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A box contains $731$ black balls and $2000$ white balls. The following process is to be repeated as long as possible. Arbitrarily select two balls from the box. If they are of the same color, throw them out and put a black ball into the box ( enough extra black balls are available to do this). If they are of different color, place the white ball back into the box and throw the black ball away. Which of the following is correct?

  1. The process can be applied indefinitely without any prior bound
  2. The process will stop with a single white ball in the box
  3. The process will stop with a single black ball in the box
  4. The process will stop with the box empty
  5. None of the above
in Numerical Ability by Boss (30.8k points) | 554 views

4 Answers

+5 votes
Best answer

The total number of balls is $2371$ and in each round $1$ ball is removed. So after $2370$ steps, we will have $1$ ball in the box and we must stop.

Here, white balls are even and every time either it gets removed in a pair or remains unchanged.

So, the last ball remaining can never be a white ball.

Since at each step the number of black balls changes from odd to even or even to odd, so at the end only 1 black ball will be in the box.

Hence, option C is the correct answer.

by Boss (16.5k points)
edited by
0
In the question it is given that there are 731 black balls..
+3 votes

Answer Option 3.

by (235 points)
reshown by
0
I meant to write XOR gates are commutative instead of associative.
0
Great interpretation!
0 votes

The option will be 5 .Here is the explaination.

Let X and Y be the No of Black and white balls in the Box.

Now There could be 3 possibilities . Below table presents Possibilities and changes happend due to the each outcome.

   BB       BW          WW
 X-1,Y X-1,Y       X+1,Y-2

Now here Y=2000 . X=731.If 3rd possibility happens all the time(WW) then There will be no white ball left in the Box  beacuse 2|2000.So option 2 will be eliminated.

Again if other two possibilty happens then white ball will be as it is and no  Black ball will be left.So option 3 is eliminated.

There is possibilty that the box may be empty  beacuse y is decresing in 3rd possibility and x is decreaing in other two possibility but it may not be the case. so option 4 also eliminated.

So Option 5 will be the answer.

Correct me if i am wrong here.

by (105 points)
0
After X becomes 2, it can only reduce to 1 but not 0.
0
Sir i think box cannot be done empty..Isnt it?
0
yes. black cannot go to 0.
0
[email protected] Sir.Thanks for the clarification.It was my mistake.After X becomes 2, it can only reduce to 1 but not 0
–1 vote
I think A.

Because 3 case BB,WW,BW. So in every case either 1Black ball or 1 white Ball is added. So Every time it will contain some pair.

White ball cannot be 1 beczits going in pair only and total no of white is even.
by Active (2.4k points)
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