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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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E

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55% (hard)

Question Stats:

65% (01:33) correct 35% (02:46) wrong based on 121 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

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

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

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!