## Matrix Arrangement Questions for CAT

Download important CAT Matrix Arrangement Questions with Solutions PDF based on previously asked questions in CAT exam. Practice Matrix Arrangement Questions with Solutions for CAT exam.

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**Question 1: **Eight people carrying food baskets are going for a picnic on motorcycles.Their names are A, B, C, D, E, F, G, and H. They have 4 motorcycles M1, M2, M3 and M4 among them. They also have 4 food baskets O, P, Q and R of different sizes and shapes and each can be carried only on motorcycles M1, M2, M3 and M4 respectively.No more than 2 persons can travel on a motorcycle and no more than one basket can be carried on a motorcycle. There are 2 husband-wife pairs in this group of 8 people and each pair will ride on a motorcycle together.C cannot travel with A or B. E cannot travel with B or F. G cannot travel with F, or H, or D.The husband-wife pairs must carry baskets O and P. Q is with A and P is with D.F travels on M1 and E travels on M2 motorcycles.G is with Q, and B cannot go with R.Who is travelling with H?

a) A

b) B

c) C

d) D

**Question 2: **Persons X, Y, Z and Q live in red, green, yellow or blue coloured houses placed in a sequence on a street. Z lives in a yellow house. The green house is adjacent to the blue house. X does not live adjacent to Z. The yellow house is the only house in between the green and red houses. The colour of the house X lives in is:

a) blue

b) green

c) red

d) not possible to determine

**Question 3: **There are ten animals — two each of lions, panthers, bison, bears, and deer — in a zoo. The enclosures in the zoo are named X, Y, Z, P and Q and each enclosure is allotted to one of the following attendants: Jack, Mohan, Shalini, Suman and Rita. Two animals of different species are housed in each enclosure. A lion and a deer cannot be together. A panther cannot be with either a deer or a bison. Suman attends to animals from among bison, deer , bear and panther only. Mohan attends to a lion and a panther. Jack does not attend to deer, lion or bison. X, Y and Z are allotted to Mohan, Jack and Rita respectively. X and Q enclosures have one animal of the same species. Z and P have the same pair of animals. The animals attended by Shalini are:

a) bear & bison

b) bison & deer

c) bear & lion

d) bear & panther

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**Instructions**

Directions for the following three questions: Answer the following questions based on the statements given below:

(i) There are three houses on each side of the road.

(ii) These six houses are labeled as P, Q, R, S, T and U.

(iii) The houses are of different colours, namely, Red, Blue, Green, Orange, Yellow and White.

(iv) The houses are of different heights.

(v) T, the tallest house, is exactly opposite to the Red coloured house.

(vi) The shortest house is exactly opposite to the Green coloured house.

(vii) U, the Orange coloured house, is located between P and S.

(viii) R, the Yellow coloured house, is exactly opposite to P.

(ix) Q, the Green coloured house, is exactly opposite to U.

(x) P, the White coloured house, is taller than R, but shorter than S and Q.

**Question 4: **What is the colour of the house diagonally opposite to the Yellow coloured house?

a) White

b) Blue

c) Green

d) Red

e) none of these

**Question 5: **Which is the second tallest house?

a) P

b) S

c) Q

d) R

e) cannot be determined

**Question 6: **What is the colour of the tallest house?

a) Red

b) Blue

c) Green

d) Yellow

e) none of these

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**Answers & Solutions:**

**1) Answer (C)**

It is given in the statements, that C cannot travel with A or B. E cannot travel with B or F. G cannot travel with F, or H, or D. By formulating the table we get

Q is with A and G is with Q => G and Q are travelling together on motorcycle M3

F travels on M1 and E travels on M2 motorcycles.

D is travelling with P on M2 => D and E are traveling together on M2

B cannot go with R => F and B go together on M1

Therefore, C and H go together on M4

So, the table can formed as below :

Hence, C would be travelling with H.

**2) Answer (A)**

According to given condition the correct sequence of houses is 1st is blue, 2nd green, 3rd is yellow and last red.

Now in the yellow house i.e. 3rd Z lives and X doesnt live as a neighbour i.e. Green and Red house.

So, X lives in blue house.

**3) Answer (C)**

Correct arrange ment would be

Hence option C.

**4) Answer (D)**

Before directly trying to answer the question, it is important to gather all the information given by the question.

There are three houses on each side of the road => Draw 6 lines, 3 in each row, to accommodate P, Q, R, S, T and U.

The houses are of different colours and different heights.

T is tallest and is opposite to red house => Let’s number T as 1.

Shortest house is opposite to green house.

U is orange and is between P and S => Two cases arise here. P-U-S is one possibility and the other possibility is S-U-P.

R is yellow and is opposite to P.

Q is green and is opposite to U. We know that green house is opposite to the shortest house. This implies that U is the shortest house => Number of U is 6.

P is white and is taller than R but shorter than S and Q => Apart from T, S and Q are also taller than P => S and Q can be 2 and 3 in any order => Number of P is 4 and number of R is 5.

We know that P is opposite to R and Q is opposite to U => S is opposite to T

It is given that T is opposite to red house => S is the red house and hence T is the blue house.

So, we know the colours of all houses and heights of P, R, T and U.

In this question, we are asked to find the house that is opposite to yellow house. R is the yellow house, P is opposite to R and S is on the other corner in P’s row. Hence S is the house that is diagonally opposite to yellow house and the colour of S is Red.

**5) Answer (E)**

Before directly trying to answer the question, it is important to gather all the information given by the question.

There are three houses on each side of the road => Draw 6 lines, 3 in each row, to accommodate P, Q, R, S, T and U.

The houses are of different colours and different heights.

T is tallest and is opposite to red house => Let’s number T as 1.

Shortest house is opposite to green house.

U is orange and is between P and S => Two cases arise here. P-U-S is one possibility and the other possibility is S-U-P.

R is yellow and is opposite to P.

Q is green and is opposite to U. We know that green house is opposite to the shortest house. This implies that U is the shortest house => Number of U is 6.

P is white and is taller than R but shorter than S and Q => Apart from T, S and Q are also taller than P => S and Q can be 2 and 3 in any order => Number of P is 4 and number of R is 5.

We only know that the second tallest house is either Q or S. Hence the answer is cannot be determined.

**6) Answer (B)**

Before directly trying to answer the question, it is important to gather all the information given by the question.

There are three houses on each side of the road => Draw 6 lines, 3 in each row, to accommodate P, Q, R, S, T and U.

The houses are of different colours and different heights.

T is tallest and is opposite to red house => Let’s number T as 1.

Shortest house is opposite to green house.

U is orange and is between P and S => Two cases arise here. P-U-S is one possibility and the other possibility is S-U-P.

R is yellow and is opposite to P.

Q is green and is opposite to U. We know that green house is opposite to the shortest house. This implies that U is the shortest house => Number of U is 6.

P is white and is taller than R but shorter than S and Q => Apart from T, S and Q are also taller than P => S and Q can be 2 and 3 in any order => Number of P is 4 and number of R is 5.

We know that P is opposite to R and Q is opposite to U => S is opposite to T

It is given that T is opposite to red house => S is the red house and hence T is the blue house.

T is the tallest house and hence the colour of the tallest house is blue.

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