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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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New post 12 Dec 2010, 08:05
2
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A
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D
E

Difficulty:

  95% (hard)

Question Stats:

31% (02:30) correct 69% (02:12) wrong based on 200 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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Re: DS Algebra  [#permalink]

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New post 12 Dec 2010, 08:25
2
4
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.

Answer: D.
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Re: DS Algebra  [#permalink]

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New post 14 Dec 2010, 03:28
Nice one... Certainly tricky
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Re: DS Algebra  [#permalink]

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New post 27 Dec 2010, 02:14
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.

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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Re: DS Algebra  [#permalink]

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New post 27 Dec 2010, 02:21
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.

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


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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New post 20 Feb 2017, 22:04
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.

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

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New post 20 Feb 2017, 22:39
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.

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?


Yes. There are infinite solutions for non-integers.

For integers:

0 + 0 = 0*0
2 + 2 = 2*2.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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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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Re: If rs 0, does 1/r + 1/s = 1 (1) rs = 1 (2) s + r = 2.5 Did &nbs [#permalink] 30 May 2018, 19:39
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