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Permutation and Combination
out of the 727 members of a club, 600 decide in favour of cricket while 173 decide in favour of cricket and hockey. each member gives his opinion about at least one of the two games.
Q1) how many of them decide in favour of hockey?
a. 250
b. 300
c. 560
d. 600
Q2. how maney of them decide in favour of only hockey?
a. 127
b. 280
c. 326
d. 494
Read Solution (Total 6)
-
- P(C U H)= P(C) + P(H) - P(CnH)
727 = 600 + P(H) -173
=> P(H)= 727+173 -600= 300
So, in favour of only hockey= 300-173= 127
Ans 1) b. 300
Ans 2) a. 127 - 8 years agoHelpfull: Yes(13) No(0)
- P(C U H)= P(C) + P(H) - P(CnH)
727 = 600 + P(H) -173
=> P(H)= 727+173 -600= 300
So, in favour of only hockey= 300-173= 127
Ans 1) b. 300
Ans 2) a. 127 - 7 years agoHelpfull: Yes(3) No(0)
- 1)b Only Hockey +hockey and cricket both
2)a - 7 years agoHelpfull: Yes(0) No(1)
- members who favour cricket=600
members who favour both = 173
hence
members who favour ONLY cricket= 600-173=427
members who favour ONLY hockey= 727-600= 127
members who favour hockry = 127+173=300 - 7 years agoHelpfull: Yes(0) No(1)
- hockey =727-600+173
so hockey =300
only hockey =127
so Option b and a - 7 years agoHelpfull: Yes(0) No(0)
- Out of 727 members
600 ( Only cricket) and 173 ( Cricket+Hockey)
Then 600+173=773
773-727=46( 46 members play both Cricket and Hockey both)
hence 173-46=127(127 members plays only Hockey)
127+173=300 (300 members plays hockey out of 727 i,e which include both Cricket players and Hockey Players) - 6 years agoHelpfull: Yes(0) No(0)
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