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At a family summer party, each of the x members of the family

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At a family summer party, each of the x members of the family [#permalink]

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New post 23 Aug 2017, 22:43
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Difficulty:

  55% (hard)

Question Stats:

62% (01:29) correct 38% (02:40) wrong based on 107 sessions

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At a family summer party, each of the x members of the family chose whether or not to have a hamburger and whether or not to have a hotdog. If 1/3 chose to have a hamburger, and of those 1/7 chose to also have a hotdog, then how many family members chose NOT to have both.

A. x/21
B. x/10
C. 9x/10
D. 10x/21
E. 20x/21

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Re: At a family summer party, each of the x members of the family [#permalink]

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New post 23 Aug 2017, 22:50
Skywalker18 wrote:
At a family summer party, each of the x members of the family chose whether or not to have a hamburger and whether or not to have a hotdog. If 1/3 chose to have a hamburger, and of those 1/7 chose to also have a hotdog, then how many family members chose NOT to have both.

A. x/21
B. x/10
C. 9x/10
D. 10x/21
E. 20x/21

9x/10

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Re: At a family summer party, each of the x members of the family [#permalink]

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New post 23 Aug 2017, 23:12
Let x=21
\(\frac{1}{3}\)*21 = 7 chose to have hamburger
\(\frac{1}{7}\)of7 = 1 chose to have both
so remaining 20 (i.e.21-1) will not have both.
Option E, \(\frac{20*21}{21}\) = 20 gives that answer
So option E
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Re: At a family summer party, each of the x members of the family [#permalink]

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New post 23 Aug 2017, 23:50
The questions asks for neither x nor y. I believe that some info is missing from this problem.

We have information for both x and y and x and not y.

chetan2u , could you please help with this
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Re: At a family summer party, each of the x members of the family [#permalink]

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New post 24 Aug 2017, 07:30
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rahulkashyap wrote:
The questions asks for neither x nor y. I believe that some info is missing from this problem.

We have information for both x and y and x and not y.

chetan2u , could you please help with this


hi..

what is it asking - person who did not have BOTH.
WHAT will it include - who ate any ONE only or NONE : SAME as ALL-BOTH

ALL is x
BOTH is \(\frac{1}{7} of \frac{1}{3}\) of x=\(\frac{x}{21}\)

ans \(x-\frac{x}{21}=\frac{20x}{21}\)
E
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Re: At a family summer party, each of the x members of the family [#permalink]

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New post 21 Sep 2017, 09:40
I get how it's figured out, but what about the people who chose to have a hotdog and not a hamburger? Do we not need to know this?

X is all the people - let's call this 21
Therefore, 7 chose to have a hamburger and 14 chose not to have a hamburger.
Of this 7, 1 chose to also have a hotdog. This leaves 6 people who only had a hamburger.
Of the 14 who chose not to have a hamburger, how many chose to have a hotdog and not a hamburger? - We don't know this piece of information, so surely it's a proportion of the 20x/21?
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Re: At a family summer party, each of the x members of the family [#permalink]

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New post 06 Oct 2017, 06:42
Say, Hamburger = H & Hotdog = D
Given n(H) =X/3 , Total = X
and n(D) = X- X/3 = 2X/3
Both = X/3 * 1/7 = X/21
Only H = X/3-X/21 = 6X/21
Only D = 2X/3-X/21 =13X/21
We know that, Total = n(H) + n(D) - n(H&B) + Neither
X = X/3 + 2X/3 - X/21 + Neither
Neither = X/21
So, who eat none = only H+ only D + Neither
= 6X/21+13X/21+X/21
= 20X/21.
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Re: At a family summer party, each of the x members of the family [#permalink]

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New post 26 Nov 2017, 14:43
Skywalker18 wrote:
At a family summer party, each of the x members of the family chose whether or not to have a hamburger and whether or not to have a hotdog. If 1/3 chose to have a hamburger, and of those 1/7 chose to also have a hotdog, then how many family members chose NOT to have both.

A. x/21
B. x/10
C. 9x/10
D. 10x/21
E. 20x/21



(1/3) * (1/7)
= 1/21 = both

anything other than both
= 1 - (1/21)
20/21

thanks
cheers through the kudos button if this helps
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Re: At a family summer party, each of the x members of the family   [#permalink] 26 Nov 2017, 14:43
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