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15 Aug 2017, 01:10
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D
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Difficulty:
15%
(low)
Question Stats:
84% (01:25) correct
16%
(01:35)
wrong
based on 348
sessions
History
Date
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50 children attended a carnival where cotton candy and ice cream were sold. If 14 children ate only cotton candy, 18 children ate only ice cream, and 8 children had neither cotton candy nor ice cream, how many children ate ice cream?
A. 18
B. 20
C. 24
D. 28
E. 32
A. 18
B. 20
C. 24
D. 28
E. 32
15 Aug 2017, 01:48
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Let the children who ate cotton candy be P(CC)
Similarly, the children who ate ice-cream be P(IC)
Given data:
P(Total) = 50
P(Only CC) = 14
P(Only IC) = 18
P(Neither) =8
Substituting values,
P(Total) = P(Only CC) + P(Only IC) + P(Both) + P(Neither)
50 = 14 + 18 + P(Both) + 8
18 - 8 = P(Both)
P(Both) = 10
Therefore, P(IC) = P(Only IC) + P(Both) = 18+10 = 28(Option D)
Similarly, the children who ate ice-cream be P(IC)
Given data:
P(Total) = 50
P(Only CC) = 14
P(Only IC) = 18
P(Neither) =8
Substituting values,
P(Total) = P(Only CC) + P(Only IC) + P(Both) + P(Neither)
50 = 14 + 18 + P(Both) + 8
18 - 8 = P(Both)
P(Both) = 10
Therefore, P(IC) = P(Only IC) + P(Both) = 18+10 = 28(Option D)
15 Aug 2017, 15:48
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Bunuel
Attachment:
treatskids.jpg [ 22.68 KiB | Viewed 4715 times ]
I might be mistaken, but I think this problem requires only two calculations. Please correct me if I'm mistaken.
We know there are 50 children. The "only cotton candy and neither" give the total for no ice cream = 22. Subtract that from 50 and you're done.
The two red arrows are the two calculations I think are required.
Blue numbers are given, red are computed, and green are . . . well, I don't know why I put them there. I didn't when I was sketching.
Slightly tricky: "only cotton candy" is not a total for cotton candy -- it means they ate cotton candy and no ice cream.
Similarly, "only ice cream" is not a total -- it means they ate ice cream and no cotton candy.
Total ice cream: 28
Answer D
Hope it helps.
