CAT 2022 Slot 2 · DILR · Set 1: TRICCEK Salesmen
The full set and all 5 questions (Q1–Q5) from CAT 2022 Slot 2, with answers and solutions.
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.
Question 1 · Caselet DI
Answer & solution
Answer: 84
Solving the set
Step 1 – Lahur meets h households and sells s items each day (fact 2), so 2s = 100 ⇒ s = 50 per day. All three met the same total, N = 2h.
Step 2 – Hokli met just 1 household on Day 2 (sold 0), so on Day 1 he met N − 1 households and sold 100 items.
Step 3 – Tohri meets t on Day 1 and t + 30 on Day 2: 2t + 30 = 2h ⇒ h = t + 15. Let Tohri sell x items on Day 1 and 100 − x on Day 2.
Step 4 – Fact 5. Day 1: x/t = 2 × 50/(t + 15) ⇒ x = 100t/(t + 15). Day 2: (100 − x)/(t + 30) = 0.75 × 50/(t + 15). Since 100 − x = 1500/(t + 15), this gives 1500 = 37.5(t + 30) ⇒ t = 10, x = 40, h = 25, N = 50.
Salesman | Day 1 households | Day 1 items | Day 1 success rate | Day 2 households | Day 2 items | Day 2 success rate | Total households | Total items |
Tohri | 10 | 40 | 4.00 | 40 | 60 | 1.50 | 50 | 100 |
Hokli | 49 | 100 | 2.04 | 1 | 0 | 0 | 50 | 100 |
Lahur | 25 | 50 | 2.00 | 25 | 50 | 2.00 | 50 | 100 |
Total | 84 | 190 | 66 | 110 | 150 | 300 |
This question
From the table, Day 1 households: Tohri 10 + Hokli 49 + Lahur 25 = 84.
Hence, the answer is 84.
Question 2 · Caselet DI
Answer & solution
Answer: 40
Solving the set
Step 1 – Lahur meets h households and sells s items each day (fact 2), so 2s = 100 ⇒ s = 50 per day. All three met the same total, N = 2h.
Step 2 – Hokli met just 1 household on Day 2 (sold 0), so on Day 1 he met N − 1 households and sold 100 items.
Step 3 – Tohri meets t on Day 1 and t + 30 on Day 2: 2t + 30 = 2h ⇒ h = t + 15. Let Tohri sell x items on Day 1 and 100 − x on Day 2.
Step 4 – Fact 5. Day 1: x/t = 2 × 50/(t + 15) ⇒ x = 100t/(t + 15). Day 2: (100 − x)/(t + 30) = 0.75 × 50/(t + 15). Since 100 − x = 1500/(t + 15), this gives 1500 = 37.5(t + 30) ⇒ t = 10, x = 40, h = 25, N = 50.
Salesman | Day 1 households | Day 1 items | Day 1 success rate | Day 2 households | Day 2 items | Day 2 success rate | Total households | Total items |
Tohri | 10 | 40 | 4.00 | 40 | 60 | 1.50 | 50 | 100 |
Hokli | 49 | 100 | 2.04 | 1 | 0 | 0 | 50 | 100 |
Lahur | 25 | 50 | 2.00 | 25 | 50 | 2.00 | 50 | 100 |
Total | 84 | 190 | 66 | 110 | 150 | 300 |
This question
From Step 4, Tohri sold x = 40 items on Day 1.
Hence, the answer is 40.
Question 3 · Caselet DI
- Abetween 21 and 29
- B20 or less
- Cmore than 35
- Dbetween 30 and 35
Answer & solution
Answer: Option A
Solving the set
Step 1 – Lahur meets h households and sells s items each day (fact 2), so 2s = 100 ⇒ s = 50 per day. All three met the same total, N = 2h.
Step 2 – Hokli met just 1 household on Day 2 (sold 0), so on Day 1 he met N − 1 households and sold 100 items.
Step 3 – Tohri meets t on Day 1 and t + 30 on Day 2: 2t + 30 = 2h ⇒ h = t + 15. Let Tohri sell x items on Day 1 and 100 − x on Day 2.
Step 4 – Fact 5. Day 1: x/t = 2 × 50/(t + 15) ⇒ x = 100t/(t + 15). Day 2: (100 − x)/(t + 30) = 0.75 × 50/(t + 15). Since 100 − x = 1500/(t + 15), this gives 1500 = 37.5(t + 30) ⇒ t = 10, x = 40, h = 25, N = 50.
Salesman | Day 1 households | Day 1 items | Day 1 success rate | Day 2 households | Day 2 items | Day 2 success rate | Total households | Total items |
Tohri | 10 | 40 | 4.00 | 40 | 60 | 1.50 | 50 | 100 |
Hokli | 49 | 100 | 2.04 | 1 | 0 | 0 | 50 | 100 |
Lahur | 25 | 50 | 2.00 | 25 | 50 | 2.00 | 50 | 100 |
Total | 84 | 190 | 66 | 110 | 150 | 300 |
This question
From Step 4, Lahur met h = 25 households each day, which is between 21 and 29.
Hence, the answer is option 1.
Question 4 · Caselet DI
- Abetween 21 and 40
- Bbetween 11 and 20
- Cmore than 40
- D10 or less
Answer & solution
Answer: Option D
Solving the set
Step 1 – Lahur meets h households and sells s items each day (fact 2), so 2s = 100 ⇒ s = 50 per day. All three met the same total, N = 2h.
Step 2 – Hokli met just 1 household on Day 2 (sold 0), so on Day 1 he met N − 1 households and sold 100 items.
Step 3 – Tohri meets t on Day 1 and t + 30 on Day 2: 2t + 30 = 2h ⇒ h = t + 15. Let Tohri sell x items on Day 1 and 100 − x on Day 2.
Step 4 – Fact 5. Day 1: x/t = 2 × 50/(t + 15) ⇒ x = 100t/(t + 15). Day 2: (100 − x)/(t + 30) = 0.75 × 50/(t + 15). Since 100 − x = 1500/(t + 15), this gives 1500 = 37.5(t + 30) ⇒ t = 10, x = 40, h = 25, N = 50.
Salesman | Day 1 households | Day 1 items | Day 1 success rate | Day 2 households | Day 2 items | Day 2 success rate | Total households | Total items |
Tohri | 10 | 40 | 4.00 | 40 | 60 | 1.50 | 50 | 100 |
Hokli | 49 | 100 | 2.04 | 1 | 0 | 0 | 50 | 100 |
Lahur | 25 | 50 | 2.00 | 25 | 50 | 2.00 | 50 | 100 |
Total | 84 | 190 | 66 | 110 | 150 | 300 |
This question
From Step 4, Tohri met 10 households on Day 1 (10 or less).
Hence, the answer is option 4.
Question 5 · Caselet DI
- AAmong the three, Tohri had the highest success rate on the second day.
- BTohri had a higher success rate on the first day compared to the second day.
- CAmong the three, Tohri had the highest success rate on the first day.
- DAmong the three, Lahur had the lowest success rate on the first day.
Answer & solution
Answer: Option A
Solving the set
Step 1 – Lahur meets h households and sells s items each day (fact 2), so 2s = 100 ⇒ s = 50 per day. All three met the same total, N = 2h.
Step 2 – Hokli met just 1 household on Day 2 (sold 0), so on Day 1 he met N − 1 households and sold 100 items.
Step 3 – Tohri meets t on Day 1 and t + 30 on Day 2: 2t + 30 = 2h ⇒ h = t + 15. Let Tohri sell x items on Day 1 and 100 − x on Day 2.
Step 4 – Fact 5. Day 1: x/t = 2 × 50/(t + 15) ⇒ x = 100t/(t + 15). Day 2: (100 − x)/(t + 30) = 0.75 × 50/(t + 15). Since 100 − x = 1500/(t + 15), this gives 1500 = 37.5(t + 30) ⇒ t = 10, x = 40, h = 25, N = 50.
Salesman | Day 1 households | Day 1 items | Day 1 success rate | Day 2 households | Day 2 items | Day 2 success rate | Total households | Total items |
Tohri | 10 | 40 | 4.00 | 40 | 60 | 1.50 | 50 | 100 |
Hokli | 49 | 100 | 2.04 | 1 | 0 | 0 | 50 | 100 |
Lahur | 25 | 50 | 2.00 | 25 | 50 | 2.00 | 50 | 100 |
Total | 84 | 190 | 66 | 110 | 150 | 300 |
This question
Success rates from the table – Day 1: Tohri 4, Hokli 2.04, Lahur 2; Day 2: Tohri 1.5, Hokli 0, Lahur 2.
• 'Tohri highest on Day 2' – false (Lahur's 2 is higher).
• 'Tohri higher on Day 1 than Day 2' – true (4 > 1.5).
• 'Tohri highest on Day 1' – true. 'Lahur lowest on Day 1' – true (2 < 2.04 < 4).
Hence, the answer is option 1.
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