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23 Aug 2015, 13:15
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
66% (02:05) correct
34%
(02:21)
wrong
based on 156
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In an intercollegiate competition that lasted for 3 days, 100 students took part on day 1, 120 on day 2 and 140 on day 3. If 60 took part on day 1 and day 2 and 50 took part on day 2 and day 3 and 20 took part on all three days, how many students took part only on day 2?
A. 0
B. 10
C. 30
D. 50
E. 90
Kudos for a correct solution.
A. 0
B. 10
C. 30
D. 50
E. 90
Kudos for a correct solution.
23 Aug 2015, 22:28
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Bunuel
In an intercollegiate competition that lasted for 3 days, 100 students took part on day 1, 120 on day 2 and 140 on day 3. If 60 took part on day 1 and day 2 and 50 took part on day 2 and day 3 and 20 took part on all three days, how many students took part only on day 2?
A. 0
B. 10
C. 30
D. 50
E. 90
A. 0
B. 10
C. 30
D. 50
E. 90
Ans: C
Solution: see the attached image.
We need to find the value of B
Values given to us G=20, D+G =60 so D=40, G+E=50 so E=30
B+D+G+E=120
B=30 Ans.
Attachments
Ans.jpg [ 25.76 KiB | Viewed 3083 times ]
23 Aug 2015, 23:39
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Hi Bunnel,
As per venn diagram for 3 cases:
Day 1 & 2 = 60 (given) , Only Day 1 & 2 (60 -20 (all three days)) = 40 ,
Day 2 & 3 = 50 (given) , Only Day 2 & 3 (50 -20(all three days)) = 30
Only Day 2 = 120 (Day 2) - (40+30+20) = 30. (C)
Hope it adds some value!
As per venn diagram for 3 cases:
Day 1 & 2 = 60 (given) , Only Day 1 & 2 (60 -20 (all three days)) = 40 ,
Day 2 & 3 = 50 (given) , Only Day 2 & 3 (50 -20(all three days)) = 30
Only Day 2 = 120 (Day 2) - (40+30+20) = 30. (C)
Hope it adds some value!
