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PR – Math Bin 4 – # 4 - pg 380

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PR – Math Bin 4 – # 4 - pg 380 [#permalink] New post 29 Aug 2006, 06:33
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

17% (03:06) correct 83% (00:54) wrong based on 6 sessions
Can anyone please explain the solution to this problem? The PR explanation is not clear to me.

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 returning 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. 11
d. 10
e. 3
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 [#permalink] New post 29 Aug 2006, 07:49
J draws 1 odd and 10 even, then in order to ensure that the sum is even he needs to draw 12 cards.
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 [#permalink] New post 29 Aug 2006, 09:11
I think the answer has to be 19. since 9 odds and 3 evens can still produce an odd sum
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 [#permalink] New post 29 Aug 2006, 09:19
b. 12

If he draws one odd and then 10 even, he needs to draw another one, which will be odd, to ensure an even sum.

odd+10xeven= odd + odd = even.
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 [#permalink] New post 29 Aug 2006, 09:42
You are right...

I didn't think through my answer... :oops:

X & Y wrote:
1+2+3+4+5+6+7+8+9+10+11+12=78
1+2+3+4+5+6+7+8+9+10+11+13=79

How can you ensure an even sum with 12 numbers?
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 [#permalink] New post 29 Aug 2006, 09:42
X & Y note that 1+2+3=6 which is even. In your example you need only 3 numbers
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 [#permalink] New post 29 Aug 2006, 10:31
BG wrote:
J draws 1 odd and 10 even, then in order to ensure that the sum is even he needs to draw 12 cards.


That's correct. What was your thought process in solving this one BG?

The explanation says as the first step to solving this: for the face values to NOT ADD up to even, the first card must be ODD...why is that?

Can I not draw: 2 and then 1 and still end up with 2+1 = ODD though I did not draw the first one to be odd?
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 [#permalink] New post 29 Aug 2006, 10:38
I think that 3,10,11 and 12 doesn’t work to ensure an even sum. Suppose Jerome takes the following numbers.

1,3,5,6,7,8,9,2,4,10,12,14

1+3+5=odd
1+3+5+6+7+8+9+2+4+10=odd
1+3+5+6+7+8+9+2+4+10+12=odd
1+3+5+6+7+8+9+2+4+10+12+14=odd
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 [#permalink] New post 29 Aug 2006, 10:45
X & Y wrote:
I think that 3,10,11 and 12 doesn’t work to ensure an even sum. Suppose Jerome takes the following numbers.

1,3,5,6,7,8,9,2,4,10,12,14

1+3+5=odd
1+3+5+6+7+8+9+2+4+10=odd
1+3+5+6+7+8+9+2+4+10+12=odd
1+3+5+6+7+8+9+2+4+10+12+14=odd


Interesting. Is there a specific logic behind choosing these numbers or combinations thereof?
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 [#permalink] New post 29 Aug 2006, 23:36
X&Y please note , again, that
1+3+5+6+7+8+9+2+4+10+12+14=odd the sum of the first 5 is EVEN, so your example is irrelevant .
The logic is simple
A sum is even if E+E=E or O+O=E, now E+O= O so in order to ensure that the sum of the numbers drawn is even, one must take 12 cards. Why?
Because, the worst scenario is if he takes first card O and then 10 E the sum will be odd. When all evens are depleted he will draw O and the sum is E
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 [#permalink] New post 30 Aug 2006, 00:10
BG wrote:
X&Y please note , again, that
1+3+5+6+7+8+9+2+4+10+12+14=odd the sum of the first 5 is EVEN, so your example is irrelevant .
The logic is simple
A sum is even if E+E=E or O+O=E, now E+O= O so in order to ensure that the sum of the numbers drawn is even, one must take 12 cards. Why?
Because, the worst scenario is if he takes first card O and then 10 E the sum will be odd. When all evens are depleted he will draw O and the sum is E

You mean to say after drwaing 12 cards, he can be sure that the sum is EVEN. I am not able to understand. :roll: :roll:
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 [#permalink] New post 30 Aug 2006, 02:17
I don't see a correct answer here.

There are 10 odd numbers and 10 even numbers

a. 19 - can draw 10 odd and 9 even..sum is even. can draw 10 even and 9 odd, sum is odd X
b. 12 can draw 5 even and 7 odd, sum is odd. can draw 6 odd and 6 even, sum is even X
c. 11 can draw 5 even and 6 odd, sum is even. can draw 5 odd and 6 even sum is odd
d. 10 can draw all even sum is even, can draw 9 even and one odd sum is odd
e. 3 can draw all even sum is even, can draw all odd sum is odd.

What am I missing?
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 [#permalink] New post 30 Aug 2006, 18:26
I am kind of new here, and probablty wrong.
So we have 20 cards. We took some of them and we have number 11 total.
It means that we drew card with #2,#3 and #6 on it. So we have 11.
And answer is E.

Again, new in GMAT, so making mistakes on every step:(
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 [#permalink] New post 30 Aug 2006, 20:11
I think 12

"In order to ensure that the sum of all the cards he drew was even", the most number of cards he would need to draw is 12.

1st card = odd
the next 10 cards = even => which makes the sum odd
the next card drawn is odd because only odd cards are left, so the sum is even.
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 [#permalink] New post 15 Sep 2006, 10:15
I think the only part I didn't agree with was "why the need to draw odd first". It really doesn't matter - the point is to arrive at the most #: Draw all even possible = 10. Draw one more, this one has to be odd since all even already drawn. now the sum is E+O = odd and we are at a tally of 10+1 = 11 cards. Draw another one (odd of course since no even left) = 11+1 = 12 cards and sum is O + O = E.
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 [#permalink] New post 03 Dec 2006, 18:52
I just got this one wrong when going through the drills..and I am inclined to agree with X & P and Futuristic. The wording of the question is:

"In order to ensure that the sum of all the cards he drew was even, how many cards did Jerome have to draw?"

If he draws:

1+3+5+6+7+8+9+2+4+10+12+14=odd

Then he will not have ensured that the sum of the cards he drew was even, therefore the answer can't be B.
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 [#permalink] New post 04 Dec 2006, 06:06
saumster wrote:
I think the only part I didn't agree with was "why the need to draw odd first". It really doesn't matter - the point is to arrive at the most #: Draw all even possible = 10. Draw one more, this one has to be odd since all even already drawn. now the sum is E+O = odd and we are at a tally of 10+1 = 11 cards. Draw another one (odd of course since no even left) = 11+1 = 12 cards and sum is O + O = E.



I guess the reason for drawing the first one as odd is, if you draw the first one as even, the sum is already even, you don't need to draw another card. Does it sound plausible ? :)
I think, the solution has to be 12 - draw the first one as odd, draw 10 even and then one more odd.
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Re: PR – Math Bin 4 – # 4 - pg 380 [#permalink] New post 04 Dec 2006, 14:51
saumster wrote:
Can anyone please explain the solution to this problem? The PR explanation is not clear to me.

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 returning 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. 11
d. 10
e. 3


I think the question is asking "after drawing how many cards can you be sure that the sum will be an even number?"

So best case is you only need to draw 1 card = 2 or 4 or 6 etc
Worst case is that you draw one odd card and then 10 even, and so require one more card to make the sume odd = 12 cards

In examples used where you draw 1+2+3+4 etc, you would reach an even number at 1+2+3 = 6.
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Re: PR – Math Bin 4 – # 4 - pg 380 [#permalink] New post 09 Jan 2011, 23:41
After being confused for a while, the solution seems clear to me.

The answer is B - 12
Keeping in mind that Jerome can stop in the middle when ever he has the sum as even .

Consider he starts with even , he stops . He has the sum as even .
Consider he starts with ODD, he can not stop. At this point, he either has the bad luck of drawing all even numbers there of . He would now draw the 12th card , which would be ODD ,

O+(EVEN Series)+O = E .

Do note, IMO the question should not have been "how many cards did Jerome have to draw" but rather should be "how many maximum number of cards might Jerome have to draw" .
:shock: :shock:
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Re: PR – Math Bin 4 – # 4 - pg 380 [#permalink] New post 10 Jan 2011, 01:49
Expert's post
adisinha2000 wrote:
After being confused for a while, the solution seems clear to me.

The answer is B - 12
Keeping in mind that Jerome can stop in the middle when ever he has the sum as even .

Consider he starts with even , he stops . He has the sum as even .
Consider he starts with ODD, he can not stop. At this point, he either has the bad luck of drawing all even numbers there of . He would now draw the 12th card , which would be ODD ,

O+(EVEN Series)+O = E .

Do note, IMO the question should not have been "how many cards did Jerome have to draw" but rather should be "how many maximum number of cards might Jerome have to draw" .
:shock: :shock:


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.
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Re: PR – Math Bin 4 – # 4 - pg 380   [#permalink] 10 Jan 2011, 01:49
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