During the past week, a local medical clinic tested \(N\) in

[VJesus12]

During the past week, a local medical clinic tested \(N\) individuals for two infections. If \(\dfrac13\) of those tested had infection \(A\) and, of those with infection \(A\), \(\dfrac15\) also had infection \(B\), how many individuals did not have both infection \(A\) and \(B?\)

A. \(\dfrac{N}{15}\)
B. \(\dfrac{4N}{15}\)
C. \(\dfrac{14N}{15}\)
D. \(\dfrac{N}5\)
E. \(\dfrac{4N}5\)

[spoiler]OA=C[/spoiler]

Source: EMPOWERgmat

[[email protected]]

Hi VJesus12,

We're told that a local medical clinic tested N individuals for two infections; 1/3 of those tested had infection A and, of those with infection A, 1/5 also had infection B. We're asked for the number of individuals who did NOT have BOTH infection A and B. This question can be solved in a couple of different ways, including by TESTing VALUES.

The 5 answer choices have a common denominator of 15, so that will likely be a good number to TEST.

IF.... N = 15
then (1/3)(15) = 5 people had infection A
and (1/5)(5) = 1 person ALSO had infection B

This means that 15 - 1 = 14 people did NOT have BOTH infections, so we're looking for an answer that equals 14 when N = 15. There's only one answer that matches...

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

[Scott@TargetTestPrep]

During the past week, a local medical clinic tested \(N\) individuals for two infections. If \(\dfrac13\) of those tested had infection \(A\) and, of those with infection \(A\), \(\dfrac15\) also had infection \(B\), how many individuals did not have both infection \(A\) and \(B?\)

A. \(\dfrac{N}{15}\)
B. \(\dfrac{4N}{15}\)
C. \(\dfrac{14N}{15}\)
D. \(\dfrac{N}5\)
E. \(\dfrac{4N}5\)

[spoiler]OA=C[/spoiler]

Source: EMPOWERgmat

We know that N/3 individuals had infection A, and 1/5 of those (1/5 x N/3 = N/15) had both infections A and B. Therefore, N - N/15 = 14N/15 individuals do not have both infections.

Answer: C