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

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If p, q, and r are integers, is pq + r even? [#permalink] New post 07 Apr 2013, 02:26
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If p, q, and r are integers, is pq + r even?

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

[Reveal] Spoiler:
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?
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Apr 2013, 04:53, edited 1 time in total.
RENAMED THE TOPIC.
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Re: A proper organised way to solve this type of questions? [#permalink] New post 07 Apr 2013, 02:46
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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
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Re: A proper organised way to solve this type of questions? [#permalink] New post 08 Apr 2013, 04:13
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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.

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

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

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