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.

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:

50% (03:01) correct
50% (02:54) wrong based on 3 sessions

Q: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 Please explain your answer choices in detain including why other choices are wrong?

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.

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.

What is chance that all cards he draws is even only? In that case why cant he draw 10 even and still ahve sum even? This is where I am confused? My answer was since he can draw 10 even or odd cards that gives an even sum 10 is answer!

Can you give a scenario as you quoted above of answer less than 12 will end up odd...perhaps that will clarify things a bit.

The question rephrased should be what is the most cards that he could draw before it has to sum to an even number;

In 1-20 there are ten odds and ten evens. Thus the most cards you could draw would be if the first sum is odd and then you keep adding evens until you run out then add an odd number.

odd+even+even+even+even+even+even+even+even+even+even+even=even; actual numbers 1+2+4+6+8+10+12+14+16+18+20+3=114 or 3+2+4+6+8+10+12+14+16+18+20+1=114 or other combinations of 2 odds.

You are right in that drawing 10 Even cards gives you an even sum, as does 10 Odds. But what about if he draws the following 10 cards:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10: The total is 55 (Odd sum = bad).

Another group of 10 cards:

1, 2, 3, 4, 5, 6, 7, 8, 9, 11: The total is 56 (Even sum = good).

Looking at this as a permutation problem will just drive you crazy - far too many options and I'm not sure it can get you the answer.

The way I read the question, the key word is "ensure" and I assume that he stops when he reaches an even sum - the wording isn't great. Based on the logic of my first post, there is no way you can draw 12 cards and not have achieved an Even sum at somepoint along the way (the worst case being that it happens on the 12th card). Note that you can draw 12 cards and get an odd sum (like 1,3,5,2,4,6,8,10,12,14,16,18 = 99) but along the way he'd gotten his even sum and stopped (1,3 = 4).

You are right in that drawing 10 Even cards gives you an even sum, as does 10 Odds. But what about if he draws the following 10 cards:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10: The total is 55 (Odd sum = bad).

Another group of 10 cards:

1, 2, 3, 4, 5, 6, 7, 8, 9, 11: The total is 56 (Even sum = good).

Looking at this as a permutation problem will just drive you crazy - far too many options and I'm not sure it can get you the answer.

The way I read the question, the key word is "ensure" and I assume that he stops when he reaches an even sum - the wording isn't great. Based on the logic of my first post, there is no way you can draw 12 cards and not have achieved an Even sum at somepoint along the way (the worst case being that it happens on the 12th card). Note that you can draw 12 cards and get an odd sum (like 1,3,5,2,4,6,8,10,12,14,16,18 = 99) but along the way he'd gotten his even sum and stopped (1,3 = 4).

Did that help?

I'm not sure if your responding to my post, but I figured I would explain my reasoning. Again in this problem we want to know when he will know for sure when he will have an even number. We have to figure out the maximum amount of cards he could draw before he would have to draw an even amount. As my post described above, it would require that he the first card he draws is odd then, if he draws all the even cards subsequently the sum will always be odd. After you have exhausted your supply of 10 even cards the only thing you have left to draw would be an odd. Any odd plus and odd would be equal to even. 1+10+1=12, thus, this tells us that he can safely conclude that at 12 cards he will have drawn an even sum.

You really only need the properties of odd+odd, even+even, and odd+even to figure out this problem. You only need logic. If you are doing any permutations or combinations you are doing too much work.

You are correct in saying that it is impossible to draw 12 and not get an even, but 12 is the point at which you can be certain that you will draw an even sum. If you consider 11, the following possibility exists;

1+2+4+6+8+10+12+14+16+18+20=111 and that is 11 terms that sum to an odd. Any odd could work in place of the 1.

Hrmm... good question. I was also confused with the wording. I wasn't aware that he stops once he reaches an even sum. (because you can basically reach an odd sum with any combination of numbers, ie 7 Odds + 5 Evens gets you an odd)

Makes sense now, thanks for the clarification guys.

Please respond... Can this be a valid question??? It is told RANDOMLY in the answer.... picking up all even card is absolutely not random.... I am not sure if this is a valid question

Please respond... Can this be a valid question??? It is told RANDOMLY in the answer.... picking up all even card is absolutely not random.... I am not sure if this is a valid question

It is possible to randomly pick all the evens in a row. I may not be likely, but it is possible.