Alva and Deep
Answer and solution

Answer: (A) Bithi and one out of Alva and Deep

A missed mark equals the average of the best three other marks (one exam missed) or of the best two of the other three (two missed, both marks equal). Of the six Hindi marks, only Alva's and Deep's fit this, so they are the two who missed Hindi. By fact III, one of them missed only Hindi and the other missed Hindi and Science. Both versions fit the table. If Alva missed Hindi and Science, each is the average of the best two of 80, 70, 60, which is 75. If Deep did, each is the average of the best two of 70, 100, 80, which is 90. So the data cannot say which one. By fact I, two students missed Science, so one more is needed. Carl (100) and Foni (88) have Science as their highest mark, so it cannot be an average. Esha's 60 is far below 2603\frac{260}{3}, the best-three average of her other marks. Bithi's 85 fits: the best three of 90, 80, 55, 85 average 2553=85\frac{255}{3} = 85. Options B, C and D each treat Alva or Deep as certain, which the data does not settle. Hence, option A (Bithi and one out of Alva and Deep).

Q31TITATables & Caselets

How many out of these six students missed exactly one examination?

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

Answer and solution

Answer: 3

Count the missed exams. By fact I, exactly two students missed each of English, Hindi, Science and Social Science. By fact II, one student missed Mathematics. That is 2×4+1=92 \times 4 + 1 = 9 missed exams. Each of the six students missed one or two exams. If ss students missed one and dd missed two, then s+d=6s + d = 6 and s+2d=9s + 2d = 9, so d=3d = 3 and s=3s = 3. The table confirms this. Carl missed only Mathematics (his 90 is the best-three average of 100, 90, 80), Esha missed only English (her 80 is the best-three average of 95, 85, 60), and by fact III one of Alva and Deep missed only Hindi. Bithi (Science and Social Science), Foni (English and Social Science) and the other of Alva and Deep (Hindi and Science) missed two each. The answer is 3.

Q32TITATables & Caselets

For how many students can we be definite about which examinations they missed?

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

Answer and solution

Answer: 4

A missed mark equals the average of the best three other marks (one exam missed) or of the best two of the other three (two missed, both marks equal). Testing every mark against this: Mathematics: only Carl's 90 fits, and by fact II he missed nothing else. English (two missed, fact I): only Esha's 80 and Foni's 83 fit, so both missed it. Hindi (two missed): only Alva's 75 and Deep's 90 fit. By fact III, one missed only Hindi and the other Hindi and Science. Science (two missed): one of Alva and Deep, plus Bithi, whose 85 is the only other Science mark that fits. Social Science (two missed): the marks that fit are Bithi's 85, Carl's 90 and Foni's 83. Carl is excluded by fact II, so it is Bithi and Foni. So Carl (Mathematics), Esha (English), Bithi (Science and Social Science) and Foni (English and Social Science) are fixed. Alva's Hindi and Science are both 75 and Deep's both 90, so either could be the one who missed two, and neither can be pinned down. The answer is 4.

Data set

Set for questions 33–36

Ten musicians (A, B, C, D, E, F, G, H, I and J) are experts in at least one of the following three percussion instruments: tabla, mridangam, and ghatam. Among them, three are experts in tabla but not in mridangam or ghatam, another three are experts in mridangam but not in tabla or ghatam, and one is an expert in ghatam but not in tabla or mridangam. Further, two are experts in tabla and mridangam but not in ghatam, and one is an expert in tabla and ghatam but not in mridangam. The following facts are known about these ten musicians. 1. Both A and B are experts in mridangam, but only one of them is also an expert in tabla. 2. D is an expert in both tabla and ghatam. 3. Both F and G are experts in tabla, but only one of them is also an expert in mridangam. 4. Neither I nor J is an expert in tabla. 5. Neither H nor I is an expert in mridangam, but only one of them is an expert in ghatam.

Q33MCQSet Theory

Who among the following is DEFINITELY an expert in tabla but not in either mridangam or ghatam?
  1. F
  2. C
  3. A
  4. H
Answer and solution

Answer: (D) H

The ten split as: tabla only 3, mridangam only 3, ghatam only 1, tabla and mridangam 2, tabla and ghatam 1. Nobody pairs mridangam with ghatam. D is the one tabla-and-ghatam expert (fact 2). By fact 1, one of A and B plays tabla and mridangam, the other mridangam only. By fact 3, one of F and G plays tabla and mridangam; the other plays tabla only, since the tabla-and-ghatam place is D's. Both tabla-and-mridangam places are now filled. I plays neither tabla nor mridangam (facts 4 and 5), so I is the ghatam-only expert. By fact 5, H then plays neither ghatam nor mridangam, so H is tabla only. J plays no tabla and the ghatam-only place is taken, so J is mridangam only. C and E fill the last tabla-only and mridangam-only places, in either order. Solution figure for question 33, CAT 2020 Slot 1 So H is definitely tabla only. Option A (F) is the strongest distractor, but F may be the tabla-and-mridangam expert. C may be mridangam only, and A always plays mridangam. Hence, option D (H).

Q34MCQSet Theory

Who among the following is DEFINITELY an expert in mridangam but not in either tabla or ghatam?
  1. B
  2. J
  3. G
  4. E
Answer and solution

Answer: (B) J

The ten split as: tabla only 3, mridangam only 3, ghatam only 1, tabla and mridangam 2, tabla and ghatam 1. Nobody pairs mridangam with ghatam. D is the one tabla-and-ghatam expert (fact 2). By fact 1, one of A and B plays tabla and mridangam, the other mridangam only. By fact 3, one of F and G plays tabla and mridangam; the other plays tabla only, since the tabla-and-ghatam place is D's. Both tabla-and-mridangam places are now filled. I plays neither tabla nor mridangam (facts 4 and 5), so I is the ghatam-only expert. By fact 5, H then plays neither ghatam nor mridangam, so H is tabla only. J plays no tabla and the ghatam-only place is taken, so J is mridangam only. C and E fill the last tabla-only and mridangam-only places, in either order. Solution figure for question 34, CAT 2020 Slot 1 So J is definitely mridangam only. Option A (B) is the strongest distractor, but B may be the tabla-and-mridangam expert. Option C (G) always plays tabla, and option D (E) may be tabla only. Hence, option B (J).

Q35MCQSet Theory

Which of the following pairs CANNOT have any musician who is an expert in both tabla and mridangam but not in ghatam?
  1. F and G
  2. C and E
  3. A and B
  4. C and F
Answer and solution

Answer: (B) C and E

The ten split as: tabla only 3, mridangam only 3, ghatam only 1, tabla and mridangam 2, tabla and ghatam 1. Nobody pairs mridangam with ghatam. D is the one tabla-and-ghatam expert (fact 2). By fact 1, one of A and B plays tabla and mridangam, the other mridangam only. By fact 3, one of F and G plays tabla and mridangam; the other plays tabla only, since the tabla-and-ghatam place is D's. Both tabla-and-mridangam places are now filled. I plays neither tabla nor mridangam (facts 4 and 5), so I is the ghatam-only expert. By fact 5, H then plays neither ghatam nor mridangam, so H is tabla only. J plays no tabla and the ghatam-only place is taken, so J is mridangam only. C and E fill the last tabla-only and mridangam-only places, in either order. Solution figure for question 35, CAT 2020 Slot 1 The two tabla-and-mridangam places go to one of A, B and one of F, G. C and E only share the last tabla-only and mridangam-only places, so neither can play both. Option A (F and G) and option C (A and B) each definitely contain one such expert. Option D (C and F) is the trap: C cannot be one, but F can. Hence, option B (C and E).

Q36MCQSet Theory

If C is an expert in mridangam and F is not, then which are the three musicians who are experts in tabla but not in either mridangam or ghatam?
  1. E, F and H
  2. C, G and H
  3. E, G and H
  4. C, E and G
Answer and solution

Answer: (A) E, F and H

The ten split as: tabla only 3, mridangam only 3, ghatam only 1, tabla and mridangam 2, tabla and ghatam 1. Nobody pairs mridangam with ghatam. D is the one tabla-and-ghatam expert (fact 2). By fact 1, one of A and B plays tabla and mridangam, the other mridangam only. By fact 3, one of F and G plays tabla and mridangam; the other plays tabla only, since the tabla-and-ghatam place is D's. Both tabla-and-mridangam places are now filled. I plays neither tabla nor mridangam (facts 4 and 5), so I is the ghatam-only expert. By fact 5, H then plays neither ghatam nor mridangam, so H is tabla only. J plays no tabla and the ghatam-only place is taken, so J is mridangam only. C and E fill the last tabla-only and mridangam-only places, in either order. Solution figure for question 36, CAT 2020 Slot 1 Now add the conditions. C plays mridangam, so C takes the mridangam-only place and E is tabla only. F plays no mridangam, so F is tabla only and G plays tabla and mridangam. The three tabla-only experts are E, F and H. Option C (E, G and H) is the trap: G also plays mridangam. Options B and D include C, who plays mridangam. Hence, option A (E, F and H).

Data set

Set for questions 37–40

The local office of the APP-CAB company evaluates the performance of five cab drivers, Arun, Barun, Chandan, Damodaran, and Eman for their monthly payment based on ratings in five different parameters (P1 to P5) as given below: P1: timely arrival P2: behaviour P3: comfortable ride P4: driver's familiarity with the route P5: value for money Based on feedback from the customers, the office assigns a rating from 1 to 5 in each of these parameters. Each rating is an integer from a low value of 1 to a high value of 5. The final rating of a driver is the average of his ratings in these five parameters. The monthly payment of the drivers has two parts - a fixed payment and final rating-based bonus. If a driver gets a rating of 1 in any of the parameters, he is not eligible to get bonus. To be eligible for bonus a driver also needs to get a rating of five in at least one of the parameters. The partial information related to the ratings of the drivers in different parameters and the monthly payment structure (in rupees) is given in the table below: Data for questions 37–40, CAT 2020 Slot 1 DILR The following additional facts are known. 1. Arun and Barun have got a rating of 5 in exactly one of the parameters. Chandan has got a rating of 5 in exactly two parameters. 2. None of drivers has got the same rating in three parameters.

Q37MCQMissing Value Tables

If Damodaran does not get a bonus, what is the maximum possible value of his final rating?
  1. 3.4
  2. 3.2
  3. 3.6
  4. 3.8
Answer and solution

Answer: (C) 3.6

Damodaran has a 3 in P2. He misses the bonus if he has a 1 in some parameter, or if he has no 5 at all. By condition 2, no rating can appear three times. Case 1, no 5 and no 1: the best he can do is 4, 4, 3 (given), 3, 2, since a third 4 or a third 3 is not allowed. The total is 16, so the final rating is 16/5=3.216/5 = 3.2. Case 2, at least one 1: the other ratings can be 5, 5, 4 and the given 3 (a third 5 is not allowed). The total is 1+3+5+5+4=181 + 3 + 5 + 5 + 4 = 18, so the final rating is 18/5=3.618/5 = 3.6. The larger value is 3.6. Option B (3.2) is only the best without a 1. Option D (3.8) needs a total of 19, which with a 1 means three 5s, and that is not allowed. Hence, option C (3.6).

Q38MCQMissing Value Tables

If Eman gets a bonus, what is the minimum possible value of his final rating?
  1. 3.2
  2. 2.8
  3. 3.4
  4. 3.0
Answer and solution

Answer: (D) 3.0

Eman has a 2 in P5. To get the bonus he needs at least one 5 and no 1. To keep his final rating low, give him just one 5 and make the other ratings as small as possible. By condition 2, no rating may appear three times. So his ratings are 5, 2 (given), 2, 3, 3. A third 2 is not allowed, so the last two must be 3s. The total is 15, and the final rating is 15/5=3.015/5 = 3.0. Option B (2.8) would need a total of 14, that is 5, 2, 2, 2, 3. That uses 2 three times, which breaks condition 2. Hence, option D (3.0).

Q39MCQMissing Value Tables

If all five drivers get bonus, what is the minimum possible value of the monthly payment (in rupees) that a driver gets?
  1. 1750
  2. 1600
  3. 1740
  4. 1700
Answer and solution

Answer: (D) 1700

Monthly pay = fixed payment + bonus rate × final rating. Since everyone gets the bonus, no driver has a 1 and each has at least one 5. Arun and Barun have exactly one 5 and Chandan exactly two. No rating may appear three times. To minimise, keep each driver's remaining ratings as low as allowed. Arun (P4 = 4): 5, 4, 2, 2, 3, rating 3.2, pay 1000+250×3.2=18001000 + 250 \times 3.2 = 1800. Barun (P1 = 3): 5, 3, 2, 2, 3, rating 3, pay 1200+200×3=18001200 + 200 \times 3 = 1800. Chandan (P3 = 2): 5, 5, 2, 2, 3, rating 3.4, pay 1400+100×3.4=17401400 + 100 \times 3.4 = 1740. Damodaran (P2 = 3): 5, 3, 2, 2, 3, rating 3, pay 1300+150×3=17501300 + 150 \times 3 = 1750. Eman (P5 = 2): 5, 2, 2, 3, 3, rating 3, pay 1100+200×3=17001100 + 200 \times 3 = 1700. The smallest pay is Eman's 1700. Option C (1740) and option A (1750) are the minimum pays of Chandan and Damodaran, and both are higher than Eman's. Hence, option D (1700).

Q40MCQMissing Value Tables

If all five drivers get bonus, what is the maximum possible value of the monthly payment (in rupees) that a driver gets?
  1. 1960
  2. 2050
  3. 1950
  4. 1900
Answer and solution

Answer: (A) 1960

Monthly pay = fixed payment + bonus rate × final rating. Since everyone gets the bonus, no driver has a 1 and each has at least one 5. Arun and Barun have exactly one 5 and Chandan exactly two. No rating may appear three times. To maximise, use as many 5s and 4s as allowed. Arun (P4 = 4): 5, 4, 4, 3, 3, rating 3.8, pay 1000+250×3.8=19501000 + 250 \times 3.8 = 1950. Barun (P1 = 3): 5, 4, 4, 3, 3, rating 3.8, pay 1200+200×3.8=19601200 + 200 \times 3.8 = 1960. Chandan (P3 = 2): 5, 5, 4, 4, 2, rating 4, pay 1400+100×4=18001400 + 100 \times 4 = 1800. Damodaran (P2 = 3): 5, 5, 4, 4, 3, rating 4.2, pay 1300+150×4.2=19301300 + 150 \times 4.2 = 1930. Eman (P5 = 2): 5, 5, 4, 4, 2, rating 4, pay 1100+200×4=19001100 + 200 \times 4 = 1900. The largest pay is Barun's 1960. Option C (1950) is Arun's maximum, just short of it. Option B (2050) is above every driver's maximum. Hence, option A (1960).

Data set

Set for questions 41–44

1000 patients currently suffering from a disease were selected to study the effectiveness of treatment of four types of medicines — A, B, C and D. These patients were first randomly assigned into two groups of equal size, called treatment group and control group. The patients in the control group were not treated with any of these medicines; instead they were given a dummy medicine, called placebo, containing only sugar and starch. The following information is known about the patients in the treatment group. a. A total of 250 patients were treated with type A medicine and a total of 210 patients were treated with type C medicine. b. 25 patients were treated with type A medicine only. 20 patients were treated with type C medicine only. 10 patients were treated with type D medicine only. c. 35 patients were treated with type A and type D medicines only. 20 patients were treated with type A and type B medicines only. 30 patients were treated with type A and type C medicines only. 20 patients were treated with type C and type D medicines only. d. 100 patients were treated with exactly three types of medicines. e. 40 patients were treated with medicines of types A, B and C, but not with medicines of type D. 20 patients were treated with medicines of types A, C and D, but not with medicines of type B. f. 50 patients were given all the four types of medicines. 75 patients were treated with exactly one type of medicine.

Q41TITASet Theory

How many patients were treated with medicine type B?

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

Answer and solution

Answer: 340

The 500 treatment-group patients each took at least one medicine, so the regions of the four-set Venn diagram add to 500. Given: only A 25, only C 20, only D 10; A and D only 35, A and B only 20, A and C only 30, C and D only 20; A, B, C only 40; A, C, D only 20; all four 50. A total is 250, so A, B, D only =250−(25+35+20+30+40+20+50)=30= 250 - (25 + 35 + 20 + 30 + 40 + 20 + 50) = 30. Exactly three medicines is 100, so B, C, D only =100−(40+20+30)=10= 100 - (40 + 20 + 30) = 10. Exactly one medicine is 75, so only B =75−(25+20+10)=20= 75 - (25 + 20 + 10) = 20. C total is 210, so B and C only =210−(20+30+20+40+20+10+50)=20= 210 - (20 + 30 + 20 + 40 + 20 + 10 + 50) = 20. All the other regions now sum to 350, so B and D only =500−350=150= 500 - 350 = 150. Solution figure for question 41, CAT 2020 Slot 1 Patients given B =20+20+20+150+40+30+10+50=340= 20 + 20 + 20 + 150 + 40 + 30 + 10 + 50 = 340. This adds every region that contains B. The answer is 340.

Q42TITASet Theory

The number of patients who were treated with medicine types B, C and D, but not type A was:

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

Answer and solution

Answer: 10

Work within the 500 patients of the treatment group. A total is 250. The known regions inside A are only A 25, A and D only 35, A and B only 20, A and C only 30, A, B, C only 40, A, C, D only 20 and all four 50. These total 220, so A, B, D only =250−220=30= 250 - 220 = 30. Exactly three medicines is 100. The three-medicine groups are A, B, C only (40), A, C, D only (20), A, B, D only (30) and B, C, D only. So B, C, D only =100−(40+20+30)=10= 100 - (40 + 20 + 30) = 10. The completed diagram shows this region as 10, alongside every other region. Solution figure for question 42, CAT 2020 Slot 1 The answer is 10.

Q43TITASet Theory

How many patients were treated with medicine types B and D only?

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

Answer and solution

Answer: 150

The 500 treatment-group patients each took at least one medicine, so the regions of the four-set Venn diagram add to 500. Given: only A 25, only C 20, only D 10; A and D only 35, A and B only 20, A and C only 30, C and D only 20; A, B, C only 40; A, C, D only 20; all four 50. A total is 250, so A, B, D only =250−(25+35+20+30+40+20+50)=30= 250 - (25 + 35 + 20 + 30 + 40 + 20 + 50) = 30. Exactly three medicines is 100, so B, C, D only =100−(40+20+30)=10= 100 - (40 + 20 + 30) = 10. Exactly one medicine is 75, so only B =75−(25+20+10)=20= 75 - (25 + 20 + 10) = 20. C total is 210, so B and C only =210−(20+30+20+40+20+10+50)=20= 210 - (20 + 30 + 20 + 40 + 20 + 10 + 50) = 20. All the other regions now sum to 350, so B and D only =500−350=150= 500 - 350 = 150. Solution figure for question 43, CAT 2020 Slot 1 So 150 patients were treated with B and D only. The answer is 150.

Q44TITASet Theory

The number of patients who were treated with medicine type D was:

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

Answer and solution

Answer: 325

The 500 treatment-group patients each took at least one medicine, so the regions of the four-set Venn diagram add to 500. Given: only A 25, only C 20, only D 10; A and D only 35, A and B only 20, A and C only 30, C and D only 20; A, B, C only 40; A, C, D only 20; all four 50. A total is 250, so A, B, D only =250−(25+35+20+30+40+20+50)=30= 250 - (25 + 35 + 20 + 30 + 40 + 20 + 50) = 30. Exactly three medicines is 100, so B, C, D only =100−(40+20+30)=10= 100 - (40 + 20 + 30) = 10. Exactly one medicine is 75, so only B =75−(25+20+10)=20= 75 - (25 + 20 + 10) = 20. C total is 210, so B and C only =210−(20+30+20+40+20+10+50)=20= 210 - (20 + 30 + 20 + 40 + 20 + 10 + 50) = 20. All the other regions now sum to 350, so B and D only =500−350=150= 500 - 350 = 150. Solution figure for question 44, CAT 2020 Slot 1 Patients given D =10+35+20+150+20+30+10+50=325= 10 + 35 + 20 + 150 + 20 + 30 + 10 + 50 = 325. This adds every region that contains D. The answer is 325.

Data set

Set for questions 45–50

Four institutes, A, B, C, and D, had contracts with four vendors W, X, Y, and Z during the ten calendar years from 2010 to 2019. The contracts were either multi-year contracts running for several consecutive years or single-year contracts. No institute had more than one contract with the same vendor. However, in a calendar year, an institute may have had contracts with multiple vendors, and a vendor may have had contracts with multiple institutes. It is known that over the decade, the institutes each got into two contracts with two of these vendors, and each vendor got into two contracts with two of these institutes. The following facts are also known about these contracts. I. Vendor Z had at least one contract in every year. II. Vendor X had one or more contracts in every year up to 2015, but no contract in any year after that. III. Vendor Y had contracts in 2010 and 2019. Vendor W had contracts only in 2012. IV. There were five contracts in 2012. V. There were exactly four multi-year contracts. Institute B had a 7-year contract, D had a 4-year contract, and A and C had one 3-year contract each. The other four contracts were single-year contracts. VI. Institute C had one or more contracts in 2012 but did not have any contract in 2011. VII. Institutes B and D each had exactly one contract in 2012. Institute D did not have any contract in 2010.

Q45MCQLogical Puzzles

In which of the following years were there two or more contracts?
  1. 2017
  2. 2016
  3. 2015
  4. 2018
Answer and solution

Answer: (C) 2015

Each institute has one multi-year contract (B 7 years, D 4, A 3, C 3) and one single-year contract. Z covers all ten years with two contracts, so it needs B's 7-year contract and another multi-year one. X covers the six years 2010–2015 with two contracts, so it needs the other two multi-year ones. The single-year contracts are W's two (2012) and Y's two (2010, 2019). If Z held A's, X would need C or D in 2010, but D had none then and a C contract from 2010 covers 2011. If Z held D's, X would be A 2010–12 and C 2013–15, Z would be B 2010–16 and D 2016–19, and 2012 would have four contracts, not five. So Z has B 2010–16 and C 2017–19; X has A 2010–12 and D 2012–15. C needs a 2012 contract, so C has W then. B and D already have one 2012 contract each, so A has the other W. D had none in 2010, so Y goes to B (2010) and D (2019). Solution figure for question 45, CAT 2020 Slot 1 Contracts per year: 2010: 3; 2011: 2; 2012: 5; 2013–15: 2 each; 2016–18: 1 each; 2019: 2. Of the four options, only 2015 has two contracts (B–Z and D–X); options A (2017), B (2016) and D (2018) have one each. Hence, option C (2015).

Q46MCQLogical Puzzles

Which of the following is true?
  1. B had a contract with Z in 2017
  2. B had a contract with Y in 2019
  3. D had a contract with X in 2011
  4. D had a contract with Y in 2019
Answer and solution

Answer: (D) D had a contract with Y in 2019

Each institute has one multi-year contract (B 7 years, D 4, A 3, C 3) and one single-year contract. Z covers all ten years with two contracts, so it needs B's 7-year contract and another multi-year one. X covers the six years 2010–2015 with two contracts, so it needs the other two multi-year ones. The single-year contracts are W's two (2012) and Y's two (2010, 2019). If Z held A's, X would need C or D in 2010, but D had none then and a C contract from 2010 covers 2011. If Z held D's, X would be A 2010–12 and C 2013–15, Z would be B 2010–16 and D 2016–19, and 2012 would have four contracts, not five. So Z has B 2010–16 and C 2017–19; X has A 2010–12 and D 2012–15. C needs a 2012 contract, so C has W then. B and D already have one 2012 contract each, so A has the other W. D had none in 2010, so Y goes to B (2010) and D (2019). Solution figure for question 46, CAT 2020 Slot 1 D–Y is in 2019, so option D is true. Option A fails (B–Z ended in 2016), B fails (B–Y was in 2010) and C fails (D–X began in 2012). Hence, option D (D had a contract with Y in 2019).

Q47MCQLogical Puzzles

In how many years during this period was there only one contract?
  1. 3
  2. 2
  3. 4
  4. 5
Answer and solution

Answer: (A) 3

Each institute has one multi-year contract (B 7 years, D 4, A 3, C 3) and one single-year contract. Z covers all ten years with two contracts, so it needs B's 7-year contract and another multi-year one. X covers the six years 2010–2015 with two contracts, so it needs the other two multi-year ones. The single-year contracts are W's two (2012) and Y's two (2010, 2019). If Z held A's, X would need C or D in 2010, but D had none then and a C contract from 2010 covers 2011. If Z held D's, X would be A 2010–12 and C 2013–15, Z would be B 2010–16 and D 2016–19, and 2012 would have four contracts, not five. So Z has B 2010–16 and C 2017–19; X has A 2010–12 and D 2012–15. C needs a 2012 contract, so C has W then. B and D already have one 2012 contract each, so A has the other W. D had none in 2010, so Y goes to B (2010) and D (2019). Solution figure for question 47, CAT 2020 Slot 1 Contracts per year: 2010: 3; 2011: 2; 2012: 5; 2013–15: 2 each; 2016–18: 1 each; 2019: 2. Only 2016, 2017 and 2018 have one contract; every other year has at least two, so option C (4) fails. Hence, option A (3).

Q48MCQLogical Puzzles

What BEST can be concluded about the number of contracts in 2010?
  1. exactly 4
  2. exactly 3
  3. at least 3
  4. at least 4
Answer and solution

Answer: (B) exactly 3

Each institute has one multi-year contract (B 7 years, D 4, A 3, C 3) and one single-year contract. Z covers all ten years with two contracts, so it needs B's 7-year contract and another multi-year one. X covers the six years 2010–2015 with two contracts, so it needs the other two multi-year ones. The single-year contracts are W's two (2012) and Y's two (2010, 2019). If Z held A's, X would need C or D in 2010, but D had none then and a C contract from 2010 covers 2011. If Z held D's, X would be A 2010–12 and C 2013–15, Z would be B 2010–16 and D 2016–19, and 2012 would have four contracts, not five. So Z has B 2010–16 and C 2017–19; X has A 2010–12 and D 2012–15. C needs a 2012 contract, so C has W then. B and D already have one 2012 contract each, so A has the other W. D had none in 2010, so Y goes to B (2010) and D (2019). Solution figure for question 48, CAT 2020 Slot 1 In 2010 the contracts are A–X, B–Z and B–Y, and the arrangement is forced. Option C (at least 3) is true but weaker than the exact count. Hence, option B (exactly 3).

Q49MCQLogical Puzzles

Which institutes had multiple contracts during the same year?
  1. A only
  2. B and C only
  3. A and B only
  4. B only
Answer and solution

Answer: (C) A and B only

Each institute has one multi-year contract (B 7 years, D 4, A 3, C 3) and one single-year contract. Z covers all ten years with two contracts, so it needs B's 7-year contract and another multi-year one. X covers the six years 2010–2015 with two contracts, so it needs the other two multi-year ones. The single-year contracts are W's two (2012) and Y's two (2010, 2019). If Z held A's, X would need C or D in 2010, but D had none then and a C contract from 2010 covers 2011. If Z held D's, X would be A 2010–12 and C 2013–15, Z would be B 2010–16 and D 2016–19, and 2012 would have four contracts, not five. So Z has B 2010–16 and C 2017–19; X has A 2010–12 and D 2012–15. C needs a 2012 contract, so C has W then. B and D already have one 2012 contract each, so A has the other W. D had none in 2010, so Y goes to B (2010) and D (2019). Solution figure for question 49, CAT 2020 Slot 1 A has two contracts in 2012 (X and W) and B two in 2010 (Z and Y). C and D never have two at once, so option B (B and C only) fails. Hence, option C (A and B only).

Q50MCQLogical Puzzles

Which institutes and vendors had more than one contracts in any year?
  1. B, W, X, and Z
  2. A, B, W, and X
  3. A, D, W, and Z
  4. B, D, W, and X
Answer and solution

Answer: (B) A, B, W, and X

Each institute has one multi-year contract (B 7 years, D 4, A 3, C 3) and one single-year contract. Z covers all ten years with two contracts, so it needs B's 7-year contract and another multi-year one. X covers the six years 2010–2015 with two contracts, so it needs the other two multi-year ones. The single-year contracts are W's two (2012) and Y's two (2010, 2019). If Z held A's, X would need C or D in 2010, but D had none then and a C contract from 2010 covers 2011. If Z held D's, X would be A 2010–12 and C 2013–15, Z would be B 2010–16 and D 2016–19, and 2012 would have four contracts, not five. So Z has B 2010–16 and C 2017–19; X has A 2010–12 and D 2012–15. C needs a 2012 contract, so C has W then. B and D already have one 2012 contract each, so A has the other W. D had none in 2010, so Y goes to B (2010) and D (2019). Solution figure for question 50, CAT 2020 Slot 1 A has two contracts in 2012 (X, W), B two in 2010 (Z, Y), and W and X two each in 2012. Z's two contracts never overlap, so option A fails. Hence, option B (A, B, W, and X).