TITA Based Questions: Set theory | Logical Reasoning (LR) and Data Interpretation (DI) - CAT PDF Download

Ans: 202
This question involves both Number Systems and Set Theory.
Set P = {4, 8, 12, ….496} ↦ 124 elements {all elements from 1 * 4 to 124 * 4}
Set Q = {7, 21, 35, 49,……497} ↦ {7 * 1, 7 * 3, 7 * 5, ….. 7 * 71} ↦ 36 elements.
Set R = {6, 12, 18, 24, …..498} ↦ {6 * 1, 6 * 2, 6 * 3, ….. 6 * 83} ↦ 83 elements.
Sets P and R have only even numbers; set Q has only odd numbers. So,
P ∩ Q = Null set
Q ∩ R = Null set
P ∩ Q ∩ R = Null set
So, If we find P ∩ R , we can plug into the formula and get P ∪ Q ∪ R
P ∩ R = Set of all multiples of 12 less than 500 = {12, 24, 36,…..492}
= {12 * 1, 12 * 2 , 12 * 3, …12 * 41} ↦ This has 41 elements
P ∪ Q ∪ R = P + Q + R – (P ∩ Q) – ( Q ∩ R) – (R ∩ P) + (P ∩ Q ∩ R)
P ∪ Q ∪ R = 124 + 36 + 83 – 0 – 0 – 41 + 0 = 202

Ans: 420
Set A = {100, 105, 110, ….995} ↦ {5 * 20, 5 * 21, ….. 5 * 199} ↦ 180 elements.
Set B = {102, 108, 114, ……996} ↦ {6 * 17, 6 * 18, 6 * 19, ….. 6 * 166} ↦ 150 elements.
Set C = {100, 104, 108, …..996} ↦ {4 * 25, 4 * 26, ….. 4 * 249} ↦ 225 elements.
A ∩ B = {120, 150, 180, …..990} ↦ All 3-digit multiples of 30 ↦ 30 elements.
B ∩ C = {108, 120, 132, …..996} ↦ All 3-digit multiples of 12 ↦ 75 elements.
C ∩ A = {120, 140, 160, …..980} ↦ All 3-digit multiples of 20 ↦ 45 elements.
A ∩ B ∩ C = {120, 180, …..960} ↦ All 3-digit multiples of 60 ↦ 15 elements.
A ∪ B ∪ C = A + B + C – A ∩ B – B ∩ C – C ∩ A + A ∩ B ∩ C
= 180 + 150 + 225 – 30 – 75 – 45 + 15 = 420

Ans: 36
Let the total No. of days Sonu deposited the earnings in Axis bank be ‘n(a)’ and
that in SBI be ‘n(b)’
Now,
Total no of days he carried on the business = n(a) + n(b) – n(a∩b) + Neither (Days he didn’t earned)
Since on any given day he does not deposit in both the bank accounts
n(a∩b) = 0
Hence,
Total no of days he carried on the business = n(a) + n(b) + Neither (Days he didn’t earned)
Now,
A/Q
Total – n(a) = 20……………………………(1)
Total – n(b) = 24……………………………(2)
n(a) + n(b) = 28………………………………(3)
Solving the 3 equations we get,
n(a) = 16
n(b) = 12

Ans: 51
Let the total no of students doing CA be n(A), those doing actuarial be n(B) and those doing MBA be n(C).
Now, ‘CA’ is a set of all even numbered students, thus
‘Actuarial’ is a set of all the students whose number are divisible by 5, thus n(B) = 150/5 = 30
‘MBA’ is a set of all the students whose number are divisible by 7, thus n(C) = 150/7 = 21
The 10^th, 20^th, 30^th…… numbered students would be doing both CA and Actuarial 
∴ n(A∩B) = 150/10 = 5
The 14^th, 28^th, 42^nd…… numbered students would be doing both CA and MBA
∴ n(A∩C) = 150/40 = 10
The 35^th, 70^th, …… numbered students would be doing both Actuarial and MBA
∴ n(B∩C) = 150/35 = 4
​And the 70^th and 140^th students must be doing all the three.
​∴n(A∩B∩C) = 2
Now, n(A∪B∪C) = n(A)+n(B)+n(C) – n(A∩B) – n(A∩C) - n(B∩C) + n(A∩B∩C) = 99
​∴ Number of students doing nothing = 150 – 99 = 51

Ans: 124
Let total no of students who took English be x
Then students who took math, science will also be x
Now let’s draw the Venn diagram
E U M U S = 345 – 43 (Neither of the subjects)
E U M U S = E + M + S – E ∩ M - E ∩ S - S ∩ M + E ∩ M ∩ S
Thus the total no of students who took English as a subject = 124
Consequently the Venn diagram becomes

Again,
The students who has taken only one subject = E U M U S - E ∩ M - E ∩ S - S ∩ M - E ∩ M ∩ S
= 302 - 16 - 14 - 14 - 12 = 246
The students who took English and Math but not science = only E + Only M + E ∩ M
= 80 + 82 + 16 = 178
Percent of students who took English and Math but not science = 178/302 * 100 = approx. 59 %
Hence, the answer is "124".

Ans: 120

Diagrammatic Approach -

It is quite clear from the diagram that the number of people at the party were 60 + 20 + 40 = 120Mathematical approach :

Ans: 80
Since, for every person who uses ‘F’ no of facilities, there are exactly 3 person who uses atleast (F-1) no. of facilities. If we take the no. of persons who uses all the four facilities to be ‘x’ then the no. of person who uses atleast 3 facilities will be 3x and so on.
The No. of persons who use exactly 3 facilities = 3x – x = 2x
Thus, the no. of persons who opt for various facilities can be summarized as follows:

We also know that no of person who uses no facilities = twice of those who uses all the 4 facilities = 2x.
So according the above deductions we can clearly see that the number of person the hotel would be
x + 2x + 6x + 18x + 2x = 29 x = 1160
x = 40
Hence the number of person who uses exactly three science = 2x = 80

Ans: 303
Let the no. of students who participated in 0,1,2,3 games be A, B, C, D respectively.
Then from the information we have we can conclude that
C + D = 152% 0f B = 1.52 of B -----------------------(1)
Since the total no. of students the college has is 420,
A + B + C + D = 510 ---------------------------------(2)
From (1) and (2), we can conclude that
A + 2.52B = 510
B = 25/635​* (510 – A)
For A to be minimum, 510 – A should give us the largest multiple of 63. Since, 63 * 8 = 504 we will get A = 6.
So B = 200 and C + D = 1.52 B = 304
Thus for the students participating in exactly 3 games to be maximum, the number of students participating in exactly 2 games has to be minimized and made equal to 1.
Thus no of students who participated in exactly 3 games = 304 – 1 = 303.

Ans: 20
Let’s start with taking a random value for all three category. So let’s first take 40 for the all three category. Now 65 + 60 + 55 = 180, this means there is an extra count of 180 – 80 = 100.
Now as we know that the extra count occurs in the in the exactly two area and the all three area. So let’s try put the extra count in these area.
Trial 1 - Since 40 is already assumed to be in the all three area, it takes care of extra count of 40 x 2 = 80. Thus we are left with 20 as extra count which we have to place at the exactly two area
Thus in the above case our venn diagram will look as:

A close look in the figure tells us that we can further decrease the value of all the three area. A bit of logical thinking will bring us to the value 20. No value less than 20 can satisfy the conditions of the question. As there is no scope left for reallocating numbers left from one area to another in this case.
Hence the final venn diagram will look as: