CATwithKiran
Home › DILR › Papers › CAT 2020 Slot 3
CAT 2020 · Slot 3 · DILR

CAT 2020 Slot 3 DILR paper with solutions

5 sets, 24 questions, in exam order.

Set 1CAT 2020 · Slot 3 · DILR Q1–Q44 questionsLogic Puzzles

Pooled blood tests in eight vials

Sixteen patients in a hospital must undergo a blood test for a disease. It is known that exactly one of them has the disease. The hospital has only eight testing kits and has decided to pool blood samples of patients into eight vials for the tests. The patients are numbered 1 through 16, and the vials are labelled A, B, C, D, E, F, G, and H. The following table shows the vials into which each patient’s blood sample is distributed.

Data for this set (chart, table or figure)

If a patient has the disease, then each vial containing his/her blood sample will test positive. If a vial tests positive, one of the patients whose blood samples were mixed in the vial has the disease. If a vial tests negative, then none of the patients whose blood samples were mixed in the vial has the disease.

Set solution: how to crack this set

Every patient is in exactly one vial from each pair: A or B, C or D, E or F, G or H. So the four results give a unique code.

A: patients 9–16, B: 1–8. C: 5–8, 13–16, D: the rest. E: 3, 4, 7, 8, 11, 12, 15, 16, F: the rest. G: even patients, H: odd patients.

Exactly one vial in each pair tests positive.

Q1CAT 2020 · Slot 3 · DILR Q1MCQ

Suppose vial C tests positive and vials A, E and H test negative. Which patient has the disease?

Q2CAT 2020 · Slot 3 · DILR Q2MCQ

Suppose vial A tests positive and vials D and G test negative. Which of the following vials should we test next to identify the patient with the disease?

Q3CAT 2020 · Slot 3 · DILR Q3MCQ

Which of the following combinations of test results is NOT possible?

Q4CAT 2020 · Slot 3 · DILR Q4MCQ

Suppose one of the lab assistants accidentally mixed two patients' blood samples before they were distributed to the vials. Which of the following correctly represents the set of all possible numbers of positive test results out of the eight vials?

Set 2CAT 2020 · Slot 3 · DILR Q5–Q84 questionsTables & Missing Data

Grocery orders booked, delivered and lost

XYZ organization got into the business of delivering groceries to home at the beginning of the last month. They have a two-day delivery promise. However, their deliveries are unreliable. An order booked on a particular day may be delivered the next day or the day after. If the order is not delivered at the end of two days, then the order is declared as lost at the end of the second day. XYZ then does not deliver the order, but informs the customer, marks the order as lost, returns the payment and pays a penalty for non-delivery. The following table provides details about the operations of XYZ for a week of the last month. The first column gives the date, the second gives the cumulative number of orders that were booked up to and including that day. The third column represents the number of orders delivered on that day. The last column gives the cumulative number of orders that were lost up to and including that day. It is known that the numbers of orders that were booked on the 11th, 12th, and 13th of the last month that took two days to deliver were 4, 6, and 8 respectively

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

For orders booked on day dd: xdx_d delivered next day, ydy_d the day after, ldl_d lost (counted on day d+2d + 2). Deliveries on day dd =xd−1+yd−2= x_{d-1} + y_{d-2}.

Bookings (differences of the cumulative column): 14th 30, 15th 28, 16th 25, 17th 25. Losses on a day are orders booked two days earlier: l12=1l_{12} = 1, l13=2l_{13} = 2, l14=12l_{14} = 12, l15=12l_{15} = 12, l16=2l_{16} = 2, l17=9l_{17} = 9.

Given y11=4y_{11} = 4, y12=6y_{12} = 6, y13=8y_{13} = 8:

Booked onBookedNext day xxDay after yyLost
12th14761
13th312182
14th3015312
15th288812
16th2513102
17th253139

(e.g. 13th deliveries: x12+4=11x_{12} + 4 = 11; 14th: x13+6=27x_{13} + 6 = 27; 15th: x14+8=23x_{14} + 8 = 23; then y=y = booked −x−- x - lost, and so on.)

Q5CAT 2020 · Slot 3 · DILR Q5MCQ

Among the following days, the largest fraction of orders booked on which day was lost?

Q6CAT 2020 · Slot 3 · DILR Q6MCQ

On which of the following days was the number of orders booked the highest?

Q7CAT 2020 · Slot 3 · DILR Q7MCQ

The delivery ratio for a given day is defined as the ratio of the number of orders booked on that day which are delivered on the next day to the number of orders booked on that day which are delivered on the second day after booking. On which of the following days, was the delivery ratio the highest?

Q8CAT 2020 · Slot 3 · DILR Q8MCQ

The average time taken to deliver orders booked on a particular day is computed as follows. Let the number of orders delivered the next day be x and the number of orders delivered the day after be y. Then the average time to deliver order is (x+2y)(x+y)\dfrac{(x+2y)}{(x+y)}. On which of the following days was the average time taken to deliver orders booked the least?

Set 3CAT 2020 · Slot 3 · DILR Q9–Q146 questionsGrids & Matrices

Trees on twelve plots shared by four daughters

A farmer had a rectangular land containing 205 trees. He distributed that land among his four daughters - Abha, Bina, Chitra and Dipti by dividing the land into twelve plots along three rows (X,Y,Z) and four Columns (1,2,3,4) as shown in the figure below:

Data for this set (chart, table or figure)

The plots in rows X, Y, Z contained mango, teak and pine trees respectively. Each plot had trees in non-zero multiples of 3 or 4 and none of the plots had the same number of trees. Each daughter got an even number of plots. In the figure, the number mentioned in top left corner of a plot is the number of trees in that plot, while the letter in the bottom right corner is the first letter of the name of the daughter who got that plot (For example, Abha got the plot in row Y and column 1 containing 21 trees). Some information in the figure got erased, but the following is known:

1. Abha got 20 trees more than Chitra but 6 trees less than Dipti.

2. The largest number of trees in a plot was 32, but it was not with Abha.

3. The number of teak trees in Column 3 was double of that in Column 2 but was half of that in Column 4.

4. Both Abha and Bina got a higher number of plots than Dipti.

5. Only Bina, Chitra and Dipti got corner plots.

6. Dipti got two adjoining plots in the same row.

7. Bina was the only one who got a plot in each row and each column.

8. Chitra and Dipti did not get plots which were adjacent to each other (either in row / column / diagonal).

9. The number of mango trees was double the number of teak trees.

Set solution: how to crack this set

Plots: Dipti has 2 (adjoining in a row) and fewer than Abha and Bina; all even, total 12: Abha 4, Bina 4, Chitra 2 (X1, Z2), Dipti 2.

Dipti can't touch Chitra's X1 or Z2: only X3–X4 works. Corners go only to Bina, Chitra and Dipti, so Z4 is Bina's. Bina has a plot in every row and column: Z1, Z4, X2, Y3. Abha: Y1, Y2, Y4, Z3.

Teak: Y2 =t= t, Y3 =2t= 2t, Y4 =4t= 4t. Distinct multiples of 3 or 4, and 12 is taken: t=4t = 4 (16 is fine; t=8t = 8 would give Abha 32).

Abha =21+4+16+9=50= 21 + 4 + 16 + 9 = 50, so Chitra =30= 30 (Z2 =18= 18) and Dipti =56= 56.

Teak =21+28=49= 21 + 28 = 49, mango =98= 98: 12+12 + X2+56=98⇒ + 56 = 98 \Rightarrow X2 =30= 30. Pine =205−147=58= 205 - 147 = 58: Z1 =58−18−9−28=3= 58 - 18 - 9 - 28 = 3.

The 32-tree plot must be Dipti's: X3, X4 are 32 and 24 in some order.

Row1234
X (mango)12 C30 B32/24 D24/32 D
Y (teak)21 A4 A8 B16 A
Z (pine)3 B18 C9 A28 B
Q9CAT 2020 · Slot 3 · DILR Q9MCQ

How many mango trees were there in total?

Q10CAT 2020 · Slot 3 · DILR Q10MCQ

Which of the following is the correct sequence of trees received by Abha, Bina, Chitra and Dipti in that order?

Q11CAT 2020 · Slot 3 · DILR Q11MCQ

How many pine trees did Chitra receive?

Q12CAT 2020 · Slot 3 · DILR Q12MCQ

Who got the plot with the smallest number of trees and how many trees did that plot have?

Q13CAT 2020 · Slot 3 · DILR Q13MCQ

Which of the following statements is NOT true?

Q14CAT 2020 · Slot 3 · DILR Q14MCQ

Which column had the highest number of trees?

Set 4CAT 2020 · Slot 3 · DILR Q15–Q206 questionsGames & Tournaments

Hi-Lo bidding game over six rounds

The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo. The bids are made simultaneously. If all four bid Hi, then all four lose 1 point each. If three players bid Hi and one bids Lo, then the players bidding Hi gain 1 point each and the player bidding Lo loses 3 points. If two players bid Hi and two bid Lo, then the players bidding Hi gain 2 points each and the players bidding Lo lose 2 points each. If one player bids Hi and three bid Lo, then the player bidding Hi gains 3 points and the players bidding Lo lose 1 point each. If all four bid Lo, then all four gain 1 point each. Four players Arun, Bankim, Charu, and Dipak played the Hi-Lo game. The following facts are known about their game:

1. At the end of three rounds, Arun had scored 6 points, Dipak had scored 2 points, Bankim and Charu had scored -2 points each.

2. At the end of six rounds, Arun had scored 7 points, Bankim and Dipak had scored -1 point each, and Charu had scored -5 points.

3. Dipak’s score in the third round was less than his score in the first round but was more than his score in the second round.

4. In exactly two out of the six rounds, Arun was the only player who bid Hi.

Set solution: how to crack this set

Possible round results (Hi players, Lo players): 4 Hi: −1 each; 3 Hi: +1, Lo −3; 2 Hi: +2, −2; 1 Hi: +3, Lo −1; 0 Hi: +1 each.

Rounds 1–3 give (A, B, C, D) =(6,−2,−2,2)= (6, -2, -2, 2). Rounds 4–6 give (1,1,−3,−3)(1, 1, -3, -3).

Dipak: Round 2 << Round 3 << Round 1. Checking the combinations, only one works for Rounds 1–3:

RoundArunBankimCharuDipak
1Hi (+2)Lo (−2)Lo (−2)Hi (+2)
2Hi (+3)Lo (−1)Lo (−1)Lo (−1)
3Lo (+1)Lo (+1)Lo (+1)Lo (+1)

Rounds 4–6, in some order: Arun alone Hi (3,−1,−1,−1)(3, -1, -1, -1), Bankim alone Hi (−1,3,−1,−1)(-1, 3, -1, -1), all Hi (−1,−1,−1,−1)(-1, -1, -1, -1). Their sum is (1,1,−3,−3)(1, 1, -3, -3) ✓, and Arun is the only Hi in exactly two rounds (Round 2 and one of Rounds 4–6).

Q15CAT 2020 · Slot 3 · DILR Q15MCQ

What were the bids by Arun, Bankim, Charu and Dipak, respectively in the first round?

Q16CAT 2020 · Slot 3 · DILR Q16TITA

In how many rounds did Arun bid Hi?

Q17CAT 2020 · Slot 3 · DILR Q17TITA

In how many rounds did Bankim bid Lo?

Q18CAT 2020 · Slot 3 · DILR Q18TITA

In how many rounds did all four players make identical bids?

Q19CAT 2020 · Slot 3 · DILR Q19TITA

In how many rounds did Dipak gain exactly 1 point?

Q20CAT 2020 · Slot 3 · DILR Q20MCQ

In which of the following rounds, was Arun DEFINITELY the only player to bid Hi?

Set 5CAT 2020 · Slot 3 · DILR Q21–Q244 questionsSets & Venn Diagrams

Online teaching facilities in 600 schools

A survey of 600 schools in India was conducted to gather information about their online teaching learning processes (OTLP). The following four facilities were studied.

F1: Own software for OTLP

F2: Trained teachers for OTLP

F3: Training materials for OTLP

F4: All students having Laptops

The following observations were summarized from the survey.

1. 80 schools did not have any of the four facilities - F1, F2, F3, F4.

2. 40 schools had all four facilities.

3. The number of schools with only F1, only F2, only F3, and only F4 was 25, 30, 26 and 20 respectively.

4. The number of schools with exactly three of the facilities was the same irrespective of which three were considered.

5. 313 schools had F2.

6. 26 schools had only F2 and F3 (but neither F1 nor F4).

7. Among the schools having F4, 24 had only F3, and 45 had only F2.

8. 162 schools had both F1 and F2.

9. The number of schools having F1 was the same as the number of schools having F4.

Set solution: how to crack this set

520 schools have at least one facility. Only one: 25+30+26+20=10125 + 30 + 26 + 20 = 101. All four: 40. Each "exactly three" region =t= t.

Exactly two: F2F3 26, F3F4 24, F2F4 45, F1F2 aa, F1F3 bb, F1F4 cc.

F2: 30+(a+26+45)+3t+40=313⇒a+3t=17230 + (a + 26 + 45) + 3t + 40 = 313 \Rightarrow a + 3t = 172.

F1 ∩ F2: a+2t+40=162⇒a+2t=122a + 2t + 40 = 162 \Rightarrow a + 2t = 122. So t=50t = 50, a=22a = 22.

Total: 101+(22+b+c+95)+200+40=520⇒b+c=62101 + (22 + b + c + 95) + 200 + 40 = 520 \Rightarrow b + c = 62.

F1 == F4: 25+22+b+c+150+40=20+c+24+45+150+40⇒b=4225 + 22 + b + c + 150 + 40 = 20 + c + 24 + 45 + 150 + 40 \Rightarrow b = 42, c=20c = 20.

Q21CAT 2020 · Slot 3 · DILR Q21MCQ

What was the total number of schools having exactly three of the four facilities?

Q22CAT 2020 · Slot 3 · DILR Q22MCQ

What was the number of schools having facilities F2 and F4?

Q23CAT 2020 · Slot 3 · DILR Q23TITA

What was the number of schools having only facilities F1 and F3?

Q24CAT 2020 · Slot 3 · DILR Q24TITA

What was the number of schools having only facilities F1 and F4?

Preparing for CAT seriously? Structured courses, live classes and mocks by Dr. Kiran Derle at Target Campus.

Explore courses ↗