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PR Practice test problem

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PR Practice test problem

by floyd1180 » Tue Mar 06, 2007 2:01 pm
I do not think this question is correct, and their explainations are awful. I'm sure it must be right, so someone please show me the way!

Jerome 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 from the box. 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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Re: PR Practice test problem

by gabriel » Wed Mar 07, 2007 12:12 am
floyd1180 wrote:I do not think this question is correct, and their explainations are awful. I'm sure it must be right, so someone please show me the way!

Jerome 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 from the box. 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

Thanks
Some basics bfor the answer... odd+odd= even..... odd + even= odd.... even +even= even....

ok... now the answer... the task bfor jerome is to ensure that the sum of the numbers on the card is even..... that is after a certain number of cards each card number added to the sum shuld give a even number...

So think abt the worst case possible .... that is he has already picked the cards that wuld give him a odd sum .... once that is done every card after that will give a even sum...

So the worst case scenario is .... jerome picks a even numbered card at first then picks a odd numbered card so the sum of the first 2 cards ... even + odd= odd.... take another even numbered card and the sum still remains odd.... jerome keeps doing this until he picks all the even numbered card ..... therefore he has one odd numbered card and 10 even numbered card ... the sum of all of which is odd... now all he has left is odd numbered card and any one of them added to the sum will give even sum .... so my take to ensure that he has a even sum he will have to pick 12 cards
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by floyd1180 » Wed Mar 07, 2007 6:36 am
That's the same answer the book gives, but I don't think it works.

Let's look at it with numbers.

1
2
3
4
5
6
7
8
9
10
11
13

Twelve numbers, and the sum is 79. What am I missing?
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by gabriel » Thu Mar 08, 2007 3:49 am
floyd1180 wrote:That's the same answer the book gives, but I don't think it works.

Let's look at it with numbers.

1
2
3
4
5
6
7
8
9
10
11
13

Twelve numbers, and the sum is 79. What am I missing?
allright floyd ... read the q properly.... what does it ask for ?...... it asks u to pick a number of cards..... after u have done that every card picked after that when added to the sum will give u a even sum... in your example there are some cards that when added will give u a even number and some that will give a odd number.... moreover the first 4 numbers in ur example already gives u a even number...


let me know if u dont understand.... i will try to explain the q once again.....
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by Vasudha » Sun Mar 11, 2007 6:45 pm
Hi Floyd,

Look at it this way...


Like Gabriel says take the worst case scenario in this case

1. lets say he picked all the ten even numbers first
2. their sum is even
3. next he has only an odd number to pick
4. so, even + odd = odd
5. Since all the remaining numbers will be odd, he has to pick another odd number
6. this time the sum will be odd+odd= Even


Hence 10 even cards + 1 odd card ( an odd number) + another odd card (an even number) = 10+1+1= 12 cards
Last edited by Vasudha on Sun Mar 11, 2007 9:16 pm, edited 1 time in total.
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by jayhawk2001 » Sun Mar 11, 2007 8:27 pm
I still feel the question is incorrectly worded. It says "In order to ensure that the sum of all the cards he drew was even, how many cards did Jerome have to draw"

The above, atleast to me, means that we need to find a number that will
ALWAYS yield an even sum. 12 cannot always yield an even sum as Floyd
has pointed out.
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by Stacey Koprince » Sun Mar 11, 2007 10:00 pm
Not a great question. Don't worry about it.
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by beeparoo » Tue May 27, 2008 12:16 pm
Thank god others thought this question is also poorly worded. It is from the Princeton Review, and I thought it was very ambiguous.

I mean, case in point, if I follow Vasudha's logic, then by the time I've picked up two cards, I already have an even sum.
Vasudha wrote:Hi Floyd,

Look at it this way...


Like Gabriel says take the worst case scenario in this case

1. lets say he picked all the ten even numbers first
2. their sum is even
3. next he has only an odd number to pick
4. so, even + odd = odd
5. Since all the remaining numbers will be odd, he has to pick another odd number
6. this time the sum will be odd+odd= Even


Hence 10 even cards + 1 odd card ( an odd number) + another odd card (an even number) = 10+1+1= 12 cards
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