Samuel has a big drawer full of exactly 59 packets, each

Hi Chetan ,

Statement B is ok to understand .
But i have trouble understanding statement 1 .
Can you please explain statement 1 and how to interpret such statement in general

Hi

In this Questions We have total number of packets = 59
and no of black markers =23
we don't know about the number of red or blue markers.

1) he needs to draw minimum 42 packets to ensure that he has at least one marker of each colour out of red, blue, black.
Now to be sure that he has at least one marker of each colours, let initial all markers are black, Bad luck
So first 23 markers are black.
Now out of 42 , 23 are black. this means 42-23=19. in 19, 18 are of one colour and we have last marker of the third one. (we have to take the worst condition to ensure that markers of all colours have been drawn)
that means, say blue = 18 and last one is for red.
So number of red = 59-23-18=18.
(if we take second colour to be red, then we get third colour blue = 18). so its a single unique combination possible.
hence we get the number of markers of each colour.

SUFFICIENT





Hi Chetan ,

Statement B is ok to understand .
But i have trouble understanding statement 1 .
Can you please explain statement 1 and how to interpret such statement in general[/quote]

[/quote]

Hi gmatbusters,

Thanks for the explanation

Hi...

Sorry missed out on your question earlier..
So the solution would be..

There are 59 packets. 23 are black markers...
So 59-23=36 contain red or blue..

I) statement I tells us that Taking 42 out ensures that we have one of each colour..
So it gives us the max two colours to be 41 and third one was the 42nd that was picked up...
Black are 23, so the second highest =41-23=18..
So these 18 can be red or blue..
The third therefore would be 59-23-18=59-41=18
So the third also will be 18..
So the number of markers are :-
a) Black - 23
b)Red and Blue - 18 each..
Sufficient

II) statement II tells us Probability of picking red and blue is same..
So number of red and blue is same..
But red + blue = 59-23=36
So red=36/2=18
Sufficient

D

Isn't it incorrect to assume that we do not have any black markers in the remaining 36(59-23) packets? Question statement should say Out of 59, exactly 23 packets contain a black marker.

Hi! Thanks for the explanations. But do you think this is a valid question?
Why do we take the worst condition? The minimum possible number of packets to draw to have the favorable results can be 3. Why do we account for the bad luck? I think Statement 1 is not very correctly formulated. Please, correct me if I am missing something.

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