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CAT 2023 · Slot 3 · DILR

CAT 2023 Slot 3 DILR paper with solutions

4 sets, 20 questions, in exam order.

Set 1CAT 2023 · Slot 3 · DILR Q1–Q55 questionsTables & Missing Data

AC sales by four dealers

An air conditioner (AC) company has four dealers - D1, D2, D3 and D4 in a city. It is evaluating sales performances of these dealers. The company sells two variants of ACs - Window and Split. Both these variants can be either Inverter type or Non-inverter type. It is known that of the total number of ACs sold in the city, 25% were of Window variant, while the rest were of Split variant. Among the Inverter ACs sold, 20% were of Window variant.

The following information is also known:

1. Every dealer sold at least two window ACs.

2. D1 sold 13 inverter ACs, while D3 sold 5 Non-inverter ACs.

3. A total of six Window Non-inverter ACs and 36 Split Inverter ACs were sold in the city.

4. The number of Split ACs sold by D1 was twice the number of Window ACs sold by it.

5. D3 and D4 sold an equal number of Window ACs and this number was one-third of the number of similar ACs sold by D2.

6. D2 and D3 were the only ones who sold Window Non-inverter ACs. The number of these ACs sold by D2 was twice the number of these ACs sold by D3.

7. D3 and D4 sold an equal number of Split Inverter ACs. This number was half the number of similar ACs sold by D2.

Set solution: how to crack this set

Split Inverter =36=80%= 36 = 80\% of Inverter, so Inverter =45= 45 and Window Inverter =9= 9. Window =9+6=15=25%= 9 + 6 = 15 = 25\% of all, so total =60= 60, Split =45= 45, Split Non-inverter =9= 9.

Window per dealer: D3 == D4 =w= w, D2 =3w= 3w, each at least 2, total 15: w=2w = 2, so D1 5, D2 6, D3 2, D4 2. D1 sells 2×5=102 \times 5 = 10 Split.

Window Non-inverter: D2 4, D3 2. So Window Inverter: D1 5, D2 2, D3 0, D4 2.

D1 Inverter 13 means Split Inverter 8. The rest, 36−8=28=s+s+2s36 - 8 = 28 = s + s + 2s, gives s=7s = 7: D3 7, D4 7, D2 14.

D3 Non-inverter 5: Window 2 + Split 3. Split Non-inverter: D1 2, D3 3, and D2 + D4 =4= 4.

Totals: D1 15, D2 20+x20 + x, D3 12, D4 13−x13 - x (x = D2's Split Non-inverter).

Q1CAT 2023 · Slot 3 · DILR Q1TITA

How many Split Inverter ACs did D2 sell?

Q2CAT 2023 · Slot 3 · DILR Q2MCQ

What percentage of ACs sold were of Non-inverter type?

Q3CAT 2023 · Slot 3 · DILR Q3TITA

What was the total number of ACs sold by D2 and D4?

Q4CAT 2023 · Slot 3 · DILR Q4MCQ

Which of the following statements is necessarily false?

Q5CAT 2023 · Slot 3 · DILR Q5MCQ

If D3 and D4 sold an equal number of ACs, then what was the number of Non-inverter ACs sold by D2?

Set 2CAT 2023 · Slot 3 · DILR Q6–Q105 questionsRoutes & Networks

Police patrol teams on a street network

A, B, C, D, E and F are the six police stations in an area, which are connected by streets as shown below. Four teams - Team 1, Team 2, Team 3 and Team 4 - patrol these streets continuously between 09:00 hrs. and 12:00 hrs. each day.

Data for this set (chart, table or figure)

The teams need 30 minutes to cross a street connecting one police station to another. All four teams start from Station A at 09:00 hrs. and must return to Station A by 12:00 hrs. They can

also pass via Station A at any point on their journeys.

The following facts are known.

1. None of the streets has more than one team traveling along it in any direction at any point in time.

2. Teams 2 and 3 are the only ones in stations E and D respectively at 10:00 hrs.

3. Teams 1 and 3 are the only ones in station E at 10:30 hrs.

4. Teams 1 and 4 are the only ones in stations B and E respectively at 11:30 hrs.

5. Team 1 and Team 4 are the only teams that patrol the street connecting stations A and E.

6. Team 4 never passes through Stations B, D or F.

Set solution: how to crack this set

Streets: A–B, A–C, A–E, A–F, C–D, D–E, E–F (30 minutes each). Teams move every half hour, so each route has 6 steps from A back to A.

Team 4 avoids B, D and F, so it shuttles between A and C or A and E; it must be at E at 11:30.

Team 3 can't use A–E, so reaching D at 10:00 means going A–C–D; it is at E at 10:30 and must get back to A through D and C.

Team 2 reaches E at 10:00 through F. Team 1 is at E at 10:30 and at B at 11:30, so it goes E–A–B–A.

Street clashes at 9:00 force Team 4 onto A–E and Team 1 onto A–B.

Time9:3010:0010:3011:0011:3012:00
Team 1BAEABA
Team 2FEFA or EFA
Team 3CDEDCA
Team 4EACAEA
Q6CAT 2023 · Slot 3 · DILR Q6MCQ

Which one among the following stations is visited the largest number of times?

Q7CAT 2023 · Slot 3 · DILR Q7TITA

How many times do the teams pass through Station B in a day?

Q8CAT 2023 · Slot 3 · DILR Q8MCQ

Which team patrols the street connecting Stations D and E at 10:15 hrs?

Q9CAT 2023 · Slot 3 · DILR Q9TITA

How many times does Team 4 pass through Station E in a day?

Q10CAT 2023 · Slot 3 · DILR Q10MCQ

How many teams pass through Station C in a day?

Set 3CAT 2023 · Slot 3 · DILR Q11–Q155 questionsTables & Missing Data

Online and offline registrations by month

In a coaching class, some students register online, and some others register offline. No student registers both online and offline; hence the total registration number is the sum of online and offline registrations. The following facts and table pertain to these registration numbers for the five months - January to May of 2023. The table shows the minimum, maximum, median registration numbers of these five months, separately for online, offline and total number of registrations. The following additional facts are known.

1. In every month, both online and offline registration numbers were multiples of 10.

2. In January, the number of offline registrations was twice that of online registrations.

3. In April, the number of online registrations was twice that of offline registrations.

4. The number of online registrations in March was the same as the number of offline registrations in February.

5. The number of online registrations was the largest in May.

Data for this set (chart, table or figure)
Set solution: how to crack this set

January: offline =2×= 2 \times online with total 110–130, so 40 + 80 = 120. April: online =2×= 2 \times offline: 80 + 40 = 120.

May has the largest online (100), and its total can't exceed 130, so May is 100 + 30 = 130.

February and March must supply an offline 50 (median), a total of 110 (minimum), and an online pair around 80, with March online == February offline.

Only one choice works: February 80 + 50 = 130, March 50 + 60 = 110.

MonthOnlineOfflineTotal
Jan4080120
Feb8050130
Mar5060110
Apr8040120
May10030130
Q11CAT 2023 · Slot 3 · DILR Q11TITA

What was the total number of registrations in April?

Q12CAT 2023 · Slot 3 · DILR Q12TITA

What was the number of online registrations in January?

Q13CAT 2023 · Slot 3 · DILR Q13MCQ

Which of the following statements can be true?

I. The number of offline registrations was the smallest in May.

II. The total number of registrations was the smallest in February.

Q14CAT 2023 · Slot 3 · DILR Q14MCQ

What best can be concluded about the number of offline registrations in February?

Q15CAT 2023 · Slot 3 · DILR Q15MCQ

Which pair of months definitely had the same total number of registrations?

I. January and April

II. February and May

Set 4CAT 2023 · Slot 3 · DILR Q16–Q205 questionsScores & Rankings

Project and test scores of six students

There are only three female students - Amala, Koli and Rini - and only three male students - Biman, Mathew and Shyamal - in a course. The course has two evaluation components, a project and a test. The aggregate score in the course is a weighted average of the two components, with the weights being positive and adding to 1.

The projects are done in groups of two, with each group consisting of a female and a male student. Both the group members obtain the same score in the project.

The following additional facts are known about the scores in the project and the test.

1. The minimum, maximum and the average of both project and test scores were identical - 40, 80 and 60, respectively.

2. The test scores of the students were all multiples of 10; four of them were distinct and the remaining two were equal to the average test scores.

3. Amala’s score in the project was double that of Koli in the same, but Koli scored 20 more than Amala in the test. Yet Amala had the highest aggregate score.

4. Shyamal scored the second highest in the test. He scored two more than Koli, but two less than Amala in the aggregate.

5. Biman scored the second lowest in the test and the lowest in the aggregate.

6. Mathew scored more than Rini in the project, but less than her in the test.

Set solution: how to crack this set

Project: three pairs with min 40, max 80, average 60, so pair scores are 40, 60, 80. Amala =2×= 2 \times Koli means Amala 80 and Koli 40; Rini 60. Mathew beats Rini, so he's Amala's partner (80).

Test: two 60s plus four distinct values including 40 and 80 summing to 240, i.e. 40, 50, 70, 80. Shyamal 70 (second highest), Biman 50 (second lowest).

Amala − Koli aggregate =4= 4: w(At−Kt)+40(1−w)=40−60w=4⇒w=0.6w(A_t - K_t) + 40(1 - w) = 40 - 60w = 4 \Rightarrow w = 0.6.

Koli == Amala +20+ 20 in the test: (40, 60) or (60, 80). Checking Shyamal == Amala −2- 2, only Amala 60, Koli 80 works, with Shyamal paired with Rini (project 60). Then Mathew 40, Rini 60 in the test, and Biman is with Koli.

Aggregates: Amala 68, Shyamal 66, Koli 64, Rini 60, Mathew 56, Biman 46.

Q16CAT 2023 · Slot 3 · DILR Q16TITA

What was Rini’s score in the project?

Q17CAT 2023 · Slot 3 · DILR Q17MCQ

What was the weight of the test component?

Q18CAT 2023 · Slot 3 · DILR Q18MCQ

What was the maximum aggregate score obtained by the students?

Q19CAT 2023 · Slot 3 · DILR Q19TITA

What was Mathew’s score in the test?

Q20CAT 2023 · Slot 3 · DILR Q20MCQ

Which of the following pairs of students were part of the same project team?

i) Amala and Biman

ii) Koli and Mathew

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