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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
21
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A
B
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E

Difficulty:

  55% (hard)

Question Stats:

63% (02:09) correct 37% (02:46) wrong based on 228 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
1
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
1
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
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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

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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 13 Oct 2018, 08:43
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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


These kinds of questions (Variables in the Answer Choices - VIACs) can be answered algebraically or using the INPUT-OUTPUT approach.
Here's the INPUT-OUTPUT approach.

It might be useful to choose a number that works well with the fractions given in the question (1/3 and 1/7).
So, let's say there are 21 family members at the party.
In other words, we're saying that x = 21

1/3 chose to have a hamburger, and of those 1/7 chose to also have a hotdog.
1/3 of 21 is 7, so 7 people chose to have a hamburger.
1/7 of 7 = 1, so 1 person had BOTH a hamburger AND a hotdog.

How many family members chose NOT to have both?
If 1 person had BOTH a hamburger AND a hotdog, then the remaining 20 people did not have BOTH a hamburger AND a hotdog.

So, when we INPUT x = 21, the answer to the question is "20 people did not have BOTH a hamburger AND a hotdog"

Now we'll INPUT x = 21 into each answer choice and see which one yields the correct OUTPUT of 20

A. 21/21 = 1. We want an output of 20. ELIMINATE A.
B. 21/10 = 2.1. We want an output of 20. ELIMINATE B.
C. (9)(21)/10 = 18.9. We want an output of 20. ELIMINATE C.
D. (10)(21)/21 = 10. We want an output of 20. ELIMINATE D.
E. (20)(21)/21 = = 20. PERFECT!

Answer: E

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

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New post 13 Oct 2018, 19:53
How to get the missing variable? I think the question is missing some data. HI Bunuel can you please check 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 04 Nov 2018, 18:08
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


Since x * 1/3 * 1/7 = x/21 of the family members have both a hamburger and a hotdog, 20x/21 of the family members do not have both.

Answer: 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 06 Jan 2019, 12:56
HAPPYatHARVARD wrote:
How to get the missing variable? I think the question is missing some data. HI Bunuel can you please check this?


No data is missing..Que asks how many people did not have both which includes people who had one (either hamburger or hotdog) + people who had neither (No ham no hotdog)

Hope that helps.
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Re: At a family summer party, each of the x members of the family   [#permalink] 06 Jan 2019, 12:56
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