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## Each person in a room attends exactly one of the three

tagged by: M7MBA

This topic has 1 expert reply and 0 member replies

### Top Member

M7MBA Master | Next Rank: 500 Posts
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#### Each person in a room attends exactly one of the three

Sun Feb 18, 2018 4:38 am
Each person in a room attends exactly one of the three colleges A, B, and C. What is the probability that a person selected at random from the people in the room attends college B?

(1) The number of people in the room attending college B is 20 less than the sum of the numbers of people in the room attending colleges A and C.

(2) The ratio of the number people in the room attending college B to the sum of the numbers of people in the room attending colleges A and C is 2 to 3.

The OA is B.

I don't know how to use the statement (2) to get an answer. Can you help me here experts? Thanks,

### GMAT/MBA Expert

ErikaPrepScholar Master | Next Rank: 500 Posts
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GMAT Score:
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Mon Feb 19, 2018 6:13 am
The question stem gives that the total number of people in the room is A + B + C. We want to find the probability that one person selected from this total attends college B. In other words, we want to find $$\frac{B}{A+B+C}$$

Statement 1 tells us that $$B=A+C-20$$ $$B+20=A+C$$
Plugging in B+20 in for A+C in our initial fraction gives $$\frac{B}{B+B+20}$$ $$\frac{B}{2B+20}$$ At this point, we have no idea what the value of B is - if B = 10, then the probability will be 10/40=1/4; if B = 30, then the probability will be 30/80=3/8. So we can't solve definitively for the probability of selecting someone from college B. Insufficient.

Statement 2 tells us that $$\frac{B}{A+C}=\frac{2}{3}$$ $$\frac{3}{2}B=A+C$$
Plugging (3/2)B in for A+C in our initial fraction gives $$\frac{B}{B+\frac{3}{2}B}$$ $$\frac{B}{\frac{5}{2}B}$$ $$\frac{1}{\frac{5}{2}}$$ $$\frac{2}{5}$$ So the probability of picking a student from college B is 2/5. Sufficient.

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