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# Is r^2 + s^2 = 0? (1) r + s = 0 (2) 3r^2 + 4s^2 = 0 17. If x

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Is r^2 + s^2 = 0? (1) r + s = 0 (2) 3r^2 + 4s^2 = 0 17. If x [#permalink]

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02 Aug 2007, 10:01
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

15. Is r^2 + s^2 = 0?
(1) r + s = 0
(2) 3r^2 + 4s^2 = 0

17. If x and y are positive integers and y = root (9-x) , what is the value of y?
(1) x <8> 1
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Re: Equations [#permalink]

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02 Aug 2007, 12:48
jet1445 wrote:
15. Is r^2 + s^2 = 0?
(1) r + s = 0
(2) 3r^2 + 4s^2 = 0

17. If x and y are positive integers and y = root (9-x) , what is the value of y?
(1) x <8> 1

Don't understand your #17, please repost.

For #15. It is B.
For r^2 + s^2 to be equal to 0, both must equal to zero. You can tell because both are positive numbers and if you add positive and positive and equal to zero, both must be 0.

(1) INSUFFICIENT. r=-4, s=4 works as well as both equal to 0.
(2) Same reason as above. SUFFICIENT.

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02 Aug 2007, 15:30
Yeah, I'd like it if you can post the other option.

based on what I see, it appears if 1 is sufficient, if you mean 1 is , x >1 and x <8> +/- 2 and since Y is said to be positive int, y value is 2.
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02 Aug 2007, 21:23
Q1:
St1:
Case1: If r = 1/2, s = 1/2, then r^2+s^2 != 0
Case2: If r = 0, s = 0, then r^2+s^2 = 0.
Insufficient.

St2:
Case1: r=0, s=0, then r^2+s^2 = 0
Case2:If r = sqrt(3) and s = sqrt(4), then 3r^2 + 4s^2 = 0 but r+s !=0. Insufficient.

Using both st1 and st2:
Must be both r and s = 0. Sufficient.

Ans C

Q2:
Post Q2 again with HTML disabled.

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02 Aug 2007, 21:23
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# Is r^2 + s^2 = 0? (1) r + s = 0 (2) 3r^2 + 4s^2 = 0 17. If x

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