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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
OA E.
These kinds of questions (Variables in the Answer Choices - VIACs) can be answered algebraically or using the INPUT-OUTPUT approach.
The Ash Mo solved the question algebraically, so let's use 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
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
