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I was skeptical about the term "numbers" as opposed to

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Manager
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03 Aug 2006, 17:53
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I was skeptical about the term "numbers" as opposed to "integers." Does the term "numbers" = "integers?" Why haven't I seen this more often in my studies, then?

I'm just very careful about reading the questions, and I don't want being too careful to mess me up on test day (8/11!). I really hope that if I saw this on test day, the question would be worded with "integers."
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Re: sorry to go nuts, but here's another [#permalink]

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03 Aug 2006, 18:29
if you only post questions, that would be betterresponse to your questions..

here, st 1 says r<10, then it must be 8 since it is > than p and q. so suff that r = 8, q =6 and p =4..

st 2 says p = 4, q could be any even integer >4.

so A.
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03 Aug 2006, 18:35
sorry...maybe I should reword...I was just wondering about the wording. I was lead to believe that you should only trust that you are dealing with whole numbers in the presence of the word "integer." The word "number" threw me off here, not the thinking of the problem...
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03 Aug 2006, 18:40
microjohn wrote:
sorry...maybe I should reword...I was just wondering about the wording. I was lead to believe that you should only trust that you are dealing with whole numbers in the presence of the word "integer." The word "number" threw me off here, not the thinking of the problem...

even numbers = even integers = 0, 2, 4, 6, 8 and so on.

however they can be +ves or -ves.
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03 Aug 2006, 19:37
microjohn wrote:
sorry...maybe I should reword...I was just wondering about the wording. I was lead to believe that you should only trust that you are dealing with whole numbers in the presence of the word "integer." The word "number" threw me off here, not the thinking of the problem...

No non-integer is even. Question says even numbers.
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03 Aug 2006, 20:10
OK, that was my mistake and I will not forget for my test. I thought the trick was that maybe p, q, and r could be something like 9.6, 9.4, 9.2, etc.

So only integers (numbers) can have the tags "odd" or "even."

Thanks...

Hope they say "integers" on the test though!!!
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03 Aug 2006, 21:38
Give p>2

1) r < 10
Hence r can be 8,6,4
when r is 4 then q is 2, not possible as q>p>2
when r is 6, then q is 4 and p is 2, again not possible as p>2
when r is 8, q is 6 and p is 4 . Only thing that satisfies.
hence q = 6

2) p<6 and p>2

hence p =4

But q can be any even integer greater than 4. Not suff.

Hence A
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03 Aug 2006, 23:20

2<p<q<r

The lowest 3 values for p,q,r are 4,6,8

From 1:
r<10
Therefore r can only be 8. We can find p,q,r. SUFF

From 2:

p<6. Since p!=2 we know that p must be 4.However this places no bounds on q. INSUFF

Ans: A
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05 Aug 2006, 22:04
p,q,r={I,E}

(1) 2<4<6<8
q=6
(2) 2<4<q<r

q can be any even number>4 depending on r

Hence A

Heman
05 Aug 2006, 22:04
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