I have just worked on some quant problems in the 4th bin of the book "cracking the gmat 2008 edition" and came across a weird question, or rather the answer seems incorrect to me.
The question goes as follows:
"Jerome wrote each of the integers 1 through 20, inclusive, on a seperate card index. He placed the cards in a box and then randomly drew them one at a time without replacing any of them. 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
Now if he drew 3, those could have been 2 evens and an odd. If he drew 10, he could have drawn 9 odds and an even. If he drew 11, he could have drawn 9 odds and 2 evens. If he drew 12 he could have drawn 7 evens and 5 odds. And if he drew 19 he could have drawn all evens and 9 odds, so all combinations would result in an odd result.
The answer to this question is supposed to be B, going by the logic that when he draws eleven cards, the first one being odd and the rest even, he would still have an odd result. But with the 12th draw he would have to draw 2 odd balls, making the result even. But this requires that he draws 10 even balls, which is obviously not guaranteed. He could have also randomly drawn 7 evens and 5 odds, making the result odd. So the only way to guarantee an even result is if he draws all cards...
So either I'm understanding the question incorrectly or the writers have made a mistake.
Can anybody clear this up for me, please?
The question goes as follows:
"Jerome wrote each of the integers 1 through 20, inclusive, on a seperate card index. He placed the cards in a box and then randomly drew them one at a time without replacing any of them. 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
Now if he drew 3, those could have been 2 evens and an odd. If he drew 10, he could have drawn 9 odds and an even. If he drew 11, he could have drawn 9 odds and 2 evens. If he drew 12 he could have drawn 7 evens and 5 odds. And if he drew 19 he could have drawn all evens and 9 odds, so all combinations would result in an odd result.
The answer to this question is supposed to be B, going by the logic that when he draws eleven cards, the first one being odd and the rest even, he would still have an odd result. But with the 12th draw he would have to draw 2 odd balls, making the result even. But this requires that he draws 10 even balls, which is obviously not guaranteed. He could have also randomly drawn 7 evens and 5 odds, making the result odd. So the only way to guarantee an even result is if he draws all cards...
So either I'm understanding the question incorrectly or the writers have made a mistake.
Can anybody clear this up for me, please?
