Last visit was: 23 Apr 2026, 16:51 It is currently 23 Apr 2026, 16:51
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
Poor Quality|      
User avatar
vinoo7
User avatar
Joined: 06 Jun 2014
Last visit: 13 Jun 2017
Posts: 10
Own Kudos:
73
 [13]
Given Kudos: 101
Posts: 10
Kudos: 73
 [13]
In a bag there are a certain number of black balls and white balls. Th 07 Mar 2016, 06:46
1
Kudos
Add Kudos
12
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

16% (01:46) correct 84% (01:59) wrong based on 75 sessions

History

Date
Time
Result
Not Attempted Yet
In a bag there are a certain number of black balls and white balls. The probability of picking up exactly 1 white ball when 2 balls are randomly drawn, is 1/2. Which of the following is the ratio of the number of black balls to white balls in the bag

A. 1:7
B. 2:1
C. 1:1
D. 4:1
E. 1:4

This Question is Locked Due to Poor Quality
Hi there,
The question you've reached has been archived due to not meeting our community quality standards. No more replies are possible here.
Looking for better-quality questions? Check out the 'Similar Questions' block below for a list of similar but high-quality questions.
Want to join other relevant Problem Solving discussions? Visit our Problem Solving (PS) Forum for the most recent and top-quality discussions.
Thank you for understanding, and happy exploring!
ShowHide Answer
Official Answer
B
avatar
nandanchatterjee
avatar
Joined: 07 Aug 2014
Last visit: 16 May 2016
Posts: 1
Given Kudos: 1
Posts: 1
Kudos: 0
In a bag there are a certain number of black balls and white balls. Th 07 Mar 2016, 07:58
Kudos
Add Kudos
Bookmarks
Bookmark this Post
since the probability of drawing a white ball out of two picks is 1/2.

the ratio of the white ball to black balls should be 1:1

C
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 22 Apr 2026
Posts: 11,229
Own Kudos:
45,002
 [4]
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
My Reviews
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,229
Kudos: 45,002
 [4]
In a bag there are a certain number of black balls and white balls. Th 07 Mar 2016, 08:02
1
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
vinoo7
In a bag there are a certain number of black balls and white balls. The probability of picking up exactly 1 white ball when 2 balls are randomly drawn, is 1/2. Which of the following is the ratio of the number of black balls to white balls in the bag

A. 1:2
B. 2:1
C. 1:1
D. 4:1
E. 1:4
Show more

Hi vinoo7,

I do not think it is Sub -600 level by any chance, it is 700 level..

Also, the number of White and Black is interchangeable..
so ratio W/B= B/W
There can be various answers for this ..
One method could be..

Let the White be W and Black be B
so only one W in 2 balls would be \(\frac{W}{T}*\frac{B}{(T-1)}*2\)= 1/2..
So \(\frac{W}{T}*\frac{B}{(T-1)}= 1/4\)..

Lets first work on denominators


If you look at the denominator it is multiple of two consecutive numbers..
4 is not a multiple of 2 consecutive numbers....
the consecutive numbers will be..
3*4
4*5
5*6.. not possible as does not contain 4 as divisor
6*7.. not possible as does not contain 4 as divisor
7*8
8*9
9*10.. not possible as does not contain 4 as divisor

solutions now


1) If denominator = 3*4 numerator = 1*3
so \(\frac{W}{T}*\frac{B}{(T-1)}= 3/4*3\)
so WB=3 and W+B=4 so W and B can be 1 and 3..
so one ratio= 1:3 or 3:1

2)If denominator = 5*4 numerator = 1*5
so \(\frac{W}{T}*\frac{B}{(T-1)}= 5/4*5\)
so WB=5 and W+B=5 No values possible

3)If denominator = 7*8 numerator = 7*8/4=14
so \(\frac{W}{T}*\frac{B}{(T-1)}= 14/7*8\)
so WB=14 and W+B=8 No values possible

4) If denominator = 8*9 numerator = 8*9/4=18
so \(\frac{W}{T}*\frac{B}{(T-1)}= 18/8*9\)
so WB=18 and W+B=9 so W and B can be 6 and 3..
so another ratio= 6:3 or 3:6, which is 2:1 or 1:2..
so this is what we are looking for
ans can be A or B

Hope the method which we can adopt is clear
User avatar
ShashankDave
User avatar
Joined: 03 Apr 2013
Last visit: 26 Jan 2020
Posts: 215
Own Kudos:
301
 [1]
Given Kudos: 872
Location: India
Concentration: Marketing, Finance
Schools: McDonough '21 (A) Kelley '21 ISB '21 (I)
GMAT 1: 740 Q50 V41
GPA: 3
Schools: McDonough '21 (A) Kelley '21 ISB '21 (I)
GMAT 1: 740 Q50 V41
Posts: 215
Kudos: 301
 [1]
In a bag there are a certain number of black balls and white balls. Th 04 Jul 2017, 00:33
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
vinoo7
In a bag there are a certain number of black balls and white balls. The probability of picking up exactly 1 white ball when 2 balls are randomly drawn, is 1/2. Which of the following is the ratio of the number of black balls to white balls in the bag

A. 1:7
B. 2:1
C. 1:1
D. 4:1
E. 1:4
Show more

Hi, I did it with another method, it took a lot of trial and error to find a way to solve this one. At first I took the pure algebraic way to solve it, but no joy. Then I kind of tweaked my method a bit. I have taken a method similar to chetan2u. As we have been asked the ratio of Black and White balls, let's call the ratio k.

Let the number of Black balls = kx
Let the number of White balls = x

Now let's write the probability expression

\(\frac{(Number of ways of selecting one white ball*Number of ways of selecting one black ball)}{Total number of ways of selecting any two balls}\)

\(\frac{(kx*x)}{(kx + x)(kx + x - 1)} = \frac{1}{2}\)

on simplifying we get

\(x^2(k^2 + 1 - 2k) = x(k + 1)\)

Here, we know that x(number of white balls) is positive and an integer. So we can simply cut x here

\(x(k^2 + 1 - 2k) = (k + 1)\)

\(x = \frac{(k + 1)}{(k^2 + 1 - 2k)}\)

Now, we know that x has to be an integer. From here, we can try values for k from options such that x turns out to be an integer.

Only (B) satisfies.
User avatar
Lucy Phuong
User avatar
Joined: 24 Jan 2017
Last visit: 12 Aug 2021
Posts: 111
Own Kudos:
351
Given Kudos: 106
GMAT 1: 640 Q50 V25
GMAT 2: 710 Q50 V35
GPA: 3.48
Products:
My Reviews
GMAT 2: 710 Q50 V35
Posts: 111
Kudos: 351
In a bag there are a certain number of black balls and white balls. Th 05 Jul 2017, 23:35
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ShashankDave
As we have been asked the ratio of Black and White balls, let's call the ratio k.

Let the number of Black balls = kx
Let the number of White balls = x

Now let's write the probability expression

\(\frac{(Number of ways of selecting one white ball*Number of ways of selecting one black ball)}{Total number of ways of selecting any two balls}\)

\(\frac{(kx*x)}{(kx + x)(kx + x - 1)} = \frac{1}{2}\)

on simplifying we get

\(x^2(k^2 + 1 - 2k) = x(k + 1)\)

Here, we know that x(number of white balls) is positive and an integer. So we can simply cut x here

\(x(k^2 + 1 - 2k) = (k + 1)\)

\(x = \frac{(k + 1)}{(k^2 + 1 - 2k)}\)

Now, we know that x has to be an integer. From here, we can try values for k from options such that x turns out to be an integer.

Only (B) satisfies.
Show more

Hi,

Your solution using variable is awesome! This algebraic way always excites me :P

However, I've just found out an error in a half-way step, thus making the final answer not as you expected.

It runs smoothly until this step:

\(\frac{(Number of ways of selecting one white ball*Number of ways of selecting one black ball)}{Total number of ways of selecting any two balls}\)

\(\frac{(kx*x)}{(kx + x)(kx + x - 1)} = \frac{1}{2}\)

In fact, after simplifying the equation, we will get the following:

\(x^2(k^2 + 1) = x(k + 1)\)
(In your post, you just made an extra \(-2kx^2\))

Moving on, because x(number of white balls) is a positive integer, we can simply cut x:

\(x(k^2 + 1) = (k + 1)\)

\(x = \frac{(k + 1)}{(k^2 + 1)}\)

x is a positive integer only when k=1. Therefore, the ratio of the number of black balls to white balls is 1:1

(C) must be the correct answer if the above probability equation (in red color) is correct!

Still not understand why OA is (B). Any thought?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Apr 2026
Posts: 109,785
Own Kudos:
810,870
Given Kudos: 105,853
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,785
Kudos: 810,870
In a bag there are a certain number of black balls and white balls. Th 06 Jul 2017, 00:06
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Lucy Phuong
ShashankDave
As we have been asked the ratio of Black and White balls, let's call the ratio k.

Let the number of Black balls = kx
Let the number of White balls = x

Now let's write the probability expression

\(\frac{(Number of ways of selecting one white ball*Number of ways of selecting one black ball)}{Total number of ways of selecting any two balls}\)

\(\frac{(kx*x)}{(kx + x)(kx + x - 1)} = \frac{1}{2}\)

on simplifying we get

\(x^2(k^2 + 1 - 2k) = x(k + 1)\)

Here, we know that x(number of white balls) is positive and an integer. So we can simply cut x here

\(x(k^2 + 1 - 2k) = (k + 1)\)

\(x = \frac{(k + 1)}{(k^2 + 1 - 2k)}\)

Now, we know that x has to be an integer. From here, we can try values for k from options such that x turns out to be an integer.

Only (B) satisfies.

Hi,

Your solution using variable is awesome! This algebraic way always excites me :P

However, I've just found out an error in a half-way step, thus making the final answer not as you expected.

It runs smoothly until this step:

\(\frac{(Number of ways of selecting one white ball*Number of ways of selecting one black ball)}{Total number of ways of selecting any two balls}\)

\(\frac{(kx*x)}{(kx + x)(kx + x - 1)} = \frac{1}{2}\)

In fact, after simplifying the equation, we will get the following:

\(x^2(k^2 + 1) = x(k + 1)\)
(In your post, you just made an extra \(-2kx^2\))

Moving on, because x(number of white balls) is a positive integer, we can simply cut x:

\(x(k^2 + 1) = (k + 1)\)

\(x = \frac{(k + 1)}{(k^2 + 1)}\)

x is a positive integer only when k=1. Therefore, the ratio of the number of black balls to white balls is 1:1

(C) must be the correct answer if the above probability equation (in red color) is correct!

Still not understand why OA is (B). Any thought?
Show more

In the highlighted part above LHS should be multiplied by 2 because we can have BW as well as WB cases. Having said that the question is not good. chetan is right, the ratio could be 6:3 (2:1) or 3:6 (1:2) among many other cases. So, you can ignore this question and move on.

TOPIC IS LOCKED.

This Question is Locked Due to Poor Quality
Hi there,
The question you've reached has been archived due to not meeting our community quality standards. No more replies are possible here.
Looking for better-quality questions? Check out the 'Similar Questions' block below for a list of similar but high-quality questions.
Want to join other relevant Problem Solving discussions? Visit our Problem Solving (PS) Forum for the most recent and top-quality discussions.
Thank you for understanding, and happy exploring!
Moderators:
Bunuel
Math Expert
109785 posts
Krunaal
Tuck School Moderator
853 posts