any takers? :)

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any takers? :)

by Tanjello » Fri May 21, 2010 6:25 pm
If 2 integers between -5 & 3, inclusive, are chosen at random, which of the following must be true?

A) The sum of the 2 integers is even
B) The sum of the 2 integers is odd
C) The product of the 2 integers is even
D) The product of the 2 integers is odd
E) The product of the 2 integers is negative
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by liferocks » Fri May 21, 2010 6:43 pm
Are you sure this is the complete question.I don't think any of the options is correct
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by Tanjello » Fri May 21, 2010 6:52 pm
yes it is, but I got A & C both have greatest probability of 4/9 so I was confused

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by Stuart@KaplanGMAT » Fri May 21, 2010 7:03 pm
Tanjello wrote:If 2 integers between -5 & 3, inclusive, are chosen at random, which of the following must be true?

A) The sum of the 2 integers is even
B) The sum of the 2 integers is odd
C) The product of the 2 integers is even
D) The product of the 2 integers is odd
E) The product of the 2 integers is negative
The question reads "which of the following MUST be true"... none of them must be true, we can find counter examples to every answer.

Accordingly, liferocks is 100% correct, the question is incomplete/misquoted/flawed.
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by kstv » Fri May 21, 2010 8:42 pm
Tanjello wrote:If 2 integers between -5 & 3, inclusive, are chosen at random, which of the following has the highest probability ?
A) The sum of the 2 integers is even
B) The sum of the 2 integers is odd
C) The product of the 2 integers is even
D) The product of the 2 integers is odd
E) The product of the 2 integers is negative
The set of nos. are -5, -4, -3, -2, -1, 0, 1, 2, 3
Prob of picking two nos. = 9C2
Even then is it possible in under 2 min

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by Tanjello » Fri May 21, 2010 8:46 pm
Sorry, it says "which of the following is MOST LIKELY to be true."

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by Stuart@KaplanGMAT » Fri May 21, 2010 9:29 pm
Tanjello wrote:Sorry, it says "which of the following is MOST LIKELY to be true."
OK, that's doable.

Without doing any math, I'm going for (c), since to get an even product you only need 1 of the two numbers to be even (and the other can be even or odd). All of the other answers require both numbers to fit a certain criterion.
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by Tanjello » Fri May 21, 2010 9:43 pm
I agree with you about C, Stuart, but isn't A is also equally likely since:
odd + odd = even
even + odd = odd
even + even = even

Ugh, this is a mean question!

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by Rahul@gurome » Fri May 21, 2010 10:01 pm
The integers from -5 to 3 inclusive are -5,-4,-3,-2,-1,0,1,2,3 - 9 in number.
The odd integers are -5,-3,-1,1,3 - 5 in number.
The even integers are -4,-2,0,2 - 4 in number.
Consider (A) first.
This is possible if both integers are even or both are odd.
Ways of selecting two odd integers is 5C2 = 10.
Ways of selecting two even integers is 4C2 = 6.
Or ways of selecting two integers such that their sum is even is 10+6 = 16.
Ways of selecting two integers from given 9 is 9C2 = 36.
Or required probability is 16/36.
Consider (B) next.
This is possible if one integer is odd and the other is even.
Number of ways this can happen is 5C1*4C1 = 20.
Or required probability is 20/36.
Consider (C) next.
This is possible if either both integers are even or one of the integers is even.
Ways of selecting both even integers is 4C2 = 6.
Ways of selecting one even integer is 5C1*4C1 = 20.
Or ways of selecting two integers such that their product is even is 26.
So the required probability is 26/36.
Consider (D) next.
Product of two integers is odd if both integers are odd.
Ways of selecting two odd integers is 5C2 = 10.
Or required probability is 10/36
Next consider (E)
The product is negative when one integer is negative and the other integer is not zero but positive.
Ways of selecting this is 5C1*3C1 = 15.
Or required probability is 15/36.

Among all the above (C) has the maximum probability which is 26/36 = 13/18.
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by Tanjello » Fri May 21, 2010 11:27 pm
wow, I'd thank you twice if I could! But I'm concerned about the amount of time if would take. Is there no quicker method?

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by frank1 » Sat May 22, 2010 6:17 am
Tanjello wrote:wow, I'd thank you twice if I could! But I'm concerned about the amount of time if would take. Is there no quicker method?
but do you think those are all possible under 2 minutes?
(if i spend 5 minutes to solve 1 question,i may be sacrificing time of atleast 1 other question...)
rather i think logic provided by Stuart Kovinsky is much in accordance to GMAT requirement and spirit...
personal thought

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by quantskillsgmat » Sun May 23, 2010 2:36 am
there are total 9 integers, 4 are even and 5 are odd.
again 5 odd 4 even and one neutal integers are there.so any of the option doesnt hold always but there can be posibility
of answer choices. so question given in incoorect form.
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