A bag contains only red balls. Initially, twenty of these 50 red balls

Question

A bag contains only red balls. Initially, fifteen 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.

chetan2u · Accepted Answer

A 10 second answer...

There are total 50 balls, out of which 10 are white.....  another    ten balls    are chosen randomly and   replaced with white balls

So, red +  blue =  50-10 =40......
Only 8+32 add up to 40.
We also know that blue are less than red, so  blue = 8 and red = 32 ...

poojaarora1818 · Answer

I'm waiting for the official explanation to be posted and looking forward to more solutions. I guess it's a question on pure algebra and word problems, similar to some probability scenarios. Anyway just re-attempted and got the correct solution.

Madhvendrasinh · Answer

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.

poojaarora1818 · Answer

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

SomeOneUnique · Answer

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+x 
 10  
  (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)

Bismuth83 · Answer

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  .

Bismuth83 · Answer

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)

Bismuth83 · Answer

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

poojaarora1818 · Answer

Thank you for your response. I have made the necessary changes to the solution as required.

egmat · Answer

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 2  —  10  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  -  4 x =  25  + x

35  =  5 x

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.

A bag contains only red balls. Initially, twenty of these 50 red balls

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