Free DI and LR - 07 Practice Test - CAT

Whose among the following can be pet no.2?

A. T
B. P
C. Q
D. S

SOLUTION

Solution : B

From (iii) we get that Ts pet and Q's pet are immediately next to each other. From (iv), we get V's pet is no.3 or 4 or 5. From (v), we know that there are two pets between R's and V's pet and that Q's pet is adjacent to V. Furthur, since Q's pet and T's pet should be together, the order should be RTQV or VQTR, if it is RTQV, the right most position V's pet can occupy position no.5 and hence R will be the second pet. From condition (ii), S's pet should be in no.1 position, 6th pet will be P's, which will not satisfy condition (i). So, RQTV is not possible.

We will have only VQTR. Since S's pet is on the left side of R's pet, only P's and U's pet can then come to right. But then condition (i) and (ii), we need to have S's pet in the first place. Then P's and U's pets will br in 2nd and 3rd position in any order.

Thus the two possible arrangements are SPUVQTR and SUPVQTR.

P and U can be in position no.2. Hence answer option B

Which pet belongs to U?

A. no. 2
B. no. 3
C. no. 4
D. Cannot be determined

SOLUTION

Solution : D

From (iii) we get that Ts pet and Q's pet are immediately next to each other. From (iv), we get V's pet is no.3 or 4 or 5. From (v), we know that there are two pets between R's and V's pet and that Q's pet is adjacent to V. Furthur, since Q's pet and T's pet should be together, the order should be RTQV or VQTR, if it is RTQV, the right most position V's pet can occupy position no.5 and hence R will be the second pet. From condition (ii), S's pet should be in no.1 position, 6th pet will be P's, which will not satisfy condition (i). So, RQTV is not possible.

We will have only VQTR. Since S's pet is on the left side of R's pet, only P's and U's pet can then come to right. But then condition (i) and (ii), we need to have S's pet in the first place. Then P's and U's pets will br in 2nd and 3rd position in any order.

Thus the two possible arrangements are SPUVQTR and SUPVQTR.

U’s pet is either 2nd or 3rd

Choice (d)

If it is known that S’s pet is U’s pet’s immediate neighbour, then whose pet is to the immediate right of P’s pet?

A. U
B. V
C. S
D. R

SOLUTION

Solution : B

From (iii) we get that Ts pet and Q's pet are immediately next to each other. From (iv), we get V's pet is no.3 or 4 or 5. From (v), we know that there are two pets between R's and V's pet and that Q's pet is adjacent to V. Furthur, since Q's pet and T's pet should be together, the order should be RTQV or VQTR, if it is RTQV, the right most position V's pet can occupy position no.5 and hence R will be the second pet. From condition (ii), S's pet should be in no.1 position, 6th pet will be P's, which will not satisfy condition (i). So, RQTV is not possible.

We will have only VQTR. Since S's pet is on the left side of R's pet, only P's and U's pet can then come to right. But then condition (i) and (ii), we need to have S's pet in the first place. Then P's and U's pets will br in 2nd and 3rd position in any order.

Thus the two possible arrangements are SPUVQTR and SUPVQTR.

If S’s is U’s immediate pet, then we are talking of the 2nd arrangement above and hence.

Choice (b)

Among the following, whose pet is nearest to S’s pet?

A. R
B. V
C. U
D. T

SOLUTION

Solution : C

From (iii) we get that Ts pet and Q's pet are immediately next to each other. From (iv), we get V's pet is no.3 or 4 or 5. From (v), we know that there are two pets between R's and V's pet and that Q's pet is adjacent to V. Furthur, since Q's pet and T's pet should be together, the order should be RTQV or VQTR, if it is RTQV, the right most position V's pet can occupy position no.5 and hence R will be the second pet. From condition (ii), S's pet should be in no.1 position, 6th pet will be P's, which will not satisfy condition (i). So, RQTV is not possible.

We will have only VQTR. Since S's pet is on the left side of R's pet, only P's and U's pet can then come to right. But then condition (i) and (ii), we need to have S's pet in the first place. Then P's and U's pets will br in 2nd and 3rd position in any order.

Thus the two possible arrangements are SPUVQTR and SUPVQTR.

U’s pet is nearest to S’s pet

Choice (c)

Among the following, whose pet is farthest from R’s pet?

A. S
B. P
C. U
D. Cannot be determined

SOLUTION

Solution : A

From (iii) we get that T`s pet and Q's pet are immediately next to each other. From (iv), we get V's pet is no.3 or 4 or 5. From (v), we know that there are two pets between R's and V's pet and that Q's pet is adjacent to V. Furthur, since Q's pet and T's pet should be together, the order should be RTQV or VQTR, if it is RTQV, the right most position V's pet can occupy position no.5 and hence R will be the second pet. From condition (ii), S's pet should be in no.1 position, 6th pet will be P's, which will not satisfy condition (i). So, RQTV is not possible.

We will have only VQTR. Since S's pet is on the left side of R's pet, only P's and U's pet can then come to right. But then condition (i) and (ii), we need to have S's pet in the first place. Then P's and U's pets will br in 2nd and 3rd position in any order.

Thus the two possible arrangements are SPUVQTR and SUPVQTR.

S’s pet is farthest from R’s pet.

Choice (a)

If a person wants to go from A to I, by travelling through the least number of cities, then how many ways are available to him?

A. 4
B. 6
C. 8
D. None of these

SOLUTION

Solution : B

Option B is the correct answer.

If a person wants to visit all the towns, each being visited exactly once, and if he wants to start at A, then how many ways are available to him?

A. 6
B. 8
C. 10
D. None of these

SOLUTION

Solution : B

Following are the ways in which a person starts from A and visits all the cities.

Hence, there are 8 such ways.Choice (2)

In how many ways can a man reach D, starting from A such that the person does not visit the same city twice?

A. 5
B. 9
C. 7
D. None of these

SOLUTION

Solution : D

From F there are three ways to reach D.

From E there are three ways to reach D.

From A to E there are two ways; from A to F there are two ways.

By comparing the above ways we get that there are 8 ways in total.

Choices (4)

What is the longest possible distance to reach D, starting from A? Assume that no city can be visited twice.

A. 60 km
B. 70 km
C. 80 km
D. None of these

SOLUTION

Solution : B

70 Km

A – B – C – F – I – H – G – D or

A – B – C – F – I – H – E – D

Choice (2)

If A owns a blue house, then E’s house can be

B. black house Bangalore
C. black in Bombay
D. red in Bombay

SOLUTION

Solution : B

If A owns a blue house in Jaipur, then B owns a house in Bombay (because E cannot own in Bombay) and hence, E owns a house in Bangalore similarly, if A’s house is blue, then E’s house will be red or black. Hence, E will have red house or black house in Bangalore

Choice (2)

If A owns a white house in Bombay, then E’s house can be

A. red in Bangalore
B. blue in Bangalore
C. Green in Bombay
D. black in Jaipur

SOLUTION

Solution : D

If A owns white house in Bombay, then E will own a house in Jaipur (because B cannot own in Jaipur). Only choice (4) Jaipur. (Also not that if A owns a white house, then the colour of E’s house can be blue, red or black).

Choice (4)

If A’s house in Bangalore is red and D’s house is white, then E owns a

A. black in Bombay
B. blue in Jaipur
C. black in Jaipur.
D. blue in Bombay

SOLUTION

Solution : B

If A’s house is in Bangalore, then E can own only a house in Jaipur. If A’s house is red and D’s white, then E can own only a blue house (because C’s house cannot be blue). Hence, E has a blue house in Jaipur. Choice 2

If E owns a red house in Bangalore and A’s house is white, then D owns a

A. green in Delhi
B. black in Delhi
C. blue in Delhi
D. red in Delhi

SOLUTION

Solution : C

If E owns a red house and A owns a white house, then D can have only a blue house (because C cannot have a blue house).option c

If B owns a house in Bangalore and C’s house is black, then A can own a

A. white in Jaipur
B. red in Jaipur
C. blue in Bombay
D. blue in Jaipur

SOLUTION

Solution : C

If B owns a house in Bangalore, then A must own a house in Bombay (because E cannot own a house in Bombay).

(Choice (3)

(Note: If C’s house is black, then A’s house can be blue, red or white.)