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Jerome wrote each of the integers 1 through 20, inclusive [#permalink]
02 Sep 2010, 02:50
00:00
A
B
C
D
E
Difficulty:
(N/A)
Question Stats:
57% (02:12) correct
43% (01:11) wrong based on 4 sessions
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?
Re: A Quantitative question by The Princeton Review [#permalink]
02 Sep 2010, 03:20
Expert's post
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. _________________
Re: A Quantitative question by The Princeton Review [#permalink]
02 Sep 2010, 03:33
Bunuel,
12 is correct. And that is the max draws you need to definitely know that the sum of drawn cards is even. Say the first card you draw is Even... then E..O...O = Even sum E...E = Even sum E...O..E...E...E..E....9 more E cards....O....(12th card) has to odd... sum is even..
Say the first card is odd.. O...O = even O...E..E...10 more E cards max..O... = even
I can't understand how you say 20 is the answer. Definitely before the draw of 20 cards you will have a sum which is even.
I know you said you don't care about the question ..but just academically..what do you say?
Re: A Quantitative question by The Princeton Review [#permalink]
02 Sep 2010, 03:52
Expert's post
hemanthp wrote:
Bunuel,
12 is correct. And that is the max draws you need to definitely know that the sum of drawn cards is even. Say the first card you draw is Even... then E..O...O = Even sum E...E = Even sum E...O..E...E...E..E....9 more E cards....O....(12th card) has to odd... sum is even..
Say the first card is odd.. O...O = even O...E..E...10 more E cards max..O... = even
I can't understand how you say 20 is the answer. Definitely before the draw of 20 cards you will have a sum which is even.
I know you said you don't care about the question ..but just academically..what do you say?
Thanks.
Please read my solution above.
The answer depends on the interpretation of the question. If you interpret it as at which drawing the sum will definitely be even then 12 is wrong, as for example you can draw 7 even numbers and 5 odd numbers so after 12 drawings you will have an odd sum. So if it is the question then the only # of drawings that will guarantee an even sum is 20, because for every other # of drawings you could have either an even or odd sum.
Re: A Quantitative question by The Princeton Review [#permalink]
02 Sep 2010, 08:31
Bunuel wrote:
hemanthp wrote:
Bunuel,
12 is correct. And that is the max draws you need to definitely know that the sum of drawn cards is even. Say the first card you draw is Even... then E..O...O = Even sum E...E = Even sum E...O..E...E...E..E....9 more E cards....O....(12th card) has to odd... sum is even..
Say the first card is odd.. O...O = even O...E..E...10 more E cards max..O... = even
I can't understand how you say 20 is the answer. Definitely before the draw of 20 cards you will have a sum which is even.
I know you said you don't care about the question ..but just academically..what do you say?
Thanks.
Please read my solution above.
The answer depends on the interpretation of the question. If you interpret it as at which drawing the sum will definitely be even then 12 is wrong, as for example you can draw 7 even numbers and 5 odd numbers so after 12 drawings you will have an odd sum. So if it is the question then the only # of drawings that will guarantee an even sum is 20, because for every other # of drawings you could have either an even or odd sum.
Again not a good question.
If you draw 7 even numbers and 5 odd numbers you will have an even sum much before 12 cards. Do you see that? Anyway..time to move on!
Re: A Quantitative question by The Princeton Review [#permalink]
02 Sep 2010, 08:48
Expert's post
hemanthp wrote:
If you draw 7 even numbers and 5 odd numbers you will have an even sum much before 12 cards. Do you see that? Anyway..time to move on!
I don't understand what you mean by that.
But again: the question allows 2 interpretations, one gives the answer 12 and another gives the answer 20. There is no reason to believe that 1st one is better than the 2nd one. That is why this is a bad question.
You'll never see such an ambiguous question on GMAT.
Re: A Quantitative question by The Princeton Review [#permalink]
02 Sep 2010, 16:18
Yup. I understand we won't get a question like this but I was only arguing (as I said earlier) for purely academic purposes. Appreciate your responses. Ciao.
gmatclubot
Re: A Quantitative question by The Princeton Review
[#permalink]
02 Sep 2010, 16:18
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