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At a dinner of 32 guests,. 18 ate yam, 16 ate beans and 7 ate both. ·How many guests...

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christiano-cr7 | Salutatorian

Posted July 13, 2013 at 7:04 AM via web

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At a dinner of 32 guests,. 18 ate yam, 16 ate beans and 7 ate both. ·How many guests did not eat?

 

 

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted July 13, 2013 at 7:12 AM (Answer #1)

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Let the sets defined as follows.

A = guests who ate yam

B = guests who ate beans

 

According to the data given;

`n(A) = 18`

`n(B) = 16`

`n(AnnB) = 7`

 

From set theory;

`n(AuuB) = n(A)+n(B)-n(AnnB)`

`n(AuuB) = 18+16-7`

`n(AuuB) = 27`

 

`n(AuuB)` represent the amount of guests who ate something. There were 32 guests but only 27 has ate something. So the rest 5 has not eat anything from the dinner.

The answer is 5 guests.

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