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50 children attended a carnival where cotton candy and ice cream were

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50 children attended a carnival where cotton candy and ice cream were [#permalink]

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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
[Reveal] Spoiler: OA

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50 children attended a carnival where cotton candy and ice cream were [#permalink]

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New post 15 Aug 2017, 01:48
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)
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50 children attended a carnival where cotton candy and ice cream were [#permalink]

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Bunuel wrote:
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

Attachment:
treatskids.jpg
treatskids.jpg [ 22.68 KiB | Viewed 704 times ]

The double matrix works well here.

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.

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Re: 50 children attended a carnival where cotton candy and ice cream were [#permalink]

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New post 18 Aug 2017, 09:14
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Bunuel wrote:
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


We can create the following equation:

total children = # ate ice cream only + # ate cotton candy only + # ate both + # ate neither

50 = 18 + 14 + b + 8

50 = b + 40

10 = b

Since 10 children ate both ice cream and cotton candy, we add these 10 to the 18 children who ate only ice cream to get 10 = 18 = 28 children who ate ice cream.

Answer: D
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Re: 50 children attended a carnival where cotton candy and ice cream were [#permalink]

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New post 20 Aug 2017, 06:17
Bunuel wrote:
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


For me, the best way to solve the problem is Double Matrix Method.

1. 14 people only CC; 18 only IC and 8 children no CC and IC.
2. Only one left category : both ate CC and IC, which is 50 - 14 - 18 - 8 = 10.
3. Thus, children who ate IC = children ONLY ate IC + BOTH = 18 + 10 = 28.

The asnwer is D.
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Re: 50 children attended a carnival where cotton candy and ice cream were   [#permalink] 20 Aug 2017, 06:17
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