# Venn diagram MBA PYQs

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### Q1. In a town of 10000 families, it was found that 40% families buy product A, 20% buy product B and 10% buy product C, 5% buy product A and product B, 3% buy product B and product C and 4% buy product A and product C. If 2% families buy product A, B, C all. Then find the number of the families buy product A only.

Solution:
Formula:
n(A U B U C) = n(A)+n(B)+n(C)-n(A ∩ B)-n(A ∩ C)-n(B ∩ C)+n(A ∩ B ∩ C)
Given,
n(A U B U C) = 10000
n(A) = 4000
n(B) = 2000
n(C) = 1000
n(A ∩ B ∩ C) = 200
n(A ∩ B) = 500
n(A ∩ C) = 400
n(B ∩ C) = 300
Number of the families buy product A only ?

From the Venn diagram,
Number of the families by product A only = 3300.

### Q2. In a managers club, 45 play polo, out of which 30 play Polo only 28 play Snookers. 25 play Tennis of which 11 play Tennis only, 7 play Tennis and Polo, but not Snooker. 5 play Polo and Snooker, but not Tennisi) How many play all the thre sports?ii) How many play Snookers only?iii) How many members are there is the club.

Solution:
P for Polo
S for Snookers
T for Tennis

From the Venn diagram:
i) How many play all the thre sports?
Ans. 3
ii) How many play Snookers only?
Ans. 16
iii) How many members are there is the club.
Ans. 76

### Q3. In a survey of 500 T.V. viewers, 285 watched KBC, 195 watch cricket, 115 watch hockey, 45 watch KBC and hockey, 70 watch KBC and cricket, 50 watch cricket and hockey, 50 do not watch any of three games. How many watch all 3 and how many watch exactly one of three ?

Solution:
K for KBC
H for Hockey
C for Cricket
Formula:
n(K U H U C) = n(K)+n(H)+n(C)-n(K ∩ H)-n(K ∩ C)-n(H ∩ C)+n(K ∩ H ∩ C)
Given,
n(K U H U C) = 500 – 50 = 450
n(K) = 285
n(H) = 115
n(C) = 195
n(K ∩ H) = 45
n(K ∩ C) = 70
n(H ∩ C) = 50
n(K ∩ H ∩ C) = ?
How many watch all the three games ?
n(K U H U C) = n(K)+n(H)+n(C)-n(K ∩ H)-n(K ∩ C)-n(H ∩ C)+n(K ∩ H ∩ C)
450 = 285+115+195-45-70-50+n(K ∩ H ∩ C)
n(K ∩ H ∩ C) = 20
How many watch all 3 ?
Ans. 20
Now, how many watch exactly one ?

From the Venn diagram,
Watch exactly KBC = 190
Watch exactly Hockey = 40
Watch exactly Cricket = 95

### Q4. In a city there are 100000 people, 64% of them speak Greek, 55% people speak Latin, 43% people speak French, 21% people speak both Greek and Latin, 31% people speak both Greek and French, and 41% people speak both Latin and French. Determine the number of people speak all the three languages.

Solution:
G for Greek
L for Latin
F for French
Formula:
n(G U L U F) = n(G)+n(L)+n(F)-n(G ∩ L)-n(G ∩ F)-n(L ∩ F)+n(G ∩ L ∩ F)
Given,
n(G U L U F) = 100000
n(G) = 64% = 64000
n(L) = 55% = 55000
n(F) = 43% = 43000
n(G ∩ L ∩ F) = ?
n(G ∩ L) = 21% = 21000
n(G ∩ F) = 31% = 31000
n(L ∩ F) = 41% = 41000
n(G U L U F) = n(G)+n(L)+n(F)-n(G ∩ L)-n(G ∩ F)-n(L ∩ F)+n(G ∩ L ∩ F)
100000 = 64000+55000+43000-21000-31000-41000+n(G ∩ L ∩ F)
n(G ∩ L ∩ F) = 31000
Number of people speak all the three languages = 31000.

### Q5. A company studies the product preferences of 20,000 consumers. It was found that each of the products A, B and C was liked by 7020, 6230 and 5980 respectively. All products were liked by 1500. Products A and B were liked by 2580, products A and C were liked by 1200 and products B and C were liked by 1950. Prove that the study results are not correct.

Solution:
Formula:
n(A U B U C) = n(A)+n(B)+n(C)-n(A ∩ B)-n(A ∩ C)-n(B ∩ C)+n(A ∩ B ∩ C)
Given,
n(A U B U C) = 20000
n(A) = 7020
n(B) = 6230
n(C) = 5980
n(A ∩ B ∩ C) = 1500
n(A ∩ B) = 2580
n(A ∩ C) = 1200
n(B ∩ C) = 1950
n(A U B U C) = 7020+6230+5980-2580-1200-1950+1500 = 15000
But, given n(A U B U C) = 20000,
So result is not correct.
n(A U B U C) ≠ 15000
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