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Manhattan tells me I should make the table which works fine. I tried doing it without the table and that worked too. However, without the table I was less convinced and more confused because in your head it gets jumbled up. So is there another foolproof way of doing these? Or do I have to stick with the Manhattan table?

Re: A proper organised way to solve this type of questions? [#permalink]
07 Apr 2013, 02:46

1

This post received KUDOS

karmapatell wrote:

If p, q, and r are integers, is pq + r even?

(1) p + r is even. (2) q + r is odd.

Manhattan tells me I should make the table which works fine. I tried doing it without the table and that worked too. However, without the table I was less convinced and more confused because in your head it gets jumbled up. So is there another foolproof way of doing these? Or do I have to stick with the Manhattan table?

The Manhattan table works fine, another method is using real numbers .

(1) p + r is even. 3+1 = even, so is 3q+1 even? depends on q : not Sufficient (2) q + r is odd. 2+1=odd, so is p2+1 even? depends on p : not Sufficient

(1)+(2) p + r is even AND q + r is odd Example 1: 3+1=even--2+1 = odd 2*3+1=odd Example 2:2+2=even--3+2=odd 2*3+2=even Not Sufficient _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: A proper organised way to solve this type of questions? [#permalink]
08 Apr 2013, 04:13

Expert's post

karmapatell wrote:

If p, q, and r are integers, is pq + r even?

(1) p + r is even. (2) q + r is odd.

Manhattan tells me I should make the table which works fine. I tried doing it without the table and that worked too. However, without the table I was less convinced and more confused because in your head it gets jumbled up. So is there another foolproof way of doing these? Or do I have to stick with the Manhattan table?

From F.S 1, assume p=r=0, thus, we get a YES for the question stem. Now assume p=1, r=1,q = 2 we get a NO. Insufficient.

From F.S 2, assume q=0,r=1, we get a NO for the question stem.Now assume r=2,q=1 ,p=2, we get a YES. Insufficient.

Taking both together, we have p=0,r=0,q=1, and a YES. Again taking, r=1,p=1,q=0, a NO. Insufficient.

What might help you in selecting good numbers is the fact that from the F.S 1,either both p,r are even or both are odd. Similarly, from F.S 2, q and r are odd/even or even/odd.

Manhattan tells me I should make the table which works fine. I tried doing it without the table and that worked too. However, without the table I was less convinced and more confused because in your head it gets jumbled up. So is there another foolproof way of doing these? Or do I have to stick with the Manhattan table?

Odds and Evens, ok

Statement 1

Clearly Insufficient

Statement 2

Same here

Statements 1 and 2 combined

p+r = even q+r = odd

p-q = odd

Then p must be even and q odd or the other way around

If p is even then pq will be even and 'r' will be even = All even= Answer is YES if q is even then pq will again be even and 'r' will be odd= All odd = Answer is NO

Hence E is your best choice

Cheers! J

gmatclubot

Re: If p, q, and r are integers, is pq + r even?
[#permalink]
06 Jan 2014, 08:44

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