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805+ (Hard)|   Algebra|      
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rxs0005
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If rs 0, does 1/r + 1/s = 1 (1) rs = 1 (2) s + r = 2.5 Did 12 Dec 2010, 09:05
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

31% (02:21) correct 69% (02:11) wrong based on 164 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
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D
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Bunuel
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If rs 0, does 1/r + 1/s = 1 (1) rs = 1 (2) s + r = 2.5 Did 12 Dec 2010, 09:25
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rxs0005
If rs ≠ 0, does 1/r + 1/s = 1

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

Did not understand the OA
Show more

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.

Answer: D.
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prashantbacchewar
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If rs 0, does 1/r + 1/s = 1 (1) rs = 1 (2) s + r = 2.5 Did 14 Dec 2010, 04:28
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Nice one... Certainly tricky
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If rs 0, does 1/r + 1/s = 1 (1) rs = 1 (2) s + r = 2.5 Did 27 Dec 2010, 03:14
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Bunuel
rxs0005
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.

Answer: D.
Show more

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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If rs 0, does 1/r + 1/s = 1 (1) rs = 1 (2) s + r = 2.5 Did 27 Dec 2010, 03:21
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oldstudent
Bunuel
rxs0005
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.

Answer: D.

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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Answer to the question is indeed D, but I wonder what do the red parts mean in your simpler solution?
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If rs 0, does 1/r + 1/s = 1 (1) rs = 1 (2) s + r = 2.5 Did 20 Feb 2017, 23:04
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Bunuel
rxs0005
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.

Answer: D.
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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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If rs 0, does 1/r + 1/s = 1 (1) rs = 1 (2) s + r = 2.5 Did 20 Feb 2017, 23:39
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Mahmud6
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rxs0005
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.

Answer: D.

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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Yes. There are infinite solutions for non-integers.

For integers:

0 + 0 = 0*0
2 + 2 = 2*2.
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If rs 0, does 1/r + 1/s = 1 (1) rs = 1 (2) s + r = 2.5 Did 13 Jan 2026, 00:18
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