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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