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# Test: Analytical Reasoning - 3

## 20 Questions MCQ Test Logical Reasoning (LR) and Data Interpretation (DI) | Test: Analytical Reasoning - 3

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This mock test of Test: Analytical Reasoning - 3 for CAT helps you for every CAT entrance exam. This contains 20 Multiple Choice Questions for CAT Test: Analytical Reasoning - 3 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Analytical Reasoning - 3 quiz give you a good mix of easy questions and tough questions. CAT students definitely take this Test: Analytical Reasoning - 3 exercise for a better result in the exam. You can find other Test: Analytical Reasoning - 3 extra questions, long questions & short questions for CAT on EduRev as well by searching above.
QUESTION: 1

### Directions for Questions: Answer the questions on the basis of the information given below. There are 6 friends — Gurvinder, Surinder, Mahinder, Bhupinder, Harinder and Joginder. Their wives are — Sita, Rama, Dolly, Monica, Trisna and Kaveri (not in the same order as their husbands). Each of these 6 friends belong to exactly one out of Ambala, Jaipur, Guntur, Kapurthala, Noida and Jammu (not necessarily in order). Each of them plays exactly one of the games — cricket, football, volleyball, snooker, TT and badminton (again not necessarily in that order). Each of the friends is married to one lady only. — Husbands of Dolly, Trisna or Kaveri do not play football or volleyball. — The one who is from Guntur plays cricket. — Joginder plays football and is from Jammu. — Mahinder and Harinder are married to Sita and Monica respectively but are not from Guntur. — The men from Jaipur and Kapurthala are TT and volleyball players respectively. — Bhupinder is from Noida. — Rama is married to the man from Jammu. — Mahinder plays snooker. (2014) Q. Who is married to the man from Kapurthala?

Solution:

According to question,

Monica is married to Harinder and he belongs from Kapurthala.

QUESTION: 2

### Directions for Questions: Answer the questions on the basis of the information given below. There are 6 friends — Gurvinder, Surinder, Mahinder, Bhupinder, Harinder and Joginder. Their wives are — Sita, Rama, Dolly, Monica, Trisna and Kaveri (not in the same order as their husbands). Each of these 6 friends belong to exactly one out of Ambala, Jaipur, Guntur, Kapurthala, Noida and Jammu (not necessarily in order). Each of them plays exactly one of the games — cricket, football, volleyball, snooker, TT and badminton (again not necessarily in that order). Each of the friends is married to one lady only. — Husbands of Dolly, Trisna or Kaveri do not play football or volleyball. — The one who is from Guntur plays cricket. — Joginder plays football and is from Jammu. — Mahinder and Harinder are married to Sita and Monica respectively but are not from Guntur. — The men from Jaipur and Kapurthala are TT and volleyball players respectively. — Bhupinder is from Noida. — Rama is married to the man from Jammu. — Mahinder plays snooker. (2014) Q. The person who plays Snooker belongs to which place?

Solution:

According to question,

Mahinder plays Snooker and he belongs to Ambala.

QUESTION: 3

### Directions for Questions: Answer the questions on the basis of the information given below. There are 6 friends — Gurvinder, Surinder, Mahinder, Bhupinder, Harinder and Joginder. Their wives are — Sita, Rama, Dolly, Monica, Trisna and Kaveri (not in the same order as their husbands). Each of these 6 friends belong to exactly one out of Ambala, Jaipur, Guntur, Kapurthala, Noida and Jammu (not necessarily in order). Each of them plays exactly one of the games — cricket, football, volleyball, snooker, TT and badminton (again not necessarily in that order). Each of the friends is married to one lady only. — Husbands of Dolly, Trisna or Kaveri do not play football or volleyball. — The one who is from Guntur plays cricket. — Joginder plays football and is from Jammu. — Mahinder and Harinder are married to Sita and Monica respectively but are not from Guntur. — The men from Jaipur and Kapurthala are TT and volleyball players respectively. — Bhupinder is from Noida. — Rama is married to the man from Jammu. — Mahinder plays snooker. (2014) Q. Which game is being played by Bhupinder?

Solution:

According to question,

Badminton is being played by Bhupinder.

QUESTION: 4

Directions for Questions: Answer the questions on the basis of the information given below.
There are 6 friends — Gurvinder, Surinder, Mahinder, Bhupinder, Harinder and Joginder. Their wives are
— Sita, Rama, Dolly, Monica, Trisna and Kaveri (not in the same order as their husbands). Each of these 6 friends belong to exactly one out of Ambala, Jaipur, Guntur, Kapurthala, Noida and Jammu (not necessarily in order). Each of them plays exactly one of the games — cricket, football, volleyball, snooker, TT and badminton (again not necessarily in that order). Each of the friends is married to one lady only.
— Husbands of Dolly, Trisna or Kaveri do not play football or volleyball.
— The one who is from Guntur plays cricket.
— Joginder plays football and is from Jammu.
— Mahinder and Harinder are married to Sita and Monica respectively but are not from Guntur.
— The men from Jaipur and Kapurthala are TT and volleyball players respectively.
— Bhupinder is from Noida.
— Rama is married to the man from Jammu.
— Mahinder plays snooker.

(2014)

Q. Who is the wife of Surinder?

Solution:

According to question,

The wife of Surinder can't be determined because data is not sufficient.

QUESTION: 5

Directions for Questions: Answer the questions on the basis of information given below.
A farmer has 60 hens in his poultry farm. Each of these 60 hens lays one egg per day. On each day out of the eggs laid, some of the eggs are found to be rotten and some of the eggs get broken. Only the eggs that are neither rotten nor broken are taken to the market for sale but due to some unavoidable reasons some eggs are not sold. The eggs that are not sold are brought back to the poultry farm.

(2014)

I. Out of the eggs that are brought back on each day, 40% are rotten and 20% are broken the next morning (and they are always integers).
II. Out of the eggs that are brought back to the poultry farm on each day, the eggs that are neither broken nor rotten are taken along with the eggs that are laid by the hens on the next day to the market for sale.
III. On any particular day the number of eggs that are not sold is less than 20% of the number of eggs that are taken to be sold out of the eggs that are laid on each day.
IV. Out of the eggs that are laid on each day, the number of eggs that are rotten is greater than 1 but less than 5 and the number of eggs that got broken is greater than 3 and less than 7.
V. Every week has five days and each week starts from day 1 and ends on day 5. Assume there are no eggs with the farmer at the beginning of day 1 of the given week.
Q. Find the minimum possible aggregate number of eggs sold on day 4 and day 5.

Solution:

Number of eggs laid on each day = Number of hens in the poultry farm = 60.
Out of the eggs laid on each day, the number of eggs t h a t got rotten is either 2 or 3 or 4.
Out of the eggs laid on each day, the number of eggs that got broken is either 4 or 5 or 6.
Maximum possible number of eggs taken to the market for sale on day 1 = 60 – (2 + 4)
= 54.
Minimum possible number of eggs taken to the market for sale on day 1 = 60 – (4 + 6)
= 50.
The minimum number of eggs that are left unsold each day must be 5, as the number of eggs that are rotten and broken among them needs to be an integer. It can be at max 10, since number of egg left unsold on any day is lessthan 20% ofthe number of eggs laid on each day, i.e. 20% of 60 = 12.
So, the number of eggs that are sold on day 1 ranges from (50 – 10 = 40) to (54 – 5 = 49), (both inclusive).
On the next day again 60 eggs are laid, so from the above logic the range of number of eggs sold should again come out to be from 42 to 49 (both inclusive), but there are eggs that remain unsold at the end of the previous day.
Minimum possible number of eggs that are left over from the previous day and are taken along with the eggs laid on a day to the market for sale = 5 – (40% of 5) – (2% of 5) = 2.
Maximum possible number of eggsthat are left over from the previous day and are taken along with the eggs laid on a day to the market for sale = 10 – (40% of 10) – (20% of 10) = 4.
So, the range of number of eggs that are sold on day 2 varies from (40 + 2 = 42) to (49 + 4 = 53) (both inclusive) and this holds true for day 3, day 4 and day 5 also,

The minimum possible number of eggs that were sold on day 4 can be 42, 42 eggs are sold in the scenorio when 10 eggs are left unsold.
So, the next day i.e. day 5, the minimum number of eggs that were sold can be calculated as Out of the 60 eggs that were laid – maximum rotten and broken eggs can be removed which are 4 and 6 respectively. Thus, left with 50 eggs. Also, from the 10 eggs of the previous day maximum rotten and broken can be removed which are 4 and 2 respectively, thus left with 4 eggs only. So, out of the total 54 eggs, a maximum of only 10 eggs can be left unsold. Therefore, the minimum eggs thatwere sold on day 5 were 54 – 10 = 44.
Hence, aggregate sum of eggs is 42 + 44 = 86.
Note: Most of the students will make a mistake of considering 42 eggs for both the days but this is not possible on any two consecutive days simultaneously.

QUESTION: 6

Directions for Questions: Answer the questions on the basis of information given below.
A farmer has 60 hens in his poultry farm. Each of these 60 hens lays one egg per day. On each day out of the eggs laid, some of the eggs are found to be rotten and some of the eggs get broken. Only the eggs that are neither rotten nor broken are taken to the market for sale but due to some unavoidable reasons some eggs are not sold. The eggs that are not sold are brought back to the poultry farm.

(2014)

I. Out of the eggs that are brought back on each day, 40% are rotten and 20% are broken the next morning (and they are always integers).
II. Out of the eggs that are brought back to the poultry farm on each day, the eggs that are neither broken nor rotten are taken along with the eggs that are laid by the hens on the next day to the market for sale.
III. On any particular day the number of eggs that are not sold is less than 20% of the number of eggs that are taken to be sold out of the eggs that are laid on each day.
IV. Out of the eggs that are laid on each day, the number of eggs that are rotten is greater than 1 but less than 5 and the number of eggs that got broken is greater than 3 and less than 7.
V. Every week has five days and each week starts from day 1 and ends on day 5. Assume there are no eggs with the farmer at the beginning of day 1 of the given week.
Q. If the number of eggs that got rotten and broken is maximum possible in a particular week, then find the difference between the total number of eggs that got rotten and the total number of eggs that got broken across all the five days of the week.

Solution:

Number of eggs laid on each day = Number of hens in the poultry farm = 60.
Out of the eggs laid on each day, the number of eggs t h a t got rotten is either 2 or 3 or 4.
Out of the eggs laid on each day, the number of eggs that got broken is either 4 or 5 or 6.
Maximum possible number of eggs taken to the market for sale on day 1 = 60 – (2 + 4)
= 54.
Minimum possible number of eggs taken to the market for sale on day 1 = 60 – (4 + 6)
= 50.
The minimum number of eggs that are left unsold each day must be 5, as the number of eggs that are rotten and broken among them needs to be an integer. It can be at max 10, since number of egg left unsold on any day is lessthan 20% ofthe number of eggs laid on each day, i.e. 20% of 60 = 12.
So, the number of eggs that are sold on day 1 ranges from (50 – 10 = 40) to (54 – 5 = 49), (both inclusive).
On the next day again 60 eggs are laid, so from the above logic the range of number of eggs sold should again come out to be from 42 to 49 (both inclusive), but there are eggs that remain unsold at the end of the previous day.
Minimum possible number of eggs that are left over from the previous day and are taken along with the eggs laid on a day to the market for sale = 5 – (40% of 5) – (2% of 5) = 2.
Maximum possible number of eggsthat are left over from the previous day and are taken along with the eggs laid on a day to the market for sale = 10 – (40% of 10) – (20% of 10) = 4.
So, the range of number of eggs that are sold on day 2 varies from (40 + 2 = 42) to (49 + 4 = 53) (both inclusive) and this holds true for day 3, day 4 and day 5 also,

Number of eggs that got rotten and broken is maximum possible.
So, assume that on each day 10 eggs remain unsold at the end of each day.
So, out of these 10 eggs that remain unsold at the end of each day, the number of eggs that get rotten and broken on the next day is 4 and 2 respectively.
The maximum number of eggs that got rotten and broken from among the eggs laid each day is 4 and 6 respectively.
So, the number of eggs that got broken across all the five days = 6 × 5 + 2 × 4
= 38.
Number of eggs that got rotten across all the five days = 4 × 5 + 4 × 4 = 36.
Required difference = 38 – 36 = 2.

QUESTION: 7

Directions for Questions: Answer the questions on the basis of information given below.
A farmer has 60 hens in his poultry farm. Each of these 60 hens lays one egg per day. On each day out of the eggs laid, some of the eggs are found to be rotten and some of the eggs get broken. Only the eggs that are neither rotten nor broken are taken to the market for sale but due to some unavoidable reasons some eggs are not sold. The eggs that are not sold are brought back to the poultry farm.

(2014)

I. Out of the eggs that are brought back on each day, 40% are rotten and 20% are broken the next morning (and they are always integers).
II. Out of the eggs that are brought back to the poultry farm on each day, the eggs that are neither broken nor rotten are taken along with the eggs that are laid by the hens on the next day to the market for sale.
III. On any particular day the number of eggs that are not sold is less than 20% of the number of eggs that are taken to be sold out of the eggs that are laid on each day.
IV. Out of the eggs that are laid on each day, the number of eggs that are rotten is greater than 1 but less than 5 and the number of eggs that got broken is greater than 3 and less than 7.
V. Every week has five days and each week starts from day 1 and ends on day 5. Assume there are no eggs with the farmer at the beginning of day 1 of the given week.
Q. What can be the maximum number of eggs (that are unbroken and not rotten) that were left unsold at the end of the fifth day?

Solution:

Number of eggs laid on each day = Number of hens in the poultry farm = 60.
Out of the eggs laid on each day, the number of eggs t h a t got rotten is either 2 or 3 or 4.
Out of the eggs laid on each day, the number of eggs that got broken is either 4 or 5 or 6.
Maximum possible number of eggs taken to the market for sale on day 1 = 60 – (2 + 4)
= 54.
Minimum possible number of eggs taken to the market for sale on day 1 = 60 – (4 + 6)
= 50.
The minimum number of eggs that are left unsold each day must be 5, as the number of eggs that are rotten and broken among them needs to be an integer. It can be at max 10, since number of egg left unsold on any day is lessthan 20% ofthe number of eggs laid on each day, i.e. 20% of 60 = 12.
So, the number of eggs that are sold on day 1 ranges from (50 – 10 = 40) to (54 – 5 = 49), (both inclusive).
On the next day again 60 eggs are laid, so from the above logic the range of number of eggs sold should again come out to be from 42 to 49 (both inclusive), but there are eggs that remain unsold at the end of the previous day.
Minimum possible number of eggs that are left over from the previous day and are taken along with the eggs laid on a day to the market for sale = 5 – (40% of 5) – (2% of 5) = 2.
Maximum possible number of eggsthat are left over from the previous day and are taken along with the eggs laid on a day to the market for sale = 10 – (40% of 10) – (20% of 10) = 4.
So, the range of number of eggs that are sold on day 2 varies from (40 + 2 = 42) to (49 + 4 = 53) (both inclusive) and this holds true for day 3, day 4 and day 5 also,

The maximum number of eggs that were left unsold at the end of the fifth day can be 10.

QUESTION: 8

Directions for Questions: Answer the questions on the basis of information given below.
A farmer has 60 hens in his poultry farm. Each of these 60 hens lays one egg per day. On each day out of the eggs laid, some of the eggs are found to be rotten and some of the eggs get broken. Only the eggs that are neither rotten nor broken are taken to the market for sale but due to some unavoidable reasons some eggs are not sold. The eggs that are not sold are brought back to the poultry farm.

(2014)

I. Out of the eggs that are brought back on each day, 40% are rotten and 20% are broken the next morning (and they are always integers).
II. Out of the eggs that are brought back to the poultry farm on each day, the eggs that are neither broken nor rotten are taken along with the eggs that are laid by the hens on the next day to the market for sale.
III. On any particular day the number of eggs that are not sold is less than 20% of the number of eggs that are taken to be sold out of the eggs that are laid on each day.
IV. Out of the eggs that are laid on each day, the number of eggs that are rotten is greater than 1 but less than 5 and the number of eggs that got broken is greater than 3 and less than 7.
V. Every week has five days and each week starts from day 1 and ends on day 5. Assume there are no eggs with the farmer at the beginning of day 1 of the given week.
Q. If the number of eggs sold on any day of the week is maximum possible then what is the aggregate sum of the total number of eggs sold in the entire week?

Solution:

Number of eggs laid on each day = Number of hens in the poultry farm = 60.
Out of the eggs laid on each day, the number of eggs t h a t got rotten is either 2 or 3 or 4.
Out of the eggs laid on each day, the number of eggs that got broken is either 4 or 5 or 6.
Maximum possible number of eggs taken to the market for sale on day 1 = 60 – (2 + 4)
= 54.
Minimum possible number of eggs taken to the market for sale on day 1 = 60 – (4 + 6)
= 50.
The minimum number of eggs that are left unsold each day must be 5, as the number of eggs that are rotten and broken among them needs to be an integer. It can be at max 10, since number of egg left unsold on any day is lessthan 20% ofthe number of eggs laid on each day, i.e. 20% of 60 = 12.
So, the number of eggs that are sold on day 1 ranges from (50 – 10 = 40) to (54 – 5 = 49), (both inclusive).
On the next day again 60 eggs are laid, so from the above logic the range of number of eggs sold should again come out to be from 42 to 49 (both inclusive), but there are eggs that remain unsold at the end of the previous day.
Minimum possible number of eggs that are left over from the previous day and are taken along with the eggs laid on a day to the market for sale = 5 – (40% of 5) – (2% of 5) = 2.
Maximum possible number of eggsthat are left over from the previous day and are taken along with the eggs laid on a day to the market for sale = 10 – (40% of 10) – (20% of 10) = 4.
So, the range of number of eggs that are sold on day 2 varies from (40 + 2 = 42) to (49 + 4 = 53) (both inclusive) and this holds true for day 3, day 4 and day 5 also,

The maximum number of eggs that were sold on day1 can be 49, in a scenario when 5 eggs were left unsold.
In this case, the maximum number of eggs that were sold on day 2 can be calculated as 60 (laid on day 2) –2 (minimum rotten out of 60) – 4 (minimum broken out of 60) + 5 (previous days unsold eggs) – 2(rotten out of 5) – 1(broken out of 5)  – 5(minimum unsold) = 51 Since, on day 2 also. 5 eggs were left unsold the maximum number eggs sold on day 3 will again be 51.
The same holds true for day 4 and day 5.
So, the maximum number of eggs that can be sold in the entire week can be 49 + 51 × 4 = 253.
Note: Most of the students will make a mistake of considering 53 eggs for any two consecutive days but this is not possible on any two consecutive days simultaneously.

QUESTION: 9

Directions for Questions: Answer the questions on the basis of the information given below.
A Cricket team of 11 players is to be formed from a group of 15 players—A, B, C, D, E, F, G, H, I, J, K, L, M, N and O. Among the players A, D, K, L, M, N and O are batsmen; B, C, E, F, G and H are bowlers; I and J are wicketkeepers. It is also known that:
I. The team must have at least 5 batsmen and exactly 1 wicketkeeper.
II. H can be selected only if B is selected.
III. F can be selected only if both G and N are selected.
IV. If I is selected, then F is also selected.
V. K and M cannot be selected together for the team. The same is true for B and G.

(2013)

Q. In how many ways can the team be formed?

Solution:

From statement (V), B and G cannot be together in the team. Therefore, there are three possible cases.
Case I: When B is selected.
F cannot be selected as F can only be selected when both G and N are selected. Thus, when B is selected, the team comprises exactly four bowlers. Also, J must be the wicketkeeper in the team, as selection of Iensures selection of F. Following table gives the possible compositions for the team.

The number of ways in which the team can be formed = 2 × 1 × 1 = 2
Case II: When G is selected.
H cannot be selected as H can only be selected when B is selected. Thus, when G is selected, then again the team comprises exactly four bowlers.
Following table gives the possible compositions for the team.

The number of ways in which the team can be formed = 2 × 1 × 2 = 4.
Case III: When neither B nor G is selected.
When both B and G are not selected, then there is no possible composition for the team.
The total number of ways in which the team can be formed = 2 + 4 = 6.

QUESTION: 10

Directions for Questions: Answer the questions on the basis of the information given below.
A Cricket team of 11 players is to be formed from a group of 15 players—A, B, C, D, E, F, G, H, I, J, K, L, M, N and O. Among the players A, D, K, L, M, N and O are batsmen; B, C, E, F, G and H are bowlers;
I and J are wicketkeepers. It is also known that: I. The team must have at least 5 batsmen and exactly 1 wicketkeeper.
II. H can be selected only if B is selected.
III. F can be selected only if both G and N are selected.
IV. If I is selected, then F is also selected.
V. K and M cannot be selected together for the team. The same is true for B and G.

(2013)

Q. If G is one of the bowlers in the team, then who will be the wicketkeeper?

Solution:

From statement (V), B and G cannot be together in the team. Therefore, there are three possible cases.
Case I: When B is selected.
F cannot be selected as F can only be selected when both G and N are selected. Thus, when B is selected, the team comprises exactly four bowlers. Also, J must be the wicketkeeper in the team, as selection of Iensures selection of F. Following table gives the possible compositions for the team.

The number of ways in which the team can be formed = 2 × 1 × 1 = 2
Case II: When G is selected.
H cannot be selected as H can only be selected when B is selected. Thus, when G is selected, then again the team comprises exactly four bowlers.
Following table gives the possible compositions for the team.

The number of ways in which the team can be formed = 2 × 1 × 2 = 4.
Case III: When neither B nor G is selected.
When both B and G are not selected, then there is no possible composition for the team.
If G is one of the bowlers in the team, then either of the two (i.e., I and J) can be selected as the wicket keeper in the team.

QUESTION: 11

Directions for Questions: Answer the questions on the basis of the information given below.
A Cricket team of 11 players is to be formed from a group of 15 players—A, B, C, D, E, F, G, H, I, J, K, L, M, N and O. Among the players A, D, K, L, M, N and O are batsmen; B, C, E, F, G and H are bowlers; I and J are wicketkeepers. It is also known that:
I. The team must have at least 5 batsmen and exactly 1 wicketkeeper.
II. H can be selected only if B is selected.
III. F can be selected only if both G and N are selected.
IV. If I is selected, then F is also selected.
V. K and M cannot be selected together for the team. The same is true for B and G.

(2013)

Q. If H is selected, then who among the following cannot be selected in the team?

Solution:

From statement (V), B and G cannot be together in the team. Therefore, there are three possible cases.
Case I: When B is selected.
F cannot be selected as F can only be selected when both G and N are selected. Thus, when B is selected, the team comprises exactly four bowlers. Also, J must be the wicketkeeper in the team, as selection of Iensures selection of F. Following table gives the possible compositions for the team.

The number of ways in which the team can be formed = 2 × 1 × 1 = 2
Case II: When G is selected.
H cannot be selected as H can only be selected when B is selected. Thus, when G is selected, then again the team comprises exactly four bowlers.
Following table gives the possible compositions for the team.

The number of ways in which the team can be formed = 2 × 1 × 2 = 4.
Case III: When neither B nor G is selected.
When both B and G are not selected, then there is no possible composition for the team.
If H is selected, G cannot be selected.

QUESTION: 12

Directions for Questions: Answer the questions on the basis of the information given below.
Eight persons are sitting at a rectangular table such that four persons are sitting along each of the longer sides of the table. Each person works in a different bank among PNB, SBI, HDFC, ICICI, CBI, BOB, BOI and Citi, and holds a different designation among IT Officer, Marketing Officer, Law Officer, Agricultural Officer, Rajbhasha Adhikari, Technical Officer, Finance Officer and HR Manager, not necessarily in the same order. It is also known that:
I. The Finance Officer, who works in BOI, is sitting to the immediate left of the HR Manager.
II. The Marketing Officer, who works in SBI, is sitting opposite the Technical Officer.
III. The Agricultural Officer, who works in ICICI, is sitting diagonally opposite the Rajbhasha Adhikari, who is sitting to the immediate left of the Technical Officer.
IV. The Law Officer, who works in HDFC, is sitting to the immediate left of the Marketing officer.
V. The persons who work in PNB and HDFC are sitting along the same side of the table.
VI. The Rajbhasha Adhikari, who works in CBI, is sitting at one of the extreme ends in his row.

(2013)

Q. Who is sitting to the immediate right of the Finance Officer?

Solution:

According to question,

or

HR Manager is sitting to the immediate right of the Finance officer.

QUESTION: 13

Directions for Questions: Answer the questions on the basis of the information given below.
Eight persons are sitting at a rectangular table such that four persons are sitting along each of the longer sides of the table. Each person works in a different bank among PNB, SBI, HDFC, ICICI, CBI, BOB, BOI and Citi, and holds a different designation among IT Officer, Marketing Officer, Law Officer, Agricultural Officer, Rajbhasha Adhikari, Technical Officer, Finance Officer and HR Manager, not necessarily in the same order. It is also known that:
I. The Finance Officer, who works in BOI, is sitting to the immediate left of the HR Manager.
II. The Marketing Officer, who works in SBI, is sitting opposite the Technical Officer.
III. The Agricultural Officer, who works in ICICI, is sitting diagonally opposite the Rajbhasha Adhikari, who is sitting to the immediate left of the Technical Officer.
IV. The Law Officer, who works in HDFC, is sitting to the immediate left of the Marketing officer.
V. The persons who work in PNB and HDFC are sitting along the same side of the table.
VI. The Rajbhasha Adhikari, who works in CBI, is sitting at one of the extreme ends in his row.

(2013)

Q. For how many persons can we definitely determine the banks in which they work?

Solution:

According to question,

or

For six persons, we can definitely determing the banks in which they work.

QUESTION: 14

Directions for Questions: Answer the questions on the basis of the information given below.
Eight persons are sitting at a rectangular table such that four persons are sitting along each of the longer sides of the table. Each person works in a different bank among PNB, SBI, HDFC, ICICI, CBI, BOB, BOI and Citi, and holds a different designation among IT Officer, Marketing Officer, Law Officer, Agricultural Officer, Rajbhasha Adhikari, Technical Officer, Finance Officer and HR Manager, not necessarily in the same order. It is also known that:
I. The Finance Officer, who works in BOI, is sitting to the immediate left of the HR Manager.
II. The Marketing Officer, who works in SBI, is sitting opposite the Technical Officer.
III. The Agricultural Officer, who works in ICICI, is sitting diagonally opposite the Rajbhasha Adhikari, who is sitting to the immediate left of the Technical Officer.
IV. The Law Officer, who works in HDFC, is sitting to the immediate left of the Marketing officer.
V. The persons who work in PNB and HDFC are sitting along the same side of the table.
VI. The Rajbhasha Adhikari, who works in CBI, is sitting at one of the extreme ends in his row.

(2013)

Q. The person who is sitting opposite the Law Officer works in

Solution:

According to question,

or

The person who is sitting opposite the Law Officer works in BOI.

QUESTION: 15

Directions for Questions: Answer the questions on the basis of the information given below.
Four teams—T1, T2, T3 and T4— participated in a tournament of ‘Bat and trap’, an English bat-and-ball pub game. In the tournament, each team played exactly one match with each of the other teams. The matches were played on six consecutive days of a week from Monday to Saturday. Two points were awarded to the winner of a match and no points to the loser. No match in the tournament resulted in a tie/draw. It is also known that:

(2013)

I. T1 won only one match in the tournament and it was played on Monday.
II. The match played on Thursday was won by T4.
III. T3 won against T2 on Tuesday.
IV. T2, T3 and T4 definitely did not play on Wednesday, Monday and Saturday respectively.
V. T2 and T3 ended up with the same number of points at the end of the tournament.

Q. How many points did T4 score in the tournament?

Solution:

According to question

T4 scored 2 point in the tournament.

QUESTION: 16

Directions for Questions: Answer the questions on the basis of the information given below.
Four teams—T1, T2, T3 and T4— participated in a tournament of ‘Bat and trap’, an English bat-and-ball pub game. In the tournament, each team played exactly one match with each of the other teams. The matches were played on six consecutive days of a week from Monday to Saturday. Two points were awarded to the winner of a match and no points to the loser. No match in the tournament resulted in a tie/draw. It is also known that:

(2013)

I. T1 won only one match in the tournament and it was played on Monday.
II. The match played on Thursday was won by T4.
III. T3 won against T2 on Tuesday.
IV. T2, T3 and T4 definitely did not play on Wednesday, Monday and Saturday respectively.
V. T2 and T3 ended up with the same number of points at the end of the tournament.

Q. T3 lost its match against

Solution:

According to question

T3 lost against T4 on Thursday.

QUESTION: 17

Directions for Questions: Answer the questions on the basis of the information given below.
Four teams—T1, T2, T3 and T4— participated in a tournament of ‘Bat and trap’, an English bat-and-ball pub game. In the tournament, each team played exactly one match with each of the other teams. The matches were played on six consecutive days of a week from Monday to Saturday. Two points were awarded to the winner of a match and no points to the loser. No match in the tournament resulted in a tie/draw. It is also known that:

(2013)

I. T1 won only one match in the tournament and it was played on Monday.
II. The match played on Thursday was won by T4.
III. T3 won against T2 on Tuesday.
IV. T2, T3 and T4 definitely did not play on Wednesday, Monday and Saturday respectively.
V. T2 and T3 ended up with the same number of points at the end of the tournament.
Q. The match played on Friday was between

Solution:

According to question

The match played on Friday was between T2 and T4.

QUESTION: 18

Directions for Questions: Answer the questions on the basis of the information given below.
Five friends, viz. Ashok, Amit, Ajay, Akansh and Abhishek are living in five different cities named Kunnamangalam, Joka, Vastrapur, Banerghatta and Prabandhnagar, not necessarily in that order. Their salaries are 700000, 800000, 900000, 1100000, 1300000 (INR per annum), in no particular order. Further, the following information is given about them:

(2012)

I. Akansh, who does not live in Banerghatta, earns a salary that is a prime number multiple of 100000.
II. Amit made a call to one of his four mentioned friends who lives in Prabandhnagar and earning a perfect square multiple of 100000 INR in salary.
III. Ajay’s salary is 100000 INR more than the average salary of Akansh and Ashok
IV. Amit lives in the city, which has the shortest name amongst the above cities.

Q. If Akansh lives in Vastrapur, then what is the average salary of the persons living in Banerghatta and Kunnamangalam?

Solution:

According to question,

Where 'B', 'K', 'V' and 'P' stand for Banerghatta, Kunnamangalam, Vastrapur and Prabandhnagar respectively.
Data is not sufficient

QUESTION: 19

Directions for Questions: Answer the questions on the basis of the information given below.
Five friends, viz. Ashok, Amit, Ajay, Akansh and Abhishek are living in five different cities named Kunnamangalam, Joka, Vastrapur, Banerghatta and Prabandhnagar, not necessarily in that order. Their salaries are 700000, 800000, 900000, 1100000, 1300000 (INR per annum), in no particular order. Further, the following information is given about them:

(2012)

I. Akansh, who does not live in Banerghatta, earns a salary that is a prime number multiple of 100000.
II. Amit made a call to one of his four mentioned friends who lives in Prabandhnagar and earning a perfect square multiple of 100000 INR in salary.
III. Ajay’s salary is 100000 INR more than the average salary of Akansh and Ashok
IV. Amit lives in the city, which has the shortest name amongst the above cities.

Q. Who stays in Prabandhnagar?

Solution:

According to question,

Abhishek stays in Prabandhnagar.

QUESTION: 20

Directions for Questions: Answer the questions on the basis of the information given below.
Five friends, viz. Ashok, Amit, Ajay, Akansh and Abhishek are living in five different cities named Kunnamangalam, Joka, Vastrapur, Banerghatta and Prabandhnagar, not necessarily in that order. Their salaries are 700000, 800000, 900000, 1100000, 1300000 (INR per annum), in no particular order. Further, the following information is given about them:

(2012)

I. Akansh, who does not live in Banerghatta, earns a salary that is a prime number multiple of 100000.
II. Amit made a call to one of his four mentioned friends who lives in Prabandhnagar and earning a perfect square multiple of 100000 INR in salary.
III. Ajay’s salary is 100000 INR more than the average salary of Akansh and Ashok
IV. Amit lives in the city, which has the shortest name amongst the above cities.

Q. If Amit and Ajay live in cities with names starting with consecutive alphabets, then who lives in Vastrapur?

Solution:

According to question,

Since, Amit lives in Joka, so Ajay must be living at Kunnamangalam.
So, Akansh lives in Vastrapur.