BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

A box contains red and blue balls only. If there are 8 balls

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Moderator
Posts: 7187
Joined: Thu Sep 07, 2017 4:43 pm
Followed by:23 members

A box contains red and blue balls only. If there are 8 balls

by BTGmoderatorDC » Wed Feb 06, 2019 5:40 am

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.

OA A

Source: Magoosh
Top
Source: — Data Sufficiency |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3008
Joined: Mon Aug 22, 2016 6:19 am
Location: Grand Central / New York
Thanked: 470 times
Followed by:34 members

by Jay@ManhattanReview » Wed Feb 06, 2019 10:27 pm

Timer

00:00

Your Answer

A

B

C

D

E

Global Stats

BTGmoderatorDC wrote:A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.

OA A

Source: Magoosh
We have to determine the number of red balls.

Say the number of red balls is r; thus, the number of blue balls = 8 - r.

Let's take each statement one by one.

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14

=> rC2 / 8C2 = r*(r - 1)/8.7 = 5/14 => r*(r - 1) = 20 => r*(r - 1) = 5*4 => r = 5. Sufficient.

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.

Probability of drawing the first ball being red = r/8;
Probability of drawing the second ball being blue = (8 - r) / 7; there would be only 7 balls left after the drawing the first ball.

Thus, the required probability = r/8 * (8 - r) / 7 = 5/14

=> r*(8 - r) = 15
r*(8 - r) = 3*5

r = 3 or 5. No unique answer. Insufficient.

The correct answer: A

Hope this helps!

-Jay
_________________
Manhattan Review GMAT Prep

Locations: GMAT Classes DC | GMAT Tutoring Seattle | GRE Prep Boston | TOEFL Prep Classes Houston | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.
Top
Post new topic Post Reply
• Page 1 of 1