Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

56% (00:58) correct
44% (00:53) wrong based on 6 sessions

HideShow timer Statistics

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

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

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.
_________________

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.

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:(

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 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.

Re: PR â€“ Math Bin 4 â€“ # 4 - pg 380 [#permalink]

Show Tags

04 Dec 2006, 15:51

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.

Schools: S3 Asia MBA (Fudan University, Korea University, National University of Singapore)

WE 1: Market Research

WE 2: Consulting

WE 3: Iron & Steel Retail/Trading

Jerome wrote each of the integers 1 through 20, inclusive [#permalink]

Show Tags

02 Sep 2010, 03:50

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?

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.
_________________