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
OA E
Source: EMPOWERgmat
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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
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These kinds of questions (Variables in the Answer Choices  VIACs) can be answered algebraically or using the INPUTOUTPUT approach.BTGmoderatorDC wrote: ↑Fri Jan 13, 2023 9:53 pmAt 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
OA E
Source: EMPOWERgmat
Here's the INPUTOUTPUT 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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BTGmoderatorDC wrote: ↑Fri Jan 13, 2023 9:53 pmAt 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
OA E
Source: EMPOWERgmat
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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