Daily Practice Questions CAT Previous Year Questions with Answer PDF

Let n, s, d and t be the number of students who likes none of the drinks, exactly one drink, exactly 2 drinks and all three drinks, respectively.

It is given, 

n + s + d + t = 100 ...... (1)

s + 2d + 3t = 73 + 80 + 52

s + 2d + 3t = 205 ...... (2)

(2)-(1), we get

d + 2t - n = 105

Maximum value t can take is 52, i.e. t = 52, d = 1 and n = 0

Minimum value t can take is 5, i.e. t = 5, d = 95 and n = 0 (This also satisfies equation (1))

Difference = 52 - 5 = 47

The answer is option A.

Now 23 students choose maths as one of their subject.

This means (MPC)+ (MC) + (PC)=23 where MPC denotes students who choose all the three subjects maths, physics and chemistry and so on.

So MC + PM =5 Similarly we have PC+ MP =7 

We have to find the smallest number of students choosing chemistry

For that in the first equation let PM=5 and MC=0. In the second equation this PC=2

Hence minimum number of students choosing chemistry will be (18+2)=20 Since 18 students chose all the three subjects.

From observing the data given, we find that it is a closed 3 set Venn diagram.
Let the three sports be F, T and C for Football, Tennis and Cricket respectively
n(F U T U C) = 256 , n(F) = 144, n(T) = 123, n(C) = 132, n(F ∩ T) = 58, n(C ∩ T) = 25, n(F ∩ C) = 63

We know that (AUBUC) = n(A) + n(B) +n(C) - n(A ∩ B) - n(B ∩C) - n(C ∩ A) + n(A ∩ B ∩ C)
So, 256 = 144 + 123 + 132 - 58 - 25 - 63 + n (F ∩ T ∩ C)
n (F ∩ T ∩ C) = 256 - 144 + 123 +132 - 146
n (F ∩ T ∩ C) = 256 - 253 = 3
Now, it is easy to calculate the number of students who only play tennis using a Venn diagram.
n (Students who play only Tennis) = 123 - (55 + 3 + 22) = 123 - 80
n (Students who play only Tennis) = 43 students

Given that out of 200 students, 105 like Pizza and 134 like Burger
In order to find the possible number of students who like only Burger, we need to find the range between the minimum and maximum possible number of students

Minimum possible number of Students who like only Burger:

For this, we must try to optimize the Venn diagram in such a way that it has maximum intersection
No of Students who like only Burger = 134 – 105= 29

Maximum possible number of Students who like only Burger:

For this, we must optimize the Venn diagram in such a way that it has minimum intersection
We know that total number of Students = 200
So, Minimum number of students who like both = 105 + 134 – 200 = 39 students
Students who like Burger = 134
Max. no of Students who like only Burger = 134 – 39 = 95
So range => 29 ≤ n ≤ 95
From the options, only (D) 93 lies within range

Given that for two sets A and B, A △ B denote the set of elements which belong to A or B but not both.
This is the diagram which represents A △ B ,the set of elements which belongs to A or B i.e. A intersection B subtracted from A union B
We have to find the number of elements in (P △ Q) △ (R △ S) if
P = {1 , 2 , 3 , 4} Q = {2 , 3 , 5 , 6 ,} R = {1 , 3 , 7 , 8 , 9} S = {2 , 4 , 9 , 10}
For P △ Q, we have to find the elements which do not to belong to both P and Q such that
(P △ Q) = {1, 4 , 5 , 6}
(R △ S) = {1 , 2 , 3 , 4 , 7 , 8 , 10}
(P △ Q) △ (R △ S) = {2 , 3 , 5 , 7 , 6 , 8 , 10}
The number of elements (P △ Q) △ (R △ S) is 7

Given that A = {6²ⁿ - 35n - 1 : n = 1 , 2 , 3 , ...} and B = {35(n - 1) : n = 1 , 2 , 3 , ...}
A = {6²ⁿ - 35n - 1}
When n = 1
{6² - 35 - 1} = 0
when n = 2
{6⁴ - 35(2) - 1} = 1225
when n = 3
{6⁶ - 35(3) - 1} = 46550
B = {35(n - 1) : n = 1 , 2 , 3 , ...}
When n = 1
{35(0)} = 0
when n = 2
{35(1)} = 35
when n = 3
{35(2)} = 70
So B is more predictable here since it has every multiple of 35 ie. {0, 35, 70, 10, 140, .......}
From this we can see that A is a subset of B ie. Every element of A is in B and not the vice versa
Therefore every member of A is in B and at least one member of B is not in A.