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50 children attended a carnival where cotton candy and ice

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50 children attended a carnival where cotton candy and ice

by BTGmoderatorDC » Thu Aug 08, 2019 10:45 pm

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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

OA D

Source: Veritas Prep
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by Jay@ManhattanReview » Fri Aug 09, 2019 12:43 am
BTGmoderatorDC 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

OA D

Source: Veritas Prep
We have the total number of children = T = 50;
Number of students who ate only cotton candy = C = 14;
Number of students who ate only ice cream = I = 18;
Number of students who ate both cotton candy and ice cream = B = ?;
Number of students who ate none = N = 8;

=> T = C + I + B + N

50 = 14 + 18 + B + 8 => = B = 10

Thus, the number of students who ate ice cream = I + B = 18 + 10 = 28

The correct answer: D

Hope this helps!

-Jay
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by deloitte247 » Fri Aug 09, 2019 12:01 pm
Let the children who ate cotton candy = c
Let the children who ate ice cream = i
Given that =>
Total children = 50
Children who took cotton candy only = 14
Children who took ice cream only = 18
Children who took neither cotton candy nor ice cream = 8
Total children = ( c only ) + ( i only ) + ( both i & c ) + ( neither i/c )
50 = 14 + 18 + ( both i & c ) + 8
50 = 40 + ( both i & c )
Both i & c = 50 - 40 = 10
Therefore, total children who took ice cream = ( cream only ) + ( both ice cream and candy )
18 + 10 = 28

The answer is, therefore, option D
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