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If rs 0, does 1/r + 1/s = 1 (1) rs = 1 (2) s + r = 2.5 Did

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If rs 0, does 1/r + 1/s = 1 (1) rs = 1 (2) s + r = 2.5 Did [#permalink]

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12 Dec 2010, 09:05
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95% (hard)

Question Stats:

32% (02:04) correct 68% (01:24) wrong based on 148 sessions

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If rs ≠ 0, does 1/r + 1/s = 1

(1) rs = 1
(2) s + r = 2.5

Did not understand the OA
[Reveal] Spoiler: OA

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12 Dec 2010, 09:25
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rxs0005 wrote:
If rs ≠ 0, does 1/r + 1/s = 1

(1) rs = 1
(2) s + r = 2.5

Did not understand the OA

Is $$\frac{1}{r}+\frac{1}{s}=1$$?

(1) rs = 1 --> $$s=\frac{1}{r}$$ --> question becomes: is $$\frac{1}{r}+r=1$$ --> is $$r^2-r+1=0$$, no real $$r$$ satisfies this equation (equation has no real roots), so the answer is NO. Sufficient.
(2) s + r = 2.5 --> $$s=2.5-r$$ --> question becomes: is $$\frac{1}{r}+\frac{1}{2.5-r}=1$$ --> is $$2r^2-5r+5=0$$, no real $$r$$ satisfies this equation (equation has no real roots), so the answer is NO. Sufficient.

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14 Dec 2010, 04:28
Nice one... Certainly tricky

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27 Dec 2010, 03:14
Bunuel wrote:
rxs0005 wrote:
If rs ≠ 0, does 1/r + 1/s = 1

(1) rs = 1
(2) s + r = 2.5

Did not understand the OA

Is $$\frac{1}{r}+\frac{1}{s}=1$$?

(1) rs = 1 --> $$s=\frac{1}{r}$$ --> question becomes: is $$\frac{1}{r}+r=1$$ --> is $$r^2-r+1=0$$, no real $$r$$ satisfies this equation (equation has no real roots), so the answer is NO. Sufficient.
(2) s + r = 2.5 --> $$s=2.5-r$$ --> question becomes: is $$\frac{1}{r}+\frac{1}{2.5-r}=1$$ --> is $$2r^2-5r+5=0$$, no real $$r$$ satisfies this equation (equation has no real roots), so the answer is NO. Sufficient.

A simpler approach would be:
Question is whether -> 1/r + 1/s = 1 or r+s/rs=1....(eq.1)

Stmt 1 -> rs=1 , substituting in (eq.1)

r+s=1 or 1/r+1/s = 1 -> Sufficient

Stmt 1 -> s+r=2.5 , substituting in (eq.1)

2.5/rs=1 => rs=2.5

or 1/r+1/s = 1 -> Sufficient

Ans - D
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27 Dec 2010, 03:21
oldstudent wrote:
Bunuel wrote:
rxs0005 wrote:
If rs ≠ 0, does 1/r + 1/s = 1

(1) rs = 1
(2) s + r = 2.5

Did not understand the OA

Is $$\frac{1}{r}+\frac{1}{s}=1$$?

(1) rs = 1 --> $$s=\frac{1}{r}$$ --> question becomes: is $$\frac{1}{r}+r=1$$ --> is $$r^2-r+1=0$$, no real $$r$$ satisfies this equation (equation has no real roots), so the answer is NO. Sufficient.
(2) s + r = 2.5 --> $$s=2.5-r$$ --> question becomes: is $$\frac{1}{r}+\frac{1}{2.5-r}=1$$ --> is $$2r^2-5r+5=0$$, no real $$r$$ satisfies this equation (equation has no real roots), so the answer is NO. Sufficient.

A simpler approach would be:
Question is whether -> 1/r + 1/s = 1 or r+s/rs=1....(eq.1)

Stmt 1 -> rs=1 , substituting in (eq.1)

r+s=1 or 1/r+1/s = 1 -> Sufficient

Stmt 1 -> s+r=2.5 , substituting in (eq.1)

2.5/rs=1 => rs=2.5

or 1/r+1/s = 1 -> Sufficient

Ans - D

Answer to the question is indeed D, but I wonder what do the red parts mean in your simpler solution?
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Re: If rs 0, does 1/r + 1/s = 1 (1) rs = 1 (2) s + r = 2.5 Did [#permalink]

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09 Sep 2016, 05:21
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Re: If rs 0, does 1/r + 1/s = 1 (1) rs = 1 (2) s + r = 2.5 Did [#permalink]

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20 Feb 2017, 23:04
Bunuel wrote:
rxs0005 wrote:
If rs ≠ 0, does 1/r + 1/s = 1

(1) rs = 1
(2) s + r = 2.5

Did not understand the OA

Is $$\frac{1}{r}+\frac{1}{s}=1$$?

(1) rs = 1 --> $$s=\frac{1}{r}$$ --> question becomes: is $$\frac{1}{r}+r=1$$ --> is $$r^2-r+1=0$$, no real $$r$$ satisfies this equation (equation has no real roots), so the answer is NO. Sufficient.
(2) s + r = 2.5 --> $$s=2.5-r$$ --> question becomes: is $$\frac{1}{r}+\frac{1}{2.5-r}=1$$ --> is $$2r^2-5r+5=0$$, no real $$r$$ satisfies this equation (equation has no real roots), so the answer is NO. Sufficient.

In view of the above question, I have a question in mind. Can multiple of two numbers be equal to the sum of those two numbers in any way?
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Hasan Mahmud

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Re: If rs 0, does 1/r + 1/s = 1 (1) rs = 1 (2) s + r = 2.5 Did [#permalink]

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20 Feb 2017, 23:39
Mahmud6 wrote:
Bunuel wrote:
rxs0005 wrote:
If rs ≠ 0, does 1/r + 1/s = 1

(1) rs = 1
(2) s + r = 2.5

Did not understand the OA

Is $$\frac{1}{r}+\frac{1}{s}=1$$?

(1) rs = 1 --> $$s=\frac{1}{r}$$ --> question becomes: is $$\frac{1}{r}+r=1$$ --> is $$r^2-r+1=0$$, no real $$r$$ satisfies this equation (equation has no real roots), so the answer is NO. Sufficient.
(2) s + r = 2.5 --> $$s=2.5-r$$ --> question becomes: is $$\frac{1}{r}+\frac{1}{2.5-r}=1$$ --> is $$2r^2-5r+5=0$$, no real $$r$$ satisfies this equation (equation has no real roots), so the answer is NO. Sufficient.

In view of the above question, I have a question in mind. Can multiple of two numbers be equal to the sum of those two numbers in any way?

Yes. There are infinite solutions for non-integers.

For integers:

0 + 0 = 0*0
2 + 2 = 2*2.
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Re: If rs 0, does 1/r + 1/s = 1 (1) rs = 1 (2) s + r = 2.5 Did   [#permalink] 20 Feb 2017, 23:39
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