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There are 600 boys in a hostel. Each plays either hockey or football or both. If 75% play hockey and 45% play football, how many play both?
(A) 48
(B) 60
(C) 80
(D) 120
(E) 150
(A) 48
(B) 60
(C) 80
(D) 120
(E) 150
Updated on: 23 Apr 2021, 08:47
Originally posted by ntngocanh19 on 23 Apr 2021, 08:36.
Last edited by ntngocanh19 on 23 Apr 2021, 08:47, edited 1 time in total.
Last edited by ntngocanh19 on 23 Apr 2021, 08:47, edited 1 time in total.
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Each plays either hockey or football or both.
So we have:
- 75% play hockey => 25% only play football
- 45% play football => 55% only play hockey
=> % play both football and hockey = 100-25-55 = 20%
=> Number of boys play both is 600*20%= 120
Hence choice D
(Edited: I miscalculate : ()
So we have:
- 75% play hockey => 25% only play football
- 45% play football => 55% only play hockey
=> % play both football and hockey = 100-25-55 = 20%
=> Number of boys play both is 600*20%= 120
Hence choice D
(Edited: I miscalculate : ()
23 Apr 2021, 08:43
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There are 600 boys in a hostel. Each plays either hockey or football or both. If 75% play hockey and 45% play football, how many play both?
(A) 48
(B) 60
(C) 80
(D) 120
(E) 150
This is a sets Problem
The formula is
\(AUB = A + B - A∩B \)
Total number of boys = Boys who play hockey + Boys who play Football - Boys who play both
Boys who play Hockey =\(\frac{75}{100}\) *600 =450
Boys who play Football =\(\frac{45}{100}\) *600 =270
Therefore substituting the values in the above formula
600=450+270 -Boys who play both
Boys who play both = 120 Option D
(A) 48
(B) 60
(C) 80
(D) 120
(E) 150
This is a sets Problem
The formula is
\(AUB = A + B - A∩B \)
Total number of boys = Boys who play hockey + Boys who play Football - Boys who play both
Boys who play Hockey =\(\frac{75}{100}\) *600 =450
Boys who play Football =\(\frac{45}{100}\) *600 =270
Therefore substituting the values in the above formula
600=450+270 -Boys who play both
Boys who play both = 120 Option D
