During the past week, a local medical clinic tested N

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AAPL: Moderator; Posts: 2695; Joined: Sun Oct 29, 2017 2:08 pmFollowed by: 2 members

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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Source: Beat The GMAT — Problem Solving | Thu Feb 21, 2019 3:25 am

deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

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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Scott@TargetTestPrep: GMAT Instructor; Posts: 8198; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

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

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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