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During the past week, a local medical clinic tested N

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by AAPL » Thu Feb 21, 2019 3:25 am

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EMPOWERgmat

During the past week, a local medical clinic tested N individuals for two infections. If 1/3 of those tested had infection A and, of those with infection A, 1/5 also had infection B, how many individuals did not have both infection A and B?

A. N/15
B. 4N/15
C. 14N/15
D. N/5
E. 4N/5

OA C
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by deloitte247 » Sun Feb 24, 2019 12:21 am
1/3 of our individuals tested had infections ; thus N/3 had infection A
Of those with infection A, 1/3 also had infection B ;thus
$$\frac{1}{5}\cdot\frac{N}{3}=\frac{N}{15}\ had\ both\ \inf ection\ A\ and\ B$$
Therefore,
$$\frac{N}{1}-\frac{N}{15}=\frac{\left(15N-1N\right)}{15}=\frac{14N}{15}did\ not\ have\ both\ \inf ection\ n\ A\ and\ B$$

$$answer\ is\ Option\ C$$
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by Scott@TargetTestPrep » Sun Feb 24, 2019 5:29 am
AAPL wrote:EMPOWERgmat

During the past week, a local medical clinic tested N individuals for two infections. If 1/3 of those tested had infection A and, of those with infection A, 1/5 also had infection B, how many individuals did not have both infection A and B?

A. N/15
B. 4N/15
C. 14N/15
D. N/5
E. 4N/5

OA C
The number who had both infections A and B is (N/3) x 1/5 = N/15, thus 14N/15 did not have both A and B.

Answer: C

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