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At a family summer party, each of the x members of the family
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23 Aug 2017, 22:43
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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.
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Re: At a family summer party, each of the x members of the family
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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.
Re: At a family summer party, each of the x members of the family
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23 Aug 2017, 23:12
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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
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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?
Re: At a family summer party, each of the x members of the family
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06 Oct 2017, 06:42
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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.
Re: At a family summer party, each of the x members of the family
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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
thanks cheers through the kudos button if this helps
Re: At a family summer party, each of the x members of the family
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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!
Re: At a family summer party, each of the x members of the family
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04 Nov 2018, 18:08
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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
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.
Re: At a family summer party, each of the x members of the family
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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.
gmatclubot
Re: At a family summer party, each of the x members of the family
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06 Jan 2019, 12:56