There are 20 cards in a box and each card is numbered from 1 to 20. Bi

Question

There are 20 cards in a box and each card is numbered from 1 to 20. Bill picked cards at random from the box without replacement. What is the minimum number of cards drawn that will give a guaranteed even number for the sum of all the cards drawn?

A. 1
B. 3
C. 10
D. 12
E. 20

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A. 1
B. 3
C. 10
D. 12
E. 20

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Bunuel · Accepted Answer

Consider this scenario as a game where the game ends as soon as the cumulative sum of the cards drawn becomes even. The question asks for the minimum number of card draws necessary to ensure that the game would have ended at that point. The correct answer is 12 because this is the least number of draws required to achieve an even sum at some point during the process.

Here’s why 12 is the minimum:

If we draw 11 cards, there's a possibility that we could select 1 odd card followed by 10 even cards, resulting in a continuously odd sum throughout these 11 draws. 
 However, when we extend the draw to 12 cards, an even sum is guaranteed. For instance, if the first 11 cards include 1 odd and 10 even numbers (resulting in an odd sum), the next card drawn must be odd (since there are only 10 odd and 10 even numbers in total). This additional odd card changes the cumulative sum from odd to even.

Thus, at the 12th draw, it is mathematically impossible not to achieve an even sum at some point in the process, which ensures the game would have ended by this draw.

P.S. I agree that the question could have been worded better.

yatharthdas · Answer

I think the answer is D 
1 2 4 6 8 10 12 14 16 18 20 then u will use odd no. I.e. 3(or any other odd no.) Which will give the answer as even

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rsrighosh · Answer

I think answer is D
If we consider the worst case scenario

We pick 1st as Odd then next 10 even then 1 odd... So total is 12

twobagels · Answer

If you pull 1 then every even card in a row you'll get {1,2,4,6,8,10,12,14,16,18,20} then you'll need one more to get an even sum. That set is 11 integers, so I'd guess the answer is 12, because you'd have to pull one more card and it would be odd, since there are no evens left: hence 12 cards would be pulled and guarantee an even sum.

I venture to guess D.

I venture to guess D.

Bunuel · Answer

Anyone else want to try ?

Check other  Worst Case Scenario Questions  from our  Special Questions Directory  to practice.

Hope it helps.

masondr · Answer

I'm confused with the wording. It's saying the minimum number of cards drawn where the sum is guaranteed to be even. The answer is saying 12 but if you did something like draw 9 evens and 3 odds that would be 9 evens so even + 3 odds so odd so basically even + odd = odd. The sum for 9 evens and 3 odds is odd not even, so how can it be guaranteed that for 12 cards drawn, the sum will be even? This wording makes it seem like 20 is the answer, as you can only be guaranteed that the sum will be even after all cards. If it was something like what's the minimum number of cards needed to guarantee that the sum of those n cards will have been even at least once, then it would make sense. In short, all cards drawn doesn't mean that no even sum ever occurred.

Ggt1234 · Answer

For the sum to be even, it is either e+e or o+o. To be certain, we need to exhaust either odd or even cards. Because we don’t want 10 odds and then 1 even. Nor do we want 10 evens and then one odd. These are our worst case scenarios. There are 19-1/2 +1 = 10 odd numbers. But after you have drawn 10 cards and if they have all been odd, then it can be certain that the next two are even. even + even is even, hence. 10+2 = 12 is the minimum

MielMatt · Answer

I don't understand why it is not 20, imagine he picks 12 cards, with 5 odds and 7 even, the sum would still be odd

Bunuel · Answer

Please review this post carefully: https://gmatclub.com/forum/there-are-20 ... l#p3387195 Check other Worst Case Scenario Questions from our Special Questions Directory to practice. Hope it helps.

Ggt1234 · Answer

Hi MielMatt, 
Regards to your query, by the time 2 odds out of the 5 were drawn, we already have an even sum. Same applies to the 7 evens. Or if it were alternating, odd then even then odd, then in three draws we have an even sum. So you wouldn’t have to draw 12.

swasmosylloret · Answer

What if you pull (2,4,6,8,10,12,14,16,18,3,5,7)? That adds up to an odd number; therefore, it is not guaranteed that you will get an even one at 12(?)

Bunuel · Answer

Please review this post carefully: https://gmatclub.com/forum/there-are-20 ... l#p3387195 Check other Worst Case Scenario Questions from our Special Questions Directory to practice. Hope it helps.

Ggt1234 · Answer

in your example, by the timeyou have pickd the first two cards, you already have an answer. It is the best case scenario. We are looking for the worst case scenario.

anirchat · Answer

For total 12 cards, if I pick 9 odd and 3 even cards, the sum is  still odd . I do not understand how does this  guarantee  me an even sum.

Krunaal · Answer

Did you try to check what was the order you picked these cards, somewhere way before picking 12 cards you must have got an even sum. The key here is to find the worst possible case. Let's say you pick 1 first, if you pick another odd, it will give you an even sum hence you must pick an even card => your luck has to keep picking all the even cards until even cards exhaust, and then your bad luck will give up, when you pick the 12th card that is indeed odd.

anirchat

For total 12 cards, if I pick 9 odd and 3 even cards, the sum is still odd. I do not understand how does this guarantee me an even sum.

Kinshook · Answer

There are 20 cards in a box and each card is numbered from 1 to 20. Bill picked cards at random from the box without replacement.

What is the minimum number of cards drawn that will give a guaranteed even number for the sum of all the cards drawn?

We have to consider the case when maximum number of cards are drawn that result in odd sum. 
If we draw first odd card and then successively draw all 10 even cards, the sum of cards in all cases will be odd.
Now only 9 odd cards are left and drawing one more card will result in even sum of the cards drawn.

IMO D

There are 20 cards in a box and each card is numbered from 1 to 20. Bi

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