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At a family summer party, each of the x members of the

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At a family summer party, each of the x members of the

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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

The OA is E.

Source: EMPOWERgmat

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swerve 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
Let the x family members = the product of the two denominators = 3*7 = 21.
Number who chose to have a hamburger = (1/3)(21) = 7.
Since 1/7 of the hamburger-eaters also had a hot dog, the number who had both = (1/7)(7) = 1.
Thus, the number who did NOT have both = (total members) - (number who had both) = 21-1 = 20.

The correct answer must yield a value of 20 when x=21.
Only E works:
(20/21)x = (20/21)(21) = 20.

The correct answer is E.

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swerve 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

The OA is E.

Source: EMPOWERgmat
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

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swerve 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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scott@targettestprep.com



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