From a bag containing 12 identical blue balls, y identical y

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imane81: Senior | Next Rank: 100 Posts; Posts: 40; Joined: Wed Jan 13, 2010 7:17 pmFollowed by: 1 member; Thanked: 1 times

by imane81 » Wed Feb 17, 2010 12:25 pm

Thank you for advising on this one...

From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the least number of yellow balls that must be in the bag?
A. 17
B. 18
C. 19
D. 20
E. 21

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Source: Beat The GMAT — Problem Solving | Wed Feb 17, 2010 12:25 pm

ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Wed Feb 17, 2010 12:38 pm

imane81 wrote:Thank you for advising on this one...

From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the least number of yellow balls that must be in the bag?
A. 17
B. 18
C. 19
D. 20
E. 21

12 is less than 2/5 of the total
30 is less than the total content

so the least no of yellow balls is 31-12 = 19

Always borrow money from a pessimist, he doesn't expect to be paid back.

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cat er ina: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Sat Feb 13, 2010 1:36 pm

by cat er ina » Wed Feb 17, 2010 2:21 pm

HEY AJITH IS GREAT BY I UNDERSTOOD FROM THIS LINK, GIVES A VERY GOOD EXPLANATION

www.manhattangmat.com/forums/from-a-bag ... t8493.html

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harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members; Location: India; Thanked: 64 times; GMAT Score:760

by harsh.champ » Thu Feb 18, 2010 2:10 pm

imane81 wrote:Thank you for advising on this one...

From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the least number of yellow balls that must be in the bag?
A. 17
B. 18
C. 19
D. 20
E. 21

Formal Approach:-
P(Removed ball is blue) = 12/12+y which is less than 2/5
Thus,we get that 60 < 24 + 2y
i.e. y>18.
So we can rule out option A and B.
Also,we have to find the least no. hence [spoiler]y=19 C should be the answer.[/spoiler]

I guess the detailed explanation makes it clear.
For shortcut approach,you can adopt ajith's method.

It takes time and effort to explain, so if my comment helped you please press Thanks button

Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker

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bonetlobo: Senior | Next Rank: 100 Posts; Posts: 93; Joined: Sun Feb 22, 2015 3:47 amFollowed by: 2 members; Thanked: 4 times

by bonetlobo » Wed May 13, 2015 6:13 am

Ajith's explanation is brilliant. I did it the "conventional" way:(.

Number of blue balls = 12
Number of yellow balls = y
=> Total number of balls in the bag = 12+y

=> probability that one ball randomly removed is blue ball = 12/(12+y)

However, it is given that:

12/(12+y) < 2/5

Since both sides are positive, we can do cross multiplication

Solving, we get y > 18

So, minimum value of y = 19

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Wed May 13, 2015 6:56 am

imane81 wrote:
From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the least number of yellow balls that must be in the bag?
A. 17
B. 18
C. 19
D. 20
E. 21

If you're not sure where to begin here, TESTING the answer choices will reveal the correct answer in no time.

A) 17
If there are 17 yellow balls, then the TOTAL number of balls = 12 + 17 = 29
So, P(selected ball is blue) = 12/29
12/29 is GREATER THAN 2/5, so we can eliminate A

B) 18
If there are 18 yellow balls, then the TOTAL number of balls = 12 + 18 = 30
So, P(selected ball is blue) = 12/30
12/30 is EQUAL TO 2/5, so we can eliminate B

At this point, we can see that adding 1 more yellow ball (i.e., 19 yellow balls) will make P(selected ball is blue) LESS THAN 2/5
So, the correct answer is C

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Wed May 13, 2015 6:57 am

From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the leat no. of yellow balls that must be in a bag?

A)17
B)18
C)19
D)20
E)21

Let T = the total number of marbles.
Since the probability of selecting a blue marble is less than 2/5, the 12 blue marbles in the bag must constitute less than 2/5 of the total:
12 < (2/5)T
30 < T.

Since the total number of marbles must be at least 31, the least number of yellow marbles that must be added to the 12 blue marbles is 19.

The correct answer is C.

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