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At a college, 100 seniors play any of the three games ...

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motion2020
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At a college, 100 seniors play any of the three games ... [#permalink] New post   Updated on: 30 Jul 2021, 21:31
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At a college, 100 seniors play any of the three games: 35 play soccer, 45 play basketball and 50 play hockey. 10 seniors play all three games and 40% of seniors, playing only two games, play soccer. How many seniors play only soccer game?

(A) 35
(B) 34
(C) 33
(D) 32
(E) 21

fixed OA.
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E

Originally posted by motion2020 on 29 Jul 2021, 07:24.
Last edited by motion2020 on 30 Jul 2021, 21:31, edited 1 time in total.
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KarunMendiratta
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Re: At a college, 100 seniors play any of the three games ... [#permalink] New post   30 Jul 2021, 20:05
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motion2020 wrote:
At a college, 100 seniors play any of the three games: 35 play soccer, 45 play basketball and 50 play hockey. 10 seniors play all three games and 40% of seniors, playing only two games, play soccer. How many seniors play only soccer game?

(A) 35
(B) 34
(C) 33
(D) 32
(E) 31


motion2020
10 seniors are already playing all 3 games, how can only soccer be greater than 25 (35 - 10)???
What is it that I am missing?

AUBUC = A + B + C - (d + e + f) - 2(A∩B∩C)
100 = 35 + 45 + 50 - (d + e + f) - 2(10)
(d + e + f) = 110 - 100 = 10

Now, 40% of seniors, playing only two games, play soccer.
d + f = \(\frac{40}{100} (10) = 4\)

So, seniors playing only soccer = 35 - 4 -10 = 21
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motion2020
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Re: At a college, 100 seniors play any of the three games ... [#permalink] New post   30 Jul 2021, 21:31
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KarunMendiratta wrote:
motion2020 wrote:
At a college, 100 seniors play any of the three games: 35 play soccer, 45 play basketball and 50 play hockey. 10 seniors play all three games and 40% of seniors, playing only two games, play soccer. How many seniors play only soccer game?

(A) 35
(B) 34
(C) 33
(D) 32
(E) 31


motion2020
10 seniors are already playing all 3 games, how can only soccer be greater than 25 (35 - 10)???
What is it that I am missing?

AUBUC = A + B + C - (d + e + f) - 2(A∩B∩C)
100 = 35 + 45 + 50 - (d + e + f) - 2(10)
(d + e + f) = 110 - 100 = 10

Now, 40% of seniors, playing only two games, play soccer.
d + f = \(\frac{40}{100} (10) = 4\)

So, seniors playing only soccer = 35 - 4 -10 = 21

Thanks for pointing out - you are correct.
I have edited OA.
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Re: At a college, 100 seniors play any of the three games ... [#permalink] New post   06 Nov 2022, 05:41
Why 2*10 for (A∩B∩C)
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Re: At a college, 100 seniors play any of the three games ... [#permalink] New post   07 Nov 2022, 10:13
35 Don't use brain bros
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At a college, 100 seniors play any of the three games ... [#permalink] New post   07 Nov 2022, 10:28
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sanketthosani wrote:
Why 2*10 for (A∩B∩C)


Refer to the figure Below:

L refers to leftovers i.e. Not participating in any group

S = a + d + x + f
B = b + e + x + d
H = c + e + x + f

S + B + H = a + b + c + 2e + 2d + 3x


We know,
Total - L = a + b + c + d + e + f + x

Now, if we write Total - L = A + B + C - d - e - f = a + b + c + 2e + 2d + 3x - d - e - f = a + b + c + d + e + f + 3x
We need to subtract 2x from it in order to get the correct form.

Hence, we subrtract 2x.
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