I would go with C

There are a total of 4m objects

First container quantity after the move = 1/16*4m = m/4

So m - m/4 = 3m/4 were moved

Each of four identical containers contains m balls.

This means to say that all containers had 4 balls each (1+6+5+4 = 16)

So m = 4

Now substitute 4 in the answer choices. [spoiler](C)[/spoiler] yields 3. 3 balls were removed from container 1.

cramya wrote:I would go with C

There are a total of 4m objects

First container quantity after the move = 1/16*4m = m/4

So m - m/4 = 3m/4 were moved

If you can't visualize cramya's algebraic solution, you can use POE. The ratios sum to 16 so you know total must be a multiple of 16. This eliminates D and E. Your tough choices will be between B and C. You might reason that since the ratios between the last 3 are closer, that more than half had to have been taken from the first and distributed to the other 3 for there to be such big gap between the first and second. That leaves you with C. If you couldn't come up with this also, a plain old guess could still produce C.

cramya wrote:I would go with C

There are a total of 4m objects

First container quantity after the move = 1/16*4m = m/4

So m - m/4 = 3m/4 were moved

Cramya, this algebra may have worked in this case but it is not true in general

16 16 16 16

4 20 22 18 12 removed from 1

New ratio is 2 10 11 9 sum to 32

16- 2 x 96 /32=10 but 12 were removed

The correct algebra is

x= 15M/4-3M ( Add the three new sums and subtract their original sum 3m)

X=3M/4

dtweah,
used the same logic as cramyas and it seems to work.

say initial distribution: 16:16:16:16
final distribution : 4:24:20:16

removed from A: 3/4*16=12 which is what we have (16-4=12)

thats such a short and sweet solution. very smart, Uri!

scoobydooby wrote:dtweah,
used the same logic as cramyas and it seems to work.

say initial distribution: 16:16:16:16
final distribution : 4:24:20:16

removed from A: 3/4*16=12 which is what we have (16-4=12)

It does not work in general.
16 16 16 16

4 20 22 18 12 removed from 1

New ratio is 2 10 11 9 sum to 32

16- 2 x 96 /32=10 removed but 12 were removed

X= 3 new sum -3M is the general solution.

hey dtweah,
what i mean is we would never come to a situation in which we will have the final distribution as 2 10 11 9

if we started with 16:16:16:16; total: 64
final distribution would be (1/16*64):(6/16*64):(5/16*64): (4/16*64)
=> 4:24:20:16
removed from 1st: 12 or 3/4*16

scoobydooby wrote:hey dtweah,
what i mean is we would never come to a situation in which we will have the final distribution as 2 10 11 9

if we started with 16:16:16:16; total: 64
final distribution would be (1/16*64):(6/16*64):(5/16*64): (4/16*64)
=> 4:24:20:16
removed from 1st: 12 or 3/4*16

A ratio must be put in its simplest and 4 24 20 16 is not its simplest. However, Cramya's algebra was correct becuase instead of sum being 64 I made it 96 which is why I began thinking of another algebra method.

2 10 11 9 sum to 32

So 16- 2 x 64/32 =12 that was removed. So the solution offer by Cramya is indeed General and I apologize for saying it wasn't!! if you check my first solution you will see 96. I owe Cramya One.

This was a very good question though. I always enjoy pushing one question to its limits as we have done here.

thanks guys for ur explanations. i did a silly mistake here, so couldn't get the right answer. OA is certainly C.
:)