hellosanthosh2k2 wrote:

Hi chetan2u,

shouldn't the answer be C?

First two statements are insufficient.

To count minimum number of marbles to have marbles of all possible color is (Highest + next highest + the third highest + ........ + second lowest + 1)

Combining,

Let us say Black is least of all, in which case, minimum no of marbles to draw to ensure we have marbles of all color = Blue + Red + 1 = 57 => Blue + Red = 56 (which is not possible as given sum is 59 and black is 23)

so Black is not lowest of all,

now from statement 2, Red < Blue

So Red must be lowest of all.

so, Black + Blue + 1 = 57 => 23 + Blue + 1 = 57 => Blue = 33, are we not getting number of blues?

Am i missing anything?

Please help.

Thanks

Hello

So we are talking about combining the two statements.

There are 23 black markers, rest 36 contain some red and some blue, but red markers are less, so red must be least in number. Lets say there are 'x' red markers, so '36-x' is the number of blue markers. And x < 36-x so we have x < 18.

Now, to get marbles of any one single colour, we first take the worst case scenario. Worst case scenario would be if we get 22 black markers, 'x-1' red markers and '36-x-1' blue markers (basically we have all but one marble of each color). And in this worst case scenario, now whichever next marker comes up will complete the objective - of having all marbles of any one particular colour. So lets add up 22, x-1 and 36-x-1 and 1:- this will give us the minimum number required.

22 + x-1 + 36-x-1 + 1 = 57.

So irrespective of the value of x, answer is coming to be 57 only (this answer is now independent of the number of red and blue markers).

You can check, lets say there are 23 black, 17 red, 19 blue. Then minimum no of markers = 22+16+18 + 1 = 57

Lets check now with 23 black, 10 red, 26 blue. Then minimum no of markers = 22+9+25 + 1 = 57 again.

So we cannot determine the number of red or blue markers based on this info.