Latest Miscellaneous Company Exam Aptitude Question SOLUTION: total no. of players is 99 other than me.50 plays soccer,45 plays basket ball,50 play valley ball.15 play all three. How many will play only 2 games/

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(#M40007076) MISCELLANEOUS COMPAN question Keep an EYE

total no. of players is 99 other than me.50 plays soccer,45 plays basket ball,50 play valley ball.15 play all three. How many will play only 2 games/

Asked In Miscellaneous Company Exam sankha chowdhury (12 years ago)

Unsolved Read Solution (5)

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rachana dhekane  (11 years ago)

let A=soccer players..B=basket ball players..C=vollyball players
n(A U B U C)=n(A) + n(B) + n(C) - n(A INTERSECTION B) - n(A INTERSECTION C) - n(C INTERSECTION B) + n(A INTERSECTION B - INTERSECTION C)
therefore..
100=50+45+50-intersections+15
hence..intersections=145-100+15=60play only 2 games..

Is this solution Helpfull? Yes (46) | No (12)  

aditya  (11 years ago)

s=play soccer
b=play basketball
v=play volliball

50S + 45B + 50V - 15(SBV)= no of ppl who play games atlest one or two
now,
no of ppl who play two games = no of ppl who play games atlest one or two -100 + 15 = 15

Is this solution Helpfull? Yes (8) | No (13)  

Abhilash Mishra  (10 years ago)

let A = soccer, B = basket, C = volleyball players
n(A U B U C)=n(A)+n(B)+n(C)-n(A intersection B)-n(B inter.. C)-n(A inter.. C) + n(A inter.. B inter.. C)

hence,
100=50+45+50-intersections+15

=> intersections = 60

thus, 60 players play 2 games....and out of which 15 play all 3 games.
hence, no. of players those will play only 2 games = 60-15 = 45. ANS.

Is this solution Helpfull? Yes (1) | No (1)  

Bhairav Patil  (9 years ago)

Let the people who play only two games be : x (Soccer and basketball) y(soccer and volleyball) z(Volleyball and basketball) Therefore, 100 = 50 + 45 + 50 -(x+15) -(y+15) -(z+15) + 15. Solving, we get x+y+z=15.

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sravani  (9 years ago)

total =99 + 1 =100
S + V + B +S^V^B-(S^V + V^B + S^B)=Total
50 + 45 + 50 +15 - x =100
160 - x = 100
x=60
so, 60 members play only 2 games...

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