CATin
  1. CATin
  2. CAT past papers
  3. CAT 2022 Slot 2
  4. DILR

CAT 2022 Slot 2 — DILR questions with answers

All 19 questions of the Data Interpretation & Logical Reasoning section (14 MCQs, 5 TITA, 4 sets). Try each one, then open its answer and solution. 1 question of the original section is left out while its answer key is checked.

CAT 2022 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. 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 2022 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: A few salesmen are employed to sell a product called TRICCEK among households in various housing complexes. On each day, a salesman is assigned to visit one housing complex. Once a salesman enters a housing complex, he can meet any number of households in the time available. However, if a household makes a complaint against the salesman, then he must leave the housing complex immediately and cannot meet any other household on that day. A household may buy any number of TRICCEK items or may not buy any item. The salesman needs to record the total number of TRICCEK items sold as well as the number of households met in each day. The success rate of a salesman for a day is defined as the ratio of the number of items sold to the number of households met on that day. Some details about the performances of three salesmen - Tohri, Hokli and Lahur, on two particular days are given below. 1. Over the two days, all three of them met the same total number of households, and each of them sold a total of 100 items. 2. On both days, Lahur met the same number of households and sold the same number of items. 3. Hokli could not sell any item on the second day because the first household he met on that day complained against him. 4. Tohri met 30 more households on the second day than on the first day. 5. Tohri’s success rate was twice that of Lahur’s on the first day, and it was 75% of Lahur’s on the second day.

Q25TITATables & Caselets

What was the total number of households met by Tohri, Hokli and Lahur on the first day?

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

Answer and solution

Answer: 84

Lahur met the same number of households, say xx, on each day and sold the same number of items; his total is 100, so he sold 50 each day. By statement 1, each salesman met 2x2x households in all. Hokli: the first household on day 2 complained, so on day 2 he met 1 household and sold 0. On day 1 he met 2x−12x-1 households and sold 100. Tohri: if he met aa on day 1, then a+(a+30)=2xa + (a+30) = 2x, so he met x−15x-15 on day 1 and x+15x+15 on day 2. Let him sell yy items on day 1 and 100−y100-y on day 2. Statement 5, day 1: yx−15=2×50x\frac{y}{x-15} = 2 \times \frac{50}{x}, so y=100(x−15)xy = \frac{100(x-15)}{x}. Day 2: 100−yx+15=34×50x\frac{100-y}{x+15} = \frac{3}{4} \times \frac{50}{x}, so 100−y=37.5(x+15)x100-y = \frac{37.5(x+15)}{x}. Adding: 100x=100x−1500+37.5x+562.5100x = 100x - 1500 + 37.5x + 562.5, so x=25x = 25 and y=40y = 40. Solution figure for question 25, CAT 2022 Slot 2 Day 1: Tohri 10 households, 40 items; Hokli 49, 100; Lahur 25, 50. Day 2: Tohri 40, 60; Hokli 1, 0; Lahur 25, 50. Households met on day 1: 10+49+25=8410 + 49 + 25 = 84. The answer is 84.

Q26TITATables & Caselets

How many TRICCEK items were sold by Tohri on the first day?

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

Answer and solution

Answer: 40

Lahur met the same number of households, say xx, on each day and sold the same number of items; his total is 100, so he sold 50 each day. By statement 1, each salesman met 2x2x households in all. Hokli: the first household on day 2 complained, so on day 2 he met 1 household and sold 0. On day 1 he met 2x−12x-1 households and sold 100. Tohri: if he met aa on day 1, then a+(a+30)=2xa + (a+30) = 2x, so he met x−15x-15 on day 1 and x+15x+15 on day 2. Let him sell yy items on day 1 and 100−y100-y on day 2. Statement 5, day 1: yx−15=2×50x\frac{y}{x-15} = 2 \times \frac{50}{x}, so y=100(x−15)xy = \frac{100(x-15)}{x}. Day 2: 100−yx+15=34×50x\frac{100-y}{x+15} = \frac{3}{4} \times \frac{50}{x}, so 100−y=37.5(x+15)x100-y = \frac{37.5(x+15)}{x}. Adding: 100x=100x−1500+37.5x+562.5100x = 100x - 1500 + 37.5x + 562.5, so 37.5x=937.537.5x = 937.5 and x=25x = 25. Then y=100×(25−15)25=40y = \frac{100 \times (25-15)}{25} = 40. Check day 2: Tohri sold 60 to 40 households, a rate of 1.5, which is 34\frac{3}{4} of Lahur's 2. Solution figure for question 26, CAT 2022 Slot 2 Day 1: Tohri 10 households, 40 items; Hokli 49, 100; Lahur 25, 50. Day 2: Tohri 40, 60; Hokli 1, 0; Lahur 25, 50. The answer is 40.

Q27MCQTables & Caselets

How many households did Lahur meet on the second day?
  1. between 21 and 29
  2. 20 or less
  3. more than 35
  4. between 30 and 35
Answer and solution

Answer: (A) between 21 and 29

Lahur met the same number of households, say xx, on each day and sold the same number of items; his total is 100, so he sold 50 each day. By statement 1, each salesman met 2x2x households in all. Hokli: the first household on day 2 complained, so on day 2 he met 1 household and sold 0. On day 1 he met 2x−12x-1 households and sold 100. Tohri: if he met aa on day 1, then a+(a+30)=2xa + (a+30) = 2x, so he met x−15x-15 on day 1 and x+15x+15 on day 2. Let him sell yy items on day 1 and 100−y100-y on day 2. Statement 5, day 1: yx−15=2×50x\frac{y}{x-15} = 2 \times \frac{50}{x}, so y=100(x−15)xy = \frac{100(x-15)}{x}. Day 2: 100−yx+15=34×50x\frac{100-y}{x+15} = \frac{3}{4} \times \frac{50}{x}, so 100−y=37.5(x+15)x100-y = \frac{37.5(x+15)}{x}. Adding: 100x=100x−1500+37.5x+562.5100x = 100x - 1500 + 37.5x + 562.5, so 37.5x=937.537.5x = 937.5 and x=25x = 25. Then y=40y = 40. Solution figure for question 27, CAT 2022 Slot 2 Day 1: Tohri 10 households, 40 items; Hokli 49, 100; Lahur 25, 50. Day 2: Tohri 40, 60; Hokli 1, 0; Lahur 25, 50. Lahur met x=25x = 25 households on day 2. That lies between 21 and 29; none of the other ranges (20 or less, 30 to 35, more than 35) contains 25. Hence, option A (between 21 and 29).

Q28MCQTables & Caselets

How many households did Tohri meet on the first day?
  1. between 21 and 40
  2. between 11 and 20
  3. more than 40
  4. 10 or less
Answer and solution

Answer: (D) 10 or less

Lahur met the same number of households, say xx, on each day and sold the same number of items; his total is 100, so he sold 50 each day. By statement 1, each salesman met 2x2x households in all. Hokli: the first household on day 2 complained, so on day 2 he met 1 household and sold 0. On day 1 he met 2x−12x-1 households and sold 100. Tohri: if he met aa on day 1, then a+(a+30)=2xa + (a+30) = 2x, so he met x−15x-15 on day 1 and x+15x+15 on day 2. Let him sell yy items on day 1 and 100−y100-y on day 2. Statement 5, day 1: yx−15=2×50x\frac{y}{x-15} = 2 \times \frac{50}{x}, so y=100(x−15)xy = \frac{100(x-15)}{x}. Day 2: 100−yx+15=34×50x\frac{100-y}{x+15} = \frac{3}{4} \times \frac{50}{x}, so 100−y=37.5(x+15)x100-y = \frac{37.5(x+15)}{x}. Adding: 100x=100x−1500+37.5x+562.5100x = 100x - 1500 + 37.5x + 562.5, so 37.5x=937.537.5x = 937.5 and x=25x = 25. Then y=40y = 40. Solution figure for question 28, CAT 2022 Slot 2 Day 1: Tohri 10 households, 40 items; Hokli 49, 100; Lahur 25, 50. Day 2: Tohri 40, 60; Hokli 1, 0; Lahur 25, 50. Tohri met x−15=10x - 15 = 10 households on day 1. That is '10 or less'; option B's range begins at 11. Hence, option D (10 or less).

Q29MCQTables & Caselets

Which of the following statements is FALSE?
  1. Among the three, Tohri had the highest success rate on the second day.
  2. Tohri had a higher success rate on the first day compared to the second day.
  3. Among the three, Tohri had the highest success rate on the first day.
  4. Among the three, Lahur had the lowest success rate on the first day.
Answer and solution

Answer: (A) Among the three, Tohri had the highest success rate on the second day.

Lahur met xx households each day and sold 50 each day (equal days, total 100). Each salesman met 2x2x in all. Hokli's first household on day 2 complained, so he met 1 and sold 0 that day; on day 1 he met 2x−12x-1 and sold 100. Tohri met x−15x-15 on day 1 and x+15x+15 on day 2 (30 more, total 2x2x). Let him sell yy on day 1 and 100−y100-y on day 2. Statement 5: yx−15=100x\frac{y}{x-15} = \frac{100}{x} and 100−yx+15=37.5x\frac{100-y}{x+15} = \frac{37.5}{x}. Adding yy and 100−y100-y: 100x=100(x−15)+37.5(x+15)100x = 100(x-15) + 37.5(x+15), so x=25x = 25 and y=40y = 40. Solution figure for question 29, CAT 2022 Slot 2 Success rate = items ÷ households. Day 1: Tohri 4010=4\frac{40}{10} = 4, Hokli 10049≈2.04\frac{100}{49} \approx 2.04, Lahur 5025=2\frac{50}{25} = 2. Day 2: Tohri 6040=1.5\frac{60}{40} = 1.5, Hokli 01=0\frac{0}{1} = 0, Lahur 2. A: on day 2 Lahur's 2 beats Tohri's 1.5, so A is false. B: Tohri's 4 on day 1 exceeds his 1.5 on day 2, so B is true. C: Tohri's 4 is the highest on day 1, so C is true. D: Lahur's 2 is just below Hokli's 2.04, so D is true. Hence, option A (Among the three, Tohri had the highest success rate on the second day.).

Data set

Set for questions 31–34

DIRECTIONS for questions 30-34: Every day a widget supplier supplies widgets from the warehouse (W) to four locations - Ahmednagar (A), Bikrampore (B), Chitrachak (C), and Deccan Park (D).
DILR Set 30-34 Diagram
The daily demand for widgets in each location is uncertain and independent of each other. Demands and corresponding probability values (in parenthesis) are given against each location (A, B, C, and D) in the figure below. For example, there is a 40% chance that the demand in Ahmednagar will be 50 units and a 60% chance that the demand will be 70 units. The lines in the figure connecting the locations and warehouse represent two-way roads connecting those places with the distances (in km) shown beside the line. The distances in both the directions along a road are equal. For example, the road from Ahmednagar to Bikrampore and the road from Bikrampore to Ahmednagar are both 6 km long. Every day the supplier gets the information about the demand values of the four locations and creates the travel route that starts from the warehouse and ends at a location after visiting all the locations exactly once. While making the route plan, the supplier goes to the locations in decreasing order of demand. If there is a tie for the choice of the next location, the supplier will go to the location closest to the current location. Also, while creating the route, the supplier can either follow the direct path (if available) from one location to another or can take the path via the warehouse. If both paths are available (direct and via warehouse), the supplier will choose the path with minimum distance.

Q31TITANetworks & Routes

If the total number of widgets delivered in a day is 250 units, then what is the total distance covered in the route (in km)?

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

Answer and solution

Answer: 38

Shortest distances between locations, using the direct road or the path via W, whichever is shorter: A–B 6, A–C 5+12=175+12=17 (no direct road), A–D 7 (via W, shorter than the direct 8), B–C 4, B–D 10+2=1210+2=12 (no direct road), C–D 6. From W: A 5, B 10, C 12, D 2. The largest possible total demand is 70+60+100+50=28070+60+100+50 = 280. A total of 250 is 30 less. The possible drops are A 20, B 20, C 30 and D 20; no combination of 20s makes 30, so only C drops. The demands are A 70, B 60, C 70, D 50. Order by decreasing demand: A and C tie at 70. The tie goes to the location closer to the current point, W: A is 5 km away and C is 12 km, so A comes first, then C, then B (60), then D (50). Route W–A–C–B–D: 5+17+4+12=385 + 17 + 4 + 12 = 38 km. The answer is 38.

Q32MCQNetworks & Routes

What is the chance that the total number of widgets delivered in a day is 260 units and the route ends at Bikrampore?
  1. 33.33%
  2. 10.80%
  3. 17.64%
  4. 7.56%
Answer and solution

Answer: (D) 7.56%

The maximum total demand is 70+60+100+50=28070+60+100+50 = 280. A total of 260 means exactly one location drops by 20 from its higher value. C can only drop by 30, so C = 100 and C is visited first. The drop is at A, B or D. The route ends at B only if B comes last: If A drops (A 50, B 60, D 50), the order is C, B, then A and D, so the route ends at A or D. If D drops (A 70, B 60, D 30), the order is C, A, B, D, ending at D. If B drops (A 70, B 40, D 50), the order is C, A, D, B, ending at B. So the only case is C 100, A 70, D 50, B 40. The demands are independent, so the probability is 0.7×0.6×0.6×0.3=0.0756=7.56%0.7 \times 0.6 \times 0.6 \times 0.3 = 0.0756 = 7.56\%. Option B, 10.80%, is 0.6×0.6×0.30.6 \times 0.6 \times 0.3: it leaves out the 0.7 chance that C's demand is 100. Hence, option D (7.56%).

Q33MCQNetworks & Routes

If the first location visited from the warehouse is Ahmednagar, then what is the chance that the total distance covered in the route is 40 km?
  1. 18%
  2. 5.4%
  3. 3.24%
  4. 30%
Answer and solution

Answer: (A) 18%

Shortest distances: W–A 5, A–C 17 (via W, no direct road), C–B 4, C–D 6, B–D 12 (via W, no direct road). A is visited first only if its demand is the highest. If A = 50, C (at least 70) would come first, so A = 70. C must not exceed A, so C = 70; in the tie A wins because it is closer to W (5 km against 12 km). So the condition means A = 70 and C = 70, and the route starts W–A–C: 5+17=225 + 17 = 22 km. A total of 40 km needs 18 km more. C–B–D: 4+12=164 + 12 = 16 km, total 38. C–D–B: 6+12=186 + 12 = 18 km, total 40. D comes before B only when D's demand exceeds B's, i.e. D = 50 and B = 40. The probability is 0.6×0.3=0.18=18%0.6 \times 0.3 = 0.18 = 18\%. This is a chance given that A is first, so the chances of A = 70 and C = 70 are not multiplied in. Option C, 3.24%, is 0.6×0.3×0.180.6 \times 0.3 \times 0.18, the chance without that condition. Hence, option A (18%).

Q34MCQNetworks & Routes

If Ahmednagar is not the first location to be visited in a route and the total route distance is 29 km, then which of the following is a possible number of widgets delivered on that day?
  1. 210
  2. 220
  3. 200
  4. 250
Answer and solution

Answer: (A) 210

Shortest distances: W–C 12; C–B 4, C–D 6, C–A 17 (via W); A–B 6, A–D 7 (via W, shorter than the direct 8); B–D 12 (via W). The first location has the highest demand. A is not first, and B (at most 60) and D (at most 50) are always below C (at least 70), so C is first. W to C is 12 km, leaving 29−12=1729 - 12 = 17 km for the other three. Orders from C: C–B–A–D is 4+6+7=174+6+7 = 17; C–B–D–A is 23; C–D–A–B is 19; C–D–B–A is 24; going from C straight to A already costs 17. So the order is C, B, A, D. B before A needs B ≥ A. If A = 70 it would beat B, so A = 50, and then B = 60 (40 would fall below A). D after A: D = 30, or D = 50, a tie that A wins because it is closer to B (6 km against 12 km). C is 70 or 100. Possible totals: 70+60+50+30=21070+60+50+30 = 210, then 230, 240 and 260. Only 210 is among the options; 220, 200 and 250 cannot occur. Hence, option A (210).

Data set

Set for questions 35–39

DIRECTIONS for questions 35-39: The two plots below show data for four companies code-named A, B, C, and D over three years - 2019, 2020, and 2021.
DILR Set 35-39 Diagram
DILR Set 35-39 Diagram 2
The first plot shows the revenues and costs incurred by the companies during these years. For example, in 2021, company C earned Rs.100 crores in revenue and spent Rs.30 crores. The profit of a company is defined as its revenue minus its costs The second plot shows the number of employees employed by the company (employee strength) at the start of each of these three years, as well as the number of new employees hired each year (new hires). For example, Company B had 250 employees at the start of 2021, and 30 new employees joined the company during the year.

Q35MCQPie & Scatter Charts

Considering all three years, which company had the highest annual profit?
  1. Company A
  2. Company D
  3. Company B
  4. Company C
Answer and solution

Answer: (D) Company C

Profit = revenue − cost. The values read from the first plot (Rs. crores) are: Solution figure for question 35, CAT 2022 Slot 2 Annual profits: A: 90−85=590-85=5 (2019), 90−65=2590-65=25 (2020), 60−30=3060-30=30 (2021). B: 100−75=25100-75=25, 90−40=5090-40=50, 30−30=030-30=0. C: 25−20=525-20=5, 70−60=1070-60=10, 100−30=70100-30=70. D: 50−40=1050-40=10, 20−50=−3020-50=-30, 70−70=070-70=0. The highest profit in any single year is C's 70 crores in 2021. The nearest rival is B, whose best year, 2020, gives only 50. Reading the question as total profit over the three years gives the same answer: C 85, B 75, A 60, D −20. Hence, option D (Company C).

Q36MCQPie & Scatter Charts

Which of the four companies experienced the highest annual loss in any of the years?
  1. Company C
  2. Company A
  3. Company B
  4. Company D
Answer and solution

Answer: (D) Company D

Profit = revenue − cost. The values read from the first plot (Rs. crores) are: Solution figure for question 36, CAT 2022 Slot 2 A company makes a loss only when its cost exceeds its revenue. In 2019 every company's cost is below its revenue. In 2020 only D's is above: revenue 20, cost 50. In 2021 the costs of B (30, 30) and D (70, 70) equal their revenues, so their profit is zero, not a loss. So the only loss in any year is D's loss of 50−20=3050 - 20 = 30 crores in 2020. Companies A, B and C never make a loss in any of the three years. Hence, option D (Company D).

Q37MCQPie & Scatter Charts

The ratio of a company's annual profit to its annual costs is a measure of its performance. Which of the four companies had the lowest value of this ratio in 2019?
  1. Company A
  2. Company D
  3. Company B
  4. Company C
Answer and solution

Answer: (A) Company A

The ratio is profit ÷ cost = (revenue − cost) ÷ cost. The values read from the first plot (Rs. crores) are: Solution figure for question 37, CAT 2022 Slot 2 For 2019: A: 90−8585=585≈0.06\frac{90-85}{85} = \frac{5}{85} \approx 0.06 B: 100−7575=2575≈0.33\frac{100-75}{75} = \frac{25}{75} \approx 0.33 C: 25−2020=520=0.25\frac{25-20}{20} = \frac{5}{20} = 0.25 D: 50−4040=1040=0.25\frac{50-40}{40} = \frac{10}{40} = 0.25 A has the lowest ratio. C also made a profit of only 5 crores, but on a cost of 20 rather than 85, so its ratio is much higher. Hence, option A (Company A).

Q38MCQPie & Scatter Charts

The total number of employees lost in 2019 and 2020 was the least for:
  1. Company B
  2. Company D
  3. Company A
  4. Company C
Answer and solution

Answer: (A) Company B

The employee strengths and new hires read from the second plot are: Solution figure for question 38, CAT 2022 Slot 2 Employees lost in a year = strength at the start of the year + new hires − strength at the start of the next year. A: 2019: 150+20−140=30150 + 20 - 140 = 30; 2020: 140+35−150=25140 + 35 - 150 = 25; total 55. B: 2019: 210+35−240=5210 + 35 - 240 = 5; 2020: 240+45−250=35240 + 45 - 250 = 35; total 40. C: 2019: 320+45−320=45320 + 45 - 320 = 45; 2020: 320+40−320=40320 + 40 - 320 = 40; total 85. D: 2019: 400+30−410=20400 + 30 - 410 = 20; 2020: 410+35−400=45410 + 35 - 400 = 45; total 65. B lost the fewest, 40. The closest is A, with 55. Hence, option A (Company B).

Q39MCQPie & Scatter Charts

Profit per employee is the ratio of a company's profit to its employee strength. For this purpose, the employee strength in a year is the average of the employee strength at the beginning of that year and the beginning of the next year. In 2020, which of the four companies had the highest profit per employee?
  1. Company D
  2. Company C
  3. Company B
  4. Company A
Answer and solution

Answer: (C) Company B

Profit comes from the first plot and employee strength from the second: Solution figure for question 39, CAT 2022 Slot 2 Solution figure for question 39, CAT 2022 Slot 2 2020 profit (Rs. crores): A 90−65=2590 - 65 = 25, B 90−40=5090 - 40 = 50, C 70−60=1070 - 60 = 10, D 20−50=−3020 - 50 = -30. Employee strength for 2020 = (start of 2020 + start of 2021) ÷ 2: A 140+1502=145\frac{140+150}{2} = 145, B 240+2502=245\frac{240+250}{2} = 245, C 320, D 410+4002=405\frac{410+400}{2} = 405. Profit per employee: A: 25145≈0.172\frac{25}{145} \approx 0.172 B: 50245≈0.204\frac{50}{245} \approx 0.204 C: 10320≈0.031\frac{10}{320} \approx 0.031 D: negative, since it made a loss. B is the highest. A is the closest: 25145=50290\frac{25}{145} = \frac{50}{290}, which is less than 50245\frac{50}{245}. Hence, option C (Company B).

Data set

Set for questions 40–44

DIRECTIONS for questions 40-44: A speciality supermarket sells 320 products. Each of these products was either a cosmetic product or a nutrition product. Each of these products was also either a foreign product or a domestic product. Each of these products had at least one of the two approvals - FDA or EU. The following facts are also known: 1. There were equal numbers of domestic and foreign products. 2. Half of the domestic products were FDA approved cosmetic products. 3. None of the foreign products had both the approvals, while 60 domestic products had both the approvals. 4. There were 140 nutrition products, half of them were foreign products. 5. There were 200 FDA approved products. 70 of them were foreign products and 120 of them were cosmetic products.

Q40TITASet Theory

How many foreign products were FDA approved cosmetic products?

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

Answer and solution

Answer: 40

Statement 1: there are 320 products, so domestic and foreign products are 160 each. Statement 2: half of the 160 domestic products, i.e. 80, are FDA-approved cosmetic products. Statement 5: 120 of the 200 FDA-approved products are cosmetic products. Each of these 120 is either domestic or foreign. So the foreign FDA-approved cosmetic products number 120−80=40120 - 80 = 40. Check: statement 5 also says 70 FDA-approved products are foreign, so the remaining 70−40=3070 - 40 = 30 foreign FDA-approved products are nutrition products. That fits statement 4, which gives 70 foreign nutrition products. The answer is 40.

Q41MCQSet Theory

How many cosmetic products did not have FDA approval?
  1. 10
  2. Cannot be determined
  3. 50
  4. 60
Answer and solution

Answer: (D) 60

Statement 1: domestic and foreign products are 160 each. Statement 4: of the 140 nutrition products, 70 are foreign and 70 domestic, so there are 160−70=90160 - 70 = 90 foreign and 90 domestic cosmetic products. Domestic cosmetics: statement 2 says 80 of them are FDA-approved, so 90−80=1090 - 80 = 10 have no FDA approval. Foreign cosmetics: statement 5 gives 120 FDA-approved cosmetics, 80 of them domestic, so 120−80=40120 - 80 = 40 foreign cosmetics are FDA-approved and 90−40=5090 - 40 = 50 are not. Every product has at least one approval, so these products have EU approval only. Cosmetic products without FDA approval: 10+50=6010 + 50 = 60. Option A, 10, counts only the domestic ones, and option C, 50, only the foreign ones. The count is fully determined, so option B is also wrong. Hence, option D (60).

Q42MCQSet Theory

Which among the following options best represents the number of domestic cosmetic products that had both the approvals?
  1. At least 10 and at most 60
  2. At least 10 and at most 80
  3. At least 20 and at most 70
  4. At least 20 and at most 50
Answer and solution

Answer: (A) At least 10 and at most 60

Domestic and foreign products are 160 each (statement 1). The 140 nutrition products split 70 foreign and 70 domestic (statement 4), so there are 90 domestic cosmetics, 80 of them FDA-approved (statement 2). FDA products number 200, of which 70 are foreign, so 130 are domestic. Domestic FDA nutrition products: 130−80=50130 - 80 = 50. Among domestic products, let aa = cosmetics with both approvals, bb = cosmetics with FDA only, cc = nutrition with both, dd = nutrition with FDA only. Solution figure for question 42, CAT 2022 Slot 2 Then a+b=80a + b = 80, c+d=50c + d = 50 and, by statement 3, a+c=60a + c = 60. Least aa: cc is at most 50, so a≥60−50=10a \ge 60 - 50 = 10. Greatest aa: cc can be 0, so a≤60a \le 60, with b=20b = 20. The top pair shows a=10a = 10; the bottom pair shows a=60a = 60: Solution figure for question 42, CAT 2022 Slot 2 So aa is at least 10 and at most 60. Option B's upper limit of 80 would need c=−20c = -20. Hence, option A (At least 10 and at most 60).

Q43MCQSet Theory

If 70 cosmetic products did not have EU approval, then how many nutrition products had both the approvals?
  1. 50
  2. 30
  3. 10
  4. 20
Answer and solution

Answer: (C) 10

Domestic and foreign products are 160 each, with 90 cosmetic and 70 nutrition products in each group (statements 1 and 4). 80 domestic cosmetics are FDA-approved (statement 2). Of the 120 FDA cosmetics, 120−80=40120 - 80 = 40 are foreign, and they have FDA only, since no foreign product has both approvals (statement 3). Among domestic products, let aa = cosmetics with both approvals, bb = cosmetics with FDA only, cc = nutrition with both, dd = nutrition with FDA only. Domestic FDA products number 200−70=130200 - 70 = 130, so a+b=80a + b = 80, c+d=50c + d = 50, and a+c=60a + c = 60 (statement 3). Solution figure for question 43, CAT 2022 Slot 2 Cosmetics without EU approval are those with FDA only: 40+b=7040 + b = 70, so b=30b = 30. Then a=80−30=50a = 80 - 30 = 50 and c=60−50=10c = 60 - 50 = 10. Solution figure for question 43, CAT 2022 Slot 2 Foreign products never have both approvals, so the nutrition products with both are just c=10c = 10. Option A, 50, is aa, the domestic cosmetics with both approvals, not nutrition products. Hence, option C (10).

Q44TITASet Theory

If 50 nutrition products did not have EU approval, then how many domestic cosmetic products did not have EU approval?

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

Answer and solution

Answer: 50

Domestic and foreign products are 160 each, with 90 cosmetic and 70 nutrition products in each group (statements 1 and 4). 80 domestic cosmetics are FDA-approved (statement 2). Of the 200 FDA products, 70 are foreign and 130 domestic, so 130−80=50130 - 80 = 50 domestic nutrition products are FDA-approved. Of the 120 FDA cosmetics, 120−80=40120 - 80 = 40 are foreign, so 70−40=3070 - 40 = 30 foreign nutrition products are FDA-approved, and they have FDA only, since no foreign product has both approvals. Among domestic products, let aa = cosmetics with both approvals, bb = cosmetics with FDA only, cc = nutrition with both, dd = nutrition with FDA only. Then a+b=80a + b = 80, c+d=50c + d = 50 and a+c=60a + c = 60 (statement 3). Solution figure for question 44, CAT 2022 Slot 2 Nutrition products without EU approval have FDA only: 30+d=5030 + d = 50, so d=20d = 20. Then c=50−20=30c = 50 - 20 = 30, a=60−30=30a = 60 - 30 = 30 and b=80−30=50b = 80 - 30 = 50. Solution figure for question 44, CAT 2022 Slot 2 Domestic cosmetic products without EU approval are those with FDA only: b=50b = 50. The answer is 50.