sjagad is right.
12 is indeed the right answer.
Whenever the second odd card is chosen the sum is guaranteed to be even.
The worst case scenario for the second odd card to occur is as the 12th card. (one odd plus 10 even having occurred before).
Hence only 12 cards are necessary!
Unless we have seen such problems before, it isn't possible to solve this in 2 mins.. It took more than 3 mins for me to think.
Last edited by deltasquare; 02-25-2008 at 11:08 AM. Reason: Additions;
I agree with killgmat. The question is to find out absolutely how many cards you have to draw to have an even sum NOT minimun number of cards.
If the number is 12, does that 'ensure' that the cards drew added up to an even sum?
(May be I am completely lost. Please please what is the OA?)
Yes. 12 ENSURES that the sum is even.Originally Posted by narda
12 cards will involve 2 odd cards FOR SURE. Hence it will definitely be even.
If these 2 odd cards occur earlier, then an even is obtained and the process stops. No need to continue picking cards after getting the second odd card.
So arguing that this 12 can have more than 3 odd cards is not right.
My take: "In order to ensure . . . ", We have to think about worst case scenario: first card will be odd, second one must be even (then sum is odd), next 9 cards will be still even to keep odd sum; now we don´t have any even cards, just odd ones left. The 12th card need to be odd, so the sum will be even.
Answer is 12.
Last edited by SIGLOCK; 02-26-2008 at 07:24 PM.
Sure, that works if you GET them in that order. You are drawing randomly. It is certainly possible to draw 10 even and 2 odd, but it is not certain that you will. I still think that the answer is 20.
A draw of 12 ensure that the sum will be Even.
Everyone will agree to the fact that it the worst case scenario for this problem.
There are some that have written that there can be more than 1 odd card drawn. Now, if that happens the moment the second odd card is drawn the sum of the cards will become EVEN.
But, if Jerome draws one card after the other a set of 12 cards will always ensures he gets a EVEN sum either at 12 cards or before it. But, worst case it will be 12.
Where is this question from and why haven't you posted the answers? You can never ENSURE that a sum will be even unless all the odd numbers are pulled. It's not possible to determine how many cards need to be pulled in order to reach this objective. Therefore, my answer will be: (E) it cannot be determined from the information given.
Kartikayeg,
Drawing 12 cards is the best case scenario. Not the worst. The worst case scenario is drawing a card that makes the set ODD no matter how many cards you have drawn. Worst case scenario is having an odd number at 19 cards. Thus, the only way to ENSURE you have an even number is to draw 20.
NO.
there are 3 black socks and 2 white socks in a drawer. to ensure that we get 1 white sock we have to take the worst case scenario, which is all the black socks and then 1 white sock. 4 draws
there are 20 cards, half even half odd. to ensure the sum is even, we need the worst case scenario. if we take even as the first card, that is being optimistic. we need to draw an odd first. then a string of evens and then another odd. 12 cards
If you have drawn more than 2 odd cards means than you had to have drawn 2 odd cards at some point. your sum of cards would be even at that point.
The cards are drawn one after the other. so you can say worst case having 9 odd cards. Because odd + odd is even. So when you have drawn 2 odd cards the sum would become even.
And the best case scenario is 1 draw (which would be even card).
What you can do
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