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25 May 2010, 21:33
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
42% (02:17) correct
58%
(01:51)
wrong
based on 61
sessions
History
Date
Time
Result
Not Attempted Yet
David wrote each of the integers 1 through 20, inclusive, on a separate index card. He placed the cards in a box, then drew cards one at a time randomly from the box, without returning the cards he had already drawn to the box. In order to ensure that the sum of all the cards he drew was even, how many cards did David have to draw?
(A) 19
(B) 12
(C) 11
(D) 10
(E) 3
(A) 19
(B) 12
(C) 11
(D) 10
(E) 3
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02 Sep 2010, 04:20
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Jerome wrote each of the integers 1 through 20, inclusive, on a separate index card. He placed the cards in a box, then drew one card at a time randomly from the box, without replacing the cards he had already drawn. In order to ensure that the sum of all the cards he drew was even, how many cards did Jerome have to draw?
A. 19
B. 12
C. 13
D. 10
E. 3
This is not a good question. The answer depends on how you interpret it.
The given solution suggests that they meant the maximum number of picks needed to ensure that on the way you had an even sum. In this case worst case scenario would be to draw odd number first, and then keep drawing all 10 even numbers. After 11 draws you'd still have an odd sum and need to draw 12th card, which will be an odd to get the even sum.
But if you interpret it as at which drawing the sum will definitely be even then the answer would be 20. As only on 20th draw I can say for sure that the sum is even.
I wouldn't worry about this question at all.
A. 19
B. 12
C. 13
D. 10
E. 3
This is not a good question. The answer depends on how you interpret it.
The given solution suggests that they meant the maximum number of picks needed to ensure that on the way you had an even sum. In this case worst case scenario would be to draw odd number first, and then keep drawing all 10 even numbers. After 11 draws you'd still have an odd sum and need to draw 12th card, which will be an odd to get the even sum.
But if you interpret it as at which drawing the sum will definitely be even then the answer would be 20. As only on 20th draw I can say for sure that the sum is even.
I wouldn't worry about this question at all.
General Discussion
26 May 2010, 00:52
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Well, here's how I did it:
First thing that came to mind was the number properties, particularly:
Odd + Odd = Even
Even + Even = Even
Since this gives you the result you're looking for (an even sum) after only 2 cards, the key is going to be what happens after you draw one of each.
Even + Odd = Odd.
The only way to make the sum Even is to draw another Odd - another Even will keep your sum Odd. As soon as you draw another Odd, you can stop drawing because you'll have an Even sum.
Now look at it from a worst case scenario: How many Evens can I draw before I'm assured of drawing an Odd? The answer is 9 - I've already drawn 1 and there are 10 Evens from 1 to 20. Once I'm done drawing all the Evens, I'll draw an Odd - my sum is now Even and I'm done. In total I've drawn 12 cards (the first 2, then the 9 remaining Evens, and an Odd at the end).
As for why the others are wrong, they just are
. For every answer less than 12, you can come up with ways to end up with an Odd sum from that many cards. For every answer more than 12, you'll have already reached an Even sum so they'll be no need to keep going.
First thing that came to mind was the number properties, particularly:
Odd + Odd = Even
Even + Even = Even
Since this gives you the result you're looking for (an even sum) after only 2 cards, the key is going to be what happens after you draw one of each.
Even + Odd = Odd.
The only way to make the sum Even is to draw another Odd - another Even will keep your sum Odd. As soon as you draw another Odd, you can stop drawing because you'll have an Even sum.
Now look at it from a worst case scenario: How many Evens can I draw before I'm assured of drawing an Odd? The answer is 9 - I've already drawn 1 and there are 10 Evens from 1 to 20. Once I'm done drawing all the Evens, I'll draw an Odd - my sum is now Even and I'm done. In total I've drawn 12 cards (the first 2, then the 9 remaining Evens, and an Odd at the end).
As for why the others are wrong, they just are
. For every answer less than 12, you can come up with ways to end up with an Odd sum from that many cards. For every answer more than 12, you'll have already reached an Even sum so they'll be no need to keep going. 
