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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?

Last edited by Bunuel on 12 Apr 2013, 05:53, edited 1 time in total.

Re: A proper organised way to solve this type of questions? [#permalink]

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07 Apr 2013, 03:46

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: If p, q, and r are integers, is pq + r even? [#permalink]

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10 Mar 2011, 13:52

GMATD11 wrote:

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

1) p+r is even 2) q+r is add

M getting D

OA is different. Pls confirm if answer is not D

We want to know if pq+r is even

Statement 1) says p+r is even implying that p and r are either both odd or both even. When they are both even, then irrespective of q being even or odd, pq+r will be even. When they are both odd, depending on q, pq+r can be odd or even. So, insufficient

Statement 2) says q+r is odd, implying at least one of q or r is odd and the other one is even. When r is odd and q is even, pq+r is odd. When q is odd and r is even, pq+r is even or odd depending on value of p, so insufficient.

Combining the two, when p and r are even and hence q is odd, pq+r is even

when p and r are odd and hence q is even, pq+r is odd, so again insufficient

Re: If p, q, and r are integers, is pq + r even? [#permalink]

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10 Mar 2011, 20:15

1) Insufficient p + r = even p = even r = even. The answer is YES p = odd r = odd q = even. The answer is NO

2) Insufficient q + r = odd q = odd r = even p = even. The answer is YES q = odd r = even p = odd. The answer is NO

combine 1) and 2) Insufficient p + r = even q + r = odd let r = even, p = even, q=odd. The answer is YES let r = odd, p = odd, q= even. The answer is NO

Re: A proper organised way to solve this type of questions? [#permalink]

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08 Apr 2013, 05:13

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

Re: If p, q, and r are integers, is pq + r even? [#permalink]

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18 Jun 2016, 05:03

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