CAT 2020 DI and Reasoning - slot 03
Question 01: Direction for questions 1 to 6:
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.
What were the bids by Arun, Bankim, Charu and Dipak, respectively in the first round?
a. Hi, Lo, Lo, Hi
b. Hi, Lo, Lo, Lo
c. Lo, Lo, Lo, Hi
d. Hi, Hi, Lo, Lo
In a round, if all players bid Hi then the sum points scored by all the four players together is -4 and if all players bid Lo then the sum points scored by all the four players together is +4. Except these two cases in remaining all other cases the sum points scored by all the four players together is 0.
At the end of three rounds, the sum of points scored by all the four players together is 6 – 2 – 2 + 2 = 4. It means in first three rounds, exactly one time all four of them bade Lo and each of them scored +1 point. The total score of Arun at the end of three rounds is 6 points. He must have scored +1, +2 and +3 points. So in one round all four players bade Lo, in another round he and one more player bade Hi and in one round he alone bade Hi.
Similarly Dipak must have scored +1, +2 and -1 points in first three rounds. 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 it means Dipak scored +2 points in First round, -1 point in Second round and +1 point in Third round.
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. If we subtract points scored by these players in first three rounds from this then we will get the points scored by these players in last three rounds. Hence, Arun scored + 1, Bankim scored +1, Charu scred -3 and Dipak scored -3 points in last three rounds.
The sum of points scored by all the four players together is 1 + 1 – 3 – 3 = – 4. It means in last three rounds, exactly one time all four of them bade Hi and each of them scored – 1 point. Hence Arun must have scored either -1, +1, +1 points or -1, +3, -1 points in last three rounds but In exactly two out of the six rounds, Arun was the only player who bid Hi.
So case 1 is not possible. Arun must have scored -1, +3 and -1 points in last three rounds.
The bids by Arun, Bankim, Charu and Dipak, were Hi, Lo, Lo and Hi respectively in the first round.
The correct answer is: Hi, Lo, Lo, Hi
Question 02: In how many rounds did Arun bid Hi?
In a round, if all players bid Hi then the sum points scored by all the four players together is -4 and if all players bid Lo then the sum points scored by all the four players together is +4. Except these two cases in remaining all other cases the sum points scored by all the four players together is 0.
At the end of three rounds, the sum of points scored by all the four players together is 6 – 2 – 2 + 2 = 4. It means in first three rounds, exactly one time all four of them bade Lo and each of them scored +1 point. The total score of Arun at the end of three rounds is 6 points. He must have scored +1, +2 and +3 points. So in one round all four players bade Lo, in another round he and one more player bade Hi and in one round he alone bade Hi.
Similarly Dipak must have scored +1, +2 and -1 points in first three rounds. 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 it means Dipak scored +2 points in First round, -1 point in Second round and +1 point in Third round.
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. If we subtract points scored by these players in first three rounds from this then we will get the points scored by these players in last three rounds. Hence, Arun scored + 1, Bankim scored +1, Charu scred -3 and Dipak scored -3 points in last three rounds.
The sum of points scored by all the four players together is 1 + 1 – 3 – 3 = – 4. It means in last three rounds, exactly one time all four of them bade Hi and each of them scored – 1 point. Hence Arun must have scored either -1, +1, +1 points or -1, +3, -1 points in last three rounds but In exactly two out of the six rounds, Arun was the only player who bid Hi.
So case 1 is not possible. Arun must have scored -1, +3 and -1 points in last three rounds.
Arun bade Hi in four rounds. The correct answer is: 4
Question 03: In how many rounds did Bankim bid Lo?
In a round, if all players bid Hi then the sum points scored by all the four players together is -4 and if all players bid Lo then the sum points scored by all the four players together is +4. Except these two cases in remaining all other cases the sum points scored by all the four players together is 0.
At the end of three rounds, the sum of points scored by all the four players together is 6 – 2 – 2 + 2 = 4. It means in first three rounds, exactly one time all four of them bade Lo and each of them scored +1 point. The total score of Arun at the end of three rounds is 6 points. He must have scored +1, +2 and +3 points. So in one round all four players bade Lo, in another round he and one more player bade Hi and in one round he alone bade Hi.
Similarly Dipak must have scored +1, +2 and -1 points in first three rounds. 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 it means Dipak scored +2 points in First round, -1 point in Second round and +1 point in Third round.
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. If we subtract points scored by these players in first three rounds from this then we will get the points scored by these players in last three rounds. Hence, Arun scored + 1, Bankim scored +1, Charu scred -3 and Dipak scored -3 points in last three rounds.
The sum of points scored by all the four players together is 1 + 1 – 3 – 3 = – 4. It means in last three rounds, exactly one time all four of them bade Hi and each of them scored – 1 point. Hence Arun must have scored either -1, +1, +1 points or -1, +3, -1 points in last three rounds but In exactly two out of the six rounds, Arun was the only player who bid Hi.
So case 1 is not possible. Arun must have scored -1, +3 and -1 points in last three rounds.
Bankim bade Lo in four rounds.
Question 04: In how many rounds did all four players make identical bids?
In a round, if all players bid Hi then the sum points scored by all the four players together is -4 and if all players bid Lo then the sum points scored by all the four players together is +4. Except these two cases in remaining all other cases the sum points scored by all the four players together is 0.
At the end of three rounds, the sum of points scored by all the four players together is 6 – 2 – 2 + 2 = 4. It means in first three rounds, exactly one time all four of them bade Lo and each of them scored +1 point. The total score of Arun at the end of three rounds is 6 points. He must have scored +1, +2 and +3 points. So in one round all four players bade Lo, in another round he and one more player bade Hi and in one round he alone bade Hi.
Similarly Dipak must have scored +1, +2 and -1 points in first three rounds. 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 it means Dipak scored +2 points in First round, -1 point in Second round and +1 point in Third round.
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. If we subtract points scored by these players in first three rounds from this then we will get the points scored by these players in last three rounds. Hence, Arun scored + 1, Bankim scored +1, Charu scred -3 and Dipak scored -3 points in last three rounds.
The sum of points scored by all the four players together is 1 + 1 – 3 – 3 = – 4. It means in last three rounds, exactly one time all four of them bade Hi and each of them scored – 1 point. Hence Arun must have scored either -1, +1, +1 points or -1, +3, -1 points in last three rounds but In exactly two out of the six rounds, Arun was the only player who bid Hi.
So case 1 is not possible. Arun must have scored -1, +3 and -1 points in last three rounds.
All four players make identical bids in 2 rounds.
b. Third
c. Fourth
d. First
In a round, if all players bid Hi then the sum points scored by all the four players together is -4 and if all players bid Lo then the sum points scored by all the four players together is +4. Except these two cases in remaining all other cases the sum points scored by all the four players together is 0.
At the end of three rounds, the sum of points scored by all the four players together is 6 – 2 – 2 + 2 = 4. It means in first three rounds, exactly one time all four of them bade Lo and each of them scored +1 point. The total score of Arun at the end of three rounds is 6 points. He must have scored +1, +2 and +3 points. So in one round all four players bade Lo, in another round he and one more player bade Hi and in one round he alone bade Hi.
Similarly Dipak must have scored +1, +2 and -1 points in first three rounds. 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 it means Dipak scored +2 points in First round, -1 point in Second round and +1 point in Third round.
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. If we subtract points scored by these players in first three rounds from this then we will get the points scored by these players in last three rounds. Hence, Arun scored + 1, Bankim scored +1, Charu scred -3 and Dipak scored -3 points in last three rounds.
The sum of points scored by all the four players together is 1 + 1 – 3 – 3 = – 4. It means in last three rounds, exactly one time all four of them bade Hi and each of them scored – 1 point. Hence Arun must have scored either -1, +1, +1 points or -1, +3, -1 points in last three rounds but In exactly two out of the six rounds, Arun was the only player who bid Hi.
So case 1 is not possible. Arun must have scored -1, +3 and -1 points in last three rounds.
Arun was the only player who bade Hi in rounds 2.
The correct answer is: Second
Question 06: In how many rounds did Dipak gain exactly 1 point?
In a round, if all players bid Hi then the sum points scored by all the four players together is -4 and if all players bid Lo then the sum points scored by all the four players together is +4. Except these two cases in remaining all other cases the sum points scored by all the four players together is 0.
At the end of three rounds, the sum of points scored by all the four players together is 6 – 2 – 2 + 2 = 4. It means in first three rounds, exactly one time all four of them bade Lo and each of them scored +1 point. The total score of Arun at the end of three rounds is 6 points. He must have scored +1, +2 and +3 points. So in one round all four players bade Lo, in another round he and one more player bade Hi and in one round he alone bade Hi.
Similarly Dipak must have scored +1, +2 and -1 points in first three rounds. 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 it means Dipak scored +2 points in First round, -1 point in Second round and +1 point in Third round.
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. If we subtract points scored by these players in first three rounds from this then we will get the points scored by these players in last three rounds. Hence, Arun scored + 1, Bankim scored +1, Charu scred -3 and Dipak scored -3 points in last three rounds.
The sum of points scored by all the four players together is 1 + 1 – 3 – 3 = – 4. It means in last three rounds, exactly one time all four of them bade Hi and each of them scored – 1 point. Hence Arun must have scored either -1, +1, +1 points or -1, +3, -1 points in last three rounds but In exactly two out of the six rounds, Arun was the only player who bid Hi.
So case 1 is not possible. Arun must have scored -1, +3 and -1 points in last three rounds.
Dipak gained exactly 1 point in only 1 round.
Question 07: Direction for questions 7 to 10:
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.
Among the following days, the largest fraction of orders booked on which day was lost?
a. 15th
b. 16th
c. 13th
d. 14th
Number of orders booked on a particular day is the sum of number of orders delivered the next day, the number of orders delivered the day after and the number of orders lost.
Cumulative orders lost on 13th was 91 and cumulative orders lost on 14th are 92. It means only 1 item was lost on 14th . This item was booked on 12th. Similarly two items which were lost on 15th were booked on 13th and so on.
Now 11 items, that were delivered on 13 th ,were booked on 11th and 12th . It is given that the numbers of orders that were booked on the 11th of the last month that took two days to deliver were 4. Hence 7 items, which were booked on 12th , were delivered on 13th . Similarly 27 items, that were delivered on 14th , were booked on 12th (6 items, given in question) and 13th (21 items).
The ratio of items lost to that of orders booked on 13th , 14th , 15th and 16th were 2/31, 12/30, 12/28 and 2/25 respectively. Hence the largest fraction of orders booked on 15th was lost.
The correct answer is: 15th
Question 08: On which of the following days was the number of orders booked the highest?
a. 15th
b. 14th
c. 13th
d. 12th
Question 09: 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?
a. 15th
b. 16th
c. 14th
d. 13th
Question 10: 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). On which of the following days was the average time taken to deliver orders booked the least?
a. 15th
b. 14th
c. 13th
d. 16th
Question 11: Direction for questions 11 to 16:
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:
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.
How many mango trees were there in total?
a. 126
b. 84
c. 98
d. 49
Both Abha and Bina got a higher number of plots than Dipti. Each daughter got an even number of plots. So Dipti got 2 plots and both Abha and Bina each got 4 plots. Chitra got 2 plots because total 12 plots are there. Dipti got 2 plots in row X column 3 and 4 because Dipti’s both the plots are adjoining in the same row but not adjacent to any plot of C.
Plot in row Z and column 4 is a corner plot so it must belong to Bina only because Abha cannot get corner plot and Chitra and Dipti already have their 2 plots. Plots in row X column 2 and row Y column 3 also belong to Bina because Bina must have at least one plot in each row and each column.
Remaining all plots belong to Abha.
In row Y the number of trees in Column 3 was double of that in Column 2 but was half of that in Column 4. So number of trees in row Y and column 2, 3 and 4 can be 4, 8 and 16 or 8, 16 and 32 respectively. As it is given that the number of Mango trees was double that of Teak trees, Hence the number of Teak trees in column 2, 3 and 4 has to be 4, 8 and 16 respectively.
The total number of trees that Abha got was 21 + 9 + 4 + 16 = 50. Abha got 20 trees more than Chitra but 6 trees less than Dipti that means Chitra got 30 trees, Dipti got 56 and Bina got 69 trees. We also now that there are 49 Teak, 98 Mango and 58 Pine trees.
Total 98 mango trees are there.
The correct answer is: 98
Question 12: Which of the following is the correct sequence of trees received by Abha, Bina, Chitra and Dipti in that order?
a. 50, 69, 30, 56
b. 60, 39, 40, 66
c. 44, 87, 24, 50
d. 54, 57, 34, 60
Question 13: How many pine trees did Chitra receive?
a. 18
b. 15
c. 30
d. 21
Question 14: Who got the plot with the smallest number of trees and how many trees did that plot have?
a. Bina, 3 trees
b. Abha, 4 trees
c. Dipti, 6 trees
d. Bina, 4 trees
Question 15: Which of the following statements is NOT true?
a. Bina got 32 pine trees.
b. Chitra got 12 mango trees.
c. Abha got 41 teak trees.
d. Dipti got 56 mango trees.
Question 16: Which column had the highest number of trees?
a. 4
b. 3
c. 2
d. Cannot be determined.
Both Abha and Bina got a higher number of plots than Dipti. Each daughter got an even number of plots. So Dipti got 2 plots and both Abha and Bina each got 4 plots. Chitra got 2 plots because total 12 plots are there. Dipti got 2 plots in row X column 3 and 4 because Dipti’s both the plots are adjoining in the same row but not adjacent to any plot of C.
Plot in row Z and column 4 is a corner plot so it must belong to Bina only because Abha cannot get corner plot and Chitra and Dipti already have their 2 plots. Plots in row X column 2 and row Y column 3 also belong to Bina because Bina must have at least one plot in each row and each column.
Remaining all plots belong to Abha.
In row Y the number of trees in Column 3 was double of that in Column 2 but was half of that in Column 4. So number of trees in row Y and column 2, 3 and 4 can be 4, 8 and 16 or 8, 16 and 32 respectively. As it is given that the number of Mango trees was double that of Teak trees, Hence the number of Teak trees in column 2, 3 and 4 has to be 4, 8 and 16 respectively.
The total number of trees that Abha got was 21 + 9 + 4 + 16 = 50. Abha got 20 trees more than Chitra but 6 trees less than Dipti that means Chitra got 30 trees, Dipti got 56 and Bina got 69 trees. We also now that there are 49 Teak, 98 Mango and 58 Pine trees.
Column 4 had the highest number of trees in both the cases.
The correct answer is: 4
Question 17: Direction for questions 17 to 20:
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.
What was the total number of schools having exactly three of the four facilities?
a. 80
b. 64
c. 200
d. 50
Let the number of schools with exactly three of the facilities was x and the number of schools with facilities F1 and F2 only was z. The number of schools having F1 is equal to the number of schools having F4, let it be y.
Now y + 30 + 26 + 26 + x + 24 + 45 + 20 + 80 = 600
Or x + y = 349 ………(1)
F2 = 313 or 3x + z = 172 …….(2)
F1 and F2 is 162 or 2x + z = 122 ……….(3)
Solving equations 1, 2 and 3 we get x = 50, y = 299 and z = 22.
So F4 = 299. Hence F1 and F4 only is 20.
F1 = 299. Hence F1 and F3 only is 42.
200 schools have exactly 3 facilities.
The correct answer is: 200
Question 18: What was the number of schools having facilities F2 and F4?
a. 185
b. 95
c. 85
d. 45
Let the number of schools with exactly three of the facilities was x and the number of schools with facilities F1 and F2 only was z. The number of schools having F1 is equal to the number of schools having F4, let it be y.
Now y + 30 + 26 + 26 + x + 24 + 45 + 20 + 80 = 600
Or x + y = 349 ………(1)
F2 = 313 or 3x + z = 172 …….(2)
F1 and F2 is 162 or 2x + z = 122 ……….(3)
Solving equations 1, 2 and 3 we get x = 50, y = 299 and z = 22.
So F4 = 299. Hence F1 and F4 only is 20.
F1 = 299. Hence F1 and F3 only is 42.
185 schools were having F2 and F4 facilities.
The correct answer is: 185
Question 19: What was the number of schools having only facilities F1 and F3?
Let the number of schools with exactly three of the facilities was x and the number of schools with facilities F1 and F2 only was z. The number of schools having F1 is equal to the number of schools having F4, let it be y.
Now y + 30 + 26 + 26 + x + 24 + 45 + 20 + 80 = 600
Or x + y = 349 ………(1)
F2 = 313 or 3x + z = 172 …….(2)
F1 and F2 is 162 or 2x + z = 122 ……….(3)
Solving equations 1, 2 and 3 we get x = 50, y = 299 and z = 22.
So F4 = 299. Hence F1 and F4 only is 20.
F1 = 299. Hence F1 and F3 only is 42.
42 schools were having only F1 and F3 facilities.
Question 20: What was the number of schools having only facilities F1 and F4?
Let the number of schools with exactly three of the facilities was x and the number of schools with facilities F1 and F2 only was z. The number of schools having F1 is equal to the number of schools having F4, let it be y.
Now y + 30 + 26 + 26 + x + 24 + 45 + 20 + 80 = 600
Or x + y = 349 ………(1)
F2 = 313 or 3x + z = 172 …….(2)
F1 and F2 is 162 or 2x + z = 122 ……….(3)
Solving equations 1, 2 and 3 we get x = 50, y = 299 and z = 22.
So F4 = 299. Hence F1 and F4 only is 20.
F1 = 299. Hence F1 and F3 only is 42.
20 schools were having only F1 and F4 facilities.
Question 21: Direction for questions 21 to 24:
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.
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.
Suppose vial C tests positive and vials A, E and H test negative. Which patient has the disease?
a. Patient 6
b. Patient 8
c. Patient 14
d. Patient 2
Vial A, E and H test negative means patient numbers 1, 3, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15 and 16 do not have the disease. Hence either patient number 2 or patient number 6 has the disease. Now vial C test positive means patient number 6 has the disease.
Question 22: 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?
a. Vial E
b. Vial H
c. Vial B
d. Vial C
Vials D and G test negative means patient numbers 1, 2, 3, 4, 6, 8, 9, 10, 11, 12, 14 and 16 do not have the disease. Hence one of the patients 5, 7, 13 or 15 has the disease. Vial A test positive means either patient number 13 or patient number 15 has the disease. Now we must test either vial E or F because they are not common in these two patients.
The correct answer is: Vial E
Question 23: Which of the following combinations of test results is NOT possible?
a. Vials A and E positive, vials C and D negative
b. Vial B positive, vials C, F and H negative
c. Vials A and G positive, vials D and E negative
d. Vials B and D positive, vials F and H negative
Vial C, F and H test negative means patient numbers 1, 2, 3, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15 and 16 do not have the disease. Hence either patient number 4 or patient number 12 has the disease. Now vial B test positive means patient number 4 has the disease.
Vial D and E test negative means patient numbers 1, 2, 3, 4, 7, 8, 9, 10, 11, 12, 15 and 16 do not have the disease. Hence either patient number 5, 6, 13 or patient number 14 has the disease. Now vial A and vial G test positive means patient number 5 has the disease.
Vial F and H test negative means patient numbers 1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14 and 15 do not have the disease. Hence either patient number 4, 8, 12 or patient number 16 has the disease. Now vial B and vial D test positive means patient number 16 has the disease.
Vial C and D test negative means no patient has the disease. Hence this choice is not possible.
The correct answer is: Vials A and E positive, vials C and D negative
Question 24: 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?
a. {4,5,6,7,8}
b. {4,5,6,7}
c. {4,5}
d. {5,6,7,8}
Either at least one the two patients, whose blood samples are mixed, have disease or none of them has disease. If none of these two patients have disease then only 4 vials test positive but if at least one of these two patients has disease then the positive number of vials can be 5, 6, 7 or 8.
The correct answer is: {4,5,6,7,8}