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Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
At a college, 100 seniors play any of the three games ...
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Updated on: 30 Jul 2021, 21:31
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.
(A) 35
(B) 34
(C) 33
(D) 32
(E) 21
fixed OA.
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.
Last edited by motion2020 on 30 Jul 2021, 21:31, edited 1 time in total.
Retired Moderator
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Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
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WE:Education (Education)
Re: At a college, 100 seniors play any of the three games ...
[#permalink]
30 Jul 2021, 20:05
30 Jul 2021, 20:05
2
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
(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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Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
Re: At a college, 100 seniors play any of the three games ...
[#permalink]
30 Jul 2021, 21:31
30 Jul 2021, 21:31
1
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
(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.
