Last visit was: 25 Apr 2026, 20:41 It is currently 25 Apr 2026, 20:41
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.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bismuth83
User avatar
DI Forum Moderator
Joined: 15 Sep 2024
Last visit: 01 Aug 2025
Posts: 714
Own Kudos:
3,147
 [8]
Given Kudos: 441
Expert
Expert reply
Posts: 714
Kudos: 3,147
 [8]
Kudos
Add Kudos
8
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 25 Apr 2026
Posts: 11,229
Own Kudos:
45,020
 [7]
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,229
Kudos: 45,020
 [7]
6
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
General Discussion
User avatar
poojaarora1818
Joined: 30 Jul 2019
Last visit: 25 Apr 2026
Posts: 1,624
Own Kudos:
Given Kudos: 3,821
Location: India
Concentration: General Management, Economics
GPA: 3
WE:Human Resources (Real Estate)
Products:
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Madhvendrasinh
Joined: 01 Feb 2024
Last visit: 26 Feb 2025
Posts: 110
Own Kudos:
63
 [3]
Given Kudos: 200
Posts: 110
Kudos: 63
 [3]
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bismuth83
A bag contains only red balls. Initially, twenty of these 50 red balls are replaced with blue balls. After some time, another ten balls are chosen randomly and replaced with white balls. On checking the bag, it was found that the number of blue balls was one-fourth of the number of red balls.

Select for Red the number of red balls that are left in the bag and select for Blue the number of blue balls that are left in the bag. Mark only two options, one in each column.
This question is nice and easy but the words should be changed.

Initially,
Only Red balls = 50 balls
20 replaced with blue balls

Now,
Red balls=30
Blue balls=20

the question says 10 balls are replaced with white balls

But answer indicates that total red balls are 32 and blue balls are 8 only,
this clearly indicates that actually 12 blue balls are replaced, as 10 balls to white balls and 2 balls to red balls. (this is not possible because the question stem says only 10 balls were replaced)

This part of the question should be edited to prevent any error or confusion.
Quote:
On checking the bag, it was found that the number of blue balls was one-fourth of the number of red balls.
User avatar
poojaarora1818
Joined: 30 Jul 2019
Last visit: 25 Apr 2026
Posts: 1,624
Own Kudos:
Given Kudos: 3,821
Location: India
Concentration: General Management, Economics
GPA: 3
WE:Human Resources (Real Estate)
Products:
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bismuth83
A bag contains only red balls. Initially, twenty of these 50 red balls are replaced with blue balls. After some time, another ten balls are chosen randomly and replaced with white balls. On checking the bag, it was found that the number of blue balls was one-fourth of the number of red balls.

Select for Red the number of red balls that are left in the bag and select for Blue the number of blue balls that are left in the bag. Mark only two options, one in each column.
Can you pls validate my approach to the solution?

As we know the bag contains only red balls and 50 red balls are replaced with blue balls. We can assume that

Total Balls = 50 As mentioned 50 red balls were replaced with 15 blue balls So, now red balls are 50-15 = 35 balls

We also know that ten balls are chosen and replaced with white balls So, now white balls are 10 balls.

It has also mentioned that no. of the blue balls was 1/4 of the no. of red balls. So, let's set up an equation

Red + Blue + White = 50

x + x/4 + 10 = 50 -----> x = 32 and x/4 = 8

Hence, Red balls = 32 and Blue balls = 8
User avatar
SomeOneUnique
Joined: 17 Mar 2019
Last visit: 26 Dec 2024
Posts: 119
Own Kudos:
Given Kudos: 41
Location: India
Posts: 119
Kudos: 126
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The number of total balls will always be 50.

After replacing some red balls with blue balls, by eyeballing the distribution of balls would be:
R B
30 20 (Total 50)


After replacing some balls with white balls, say x balls, the number of blue balls would be 20-(10-x) which is nothing but 10+x.
By eyeballing the distribution of all balls would be:
R B W
30-x 10+x10 (Total 50)


We are given 1 piece of information; the number of blue balls is one-fourth of the number of red balls.
10+x = (30-x)/4 (This is the only calculation needed in my approach)
x = -2

Number of red balls = 30-x = 30-(-2) = 30+2 = 32 (Option F)
Number of blue balls = 10+x = 10+(-2) = 10-2 = 8 (Option C)
User avatar
Bismuth83
User avatar
DI Forum Moderator
Joined: 15 Sep 2024
Last visit: 01 Aug 2025
Posts: 714
Own Kudos:
3,147
 [1]
Given Kudos: 441
Expert
Expert reply
Posts: 714
Kudos: 3,147
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
1. We are asked to find the number of red and blue balls that are left in the bag. Let’s denote them as R and B, respectively.

2. Let’s also write the number of red, blue, and white balls at a certain moment as (# of red, # of blue, # of white).

3. Initially we have (50, 0, 0). Then, 20 red balls are replaced with blue balls - (30, 20, 0). Finally, 10 random balls are replaced with white ones, so we have (R, B, 10). It can be noted that the total number of balls is constant (which is 50) throughout the processes.

4. We are also told that R = 4B. From the state (R, B, 10) we have that R + B + 10 = 50 -> R + B = 40 -> 4B + B = 40 -> 5B = 40 -> B = 8 -> R = 32.

4. Our answer will be 32 red balls left in the bag and 8 blue balls left in the bag.
User avatar
Bismuth83
User avatar
DI Forum Moderator
Joined: 15 Sep 2024
Last visit: 01 Aug 2025
Posts: 714
Own Kudos:
Given Kudos: 441
Expert
Expert reply
Posts: 714
Kudos: 3,147
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Madhvendrasinh
Bismuth83
A bag contains only red balls. Initially, twenty of these 50 red balls are replaced with blue balls. After some time, another ten balls are chosen randomly and replaced with white balls. On checking the bag, it was found that the number of blue balls was one-fourth of the number of red balls.

Select for Red the number of red balls that are left in the bag and select for Blue the number of blue balls that are left in the bag. Mark only two options, one in each column.
This question is nice and easy but the words should be changed.

Initially,
Only Red balls = 50 balls
20 replaced with blue balls

Now,
Red balls=30
Blue balls=20

the question says 10 balls are replaced with white balls

But answer indicates that total red balls are 32 and blue balls are 8 only,
this clearly indicates that actually 12 blue balls are replaced, as 10 balls to white balls and 2 balls to red balls. (this is not possible because the question stem says only 10 balls were replaced)

This part of the question should be edited to prevent any error or confusion.
Quote:
On checking the bag, it was found that the number of blue balls was one-fourth of the number of red balls.

Very good catch! As you said, the number of red balls can't increase. I changed the question to initially replacing 15 red balls to blue. So, we would have: (50, 0, 0) -> (35, 15, 0) -> (32, 8, 10)
User avatar
Bismuth83
User avatar
DI Forum Moderator
Joined: 15 Sep 2024
Last visit: 01 Aug 2025
Posts: 714
Own Kudos:
3,147
 [1]
Given Kudos: 441
Expert
Expert reply
Posts: 714
Kudos: 3,147
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
poojaarora1818
Bismuth83
A bag contains only red balls. Initially, twenty of these 50 red balls are replaced with blue balls. After some time, another ten balls are chosen randomly and replaced with white balls. On checking the bag, it was found that the number of blue balls was one-fourth of the number of red balls.

Select for Red the number of red balls that are left in the bag and select for Blue the number of blue balls that are left in the bag. Mark only two options, one in each column.
Can you pls validate my approach to the solution?

As we know the bag contains only red balls and 50 red balls are replaced with blue balls. We can assume that

Total Balls = 50 As mentioned 50 red balls were replaced with 20 blue balls So, now red balls are 50-20 = 30 balls

We also know that ten balls are chosen and replaced with white balls So, now white balls are 10 balls.

It has also mentioned that no. of the blue balls was 1/4 of the no. of red balls. So, let's set up an equation

Red + Blue + White = 50

x + x/4 + 10 = 50 -----> x = 32 and x/4 = 8

Hence, Red balls = 32 and Blue balls = 8
Nice solution! I changed the question a bit for the question to work (initially turning 15 red balls to blue instead of 20), so feel free to update your answer
User avatar
poojaarora1818
Joined: 30 Jul 2019
Last visit: 25 Apr 2026
Posts: 1,624
Own Kudos:
Given Kudos: 3,821
Location: India
Concentration: General Management, Economics
GPA: 3
WE:Human Resources (Real Estate)
Products:
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bismuth83
poojaarora1818
Bismuth83
A bag contains only red balls. Initially, twenty of these 50 red balls are replaced with blue balls. After some time, another ten balls are chosen randomly and replaced with white balls. On checking the bag, it was found that the number of blue balls was one-fourth of the number of red balls.

Select for Red the number of red balls that are left in the bag and select for Blue the number of blue balls that are left in the bag. Mark only two options, one in each column.
Can you pls validate my approach to the solution?

As we know the bag contains only red balls and 50 red balls are replaced with blue balls. We can assume that

Total Balls = 50 As mentioned 50 red balls were replaced with 20 blue balls So, now red balls are 50-20 = 30 balls

We also know that ten balls are chosen and replaced with white balls So, now white balls are 10 balls.

It has also mentioned that no. of the blue balls was 1/4 of the no. of red balls. So, let's set up an equation

Red + Blue + White = 50

x + x/4 + 10 = 50 -----> x = 32 and x/4 = 8

Hence, Red balls = 32 and Blue balls = 8
Nice solution! I changed the question a bit for the question to work (initially turning 15 red balls to blue instead of 20), so feel free to update your answer
Thank you for your response. I have made the necessary changes to the solution as required.
User avatar
egmat
User avatar
e-GMAT Representative
Joined: 02 Nov 2011
Last visit: 24 Apr 2026
Posts: 5,632
Own Kudos:
Given Kudos: 707
GMAT Date: 08-19-2020
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 5,632
Kudos: 33,435
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Happy to walk you through this step by step!

Start: The bag has 50 red balls.

Step 1 — Replace 15 red balls with blue balls:
- Red: 50 - 15 = 35
- Blue: 15
- Total still = 50

Step 210 random balls are replaced with white balls.
This is the key step. Among those 10 balls removed, some could be red and some could be blue. Let's call the number of blue balls removed x. Then the number of red balls removed is (10 - x).

After this replacement:
- Red: 35 - (10 - x) = 25 + x
- Blue: 15 - x
- White: 10
- Total still = 50

Step 3 — Apply the given condition: Blue = (1/4) × Red

15 - x = (1/4)(25 + x)

Multiply both sides by 4:
60 - 4x = 25 + x

35 = 5x

x = 7

So 7 blue balls and 3 red balls were among the 10 replaced with white.

Final count:
- Red: 25 + 7 = 32
- Blue: 15 - 7 = 8
- White: 10

Verification: Is Blue = (1/4) × Red? → 8 = (1/4)(32) = 8
Total: 32 + 8 + 10 = 50

Answer: Red column → 32 (Row 6), Blue column → 8 (Row 3)

The trick in this problem is setting up the unknown x for how the 10 randomly chosen balls split between red and blue. Once you define that variable, the constraint 'blue = one-fourth of red' gives you a single clean equation to solve.
Moderators:
Math Expert
109830 posts
498 posts
212 posts