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

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

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10 Mar 2011, 13:18
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56% (01:53) correct 44% (01:34) wrong based on 132 sessions

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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: OA

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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

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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

Hence E.

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

Kudos [?]: 396 [0], given: 123

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

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10 Mar 2011, 20:22
For pq+r to be even, pq = even and r = even OR pq = odd and r = odd.

From (1)
p + r is even so p is even and r is even or p is odd and r is odd

but no info about q, if q is even and p and r are odd, then pq+r is odd, otherwise even

From (2) , q + r is odd, so q is even and r is odd OR q is odd and r is even

So if p is even, then pq -> even + r (odd) = odd or even + r (even) = even

Not sufficient

Combining both (1) and (2), if p and r ar even, pq + r is even

if p and r are odd then, pq + r is odd (as q is even then)

Hence we don't have definitive answer,and the answer is E.
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If p, q, and r are integers, is pq + r even? [#permalink]

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07 Apr 2013, 03:26
1
KUDOS
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?

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

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

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07 Apr 2013, 03:46
1
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
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Re: A proper organised way to solve this type of questions? [#permalink]

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08 Apr 2013, 05:13
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.

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

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06 Jan 2014, 09:44
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

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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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Re: If p, q, and r are integers, is pq + r even?   [#permalink] 18 Jun 2016, 05:03
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