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705-805 (Hard)|   Algebra|   Inequalities|      
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If rs ≠ 0, does 1/r + 1/s = 5 ? 02 Oct 2017, 04:45
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40% (02:04) correct 60% (02:02) wrong based on 67 sessions

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If rs ≠ 0, does 1/r + 1/s = 5 ?

(1) rs > 1
(2) s < –r
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C
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If rs ≠ 0, does 1/r + 1/s = 5 ? Updated on: 14 Oct 2017, 20:30
Originally posted by niks18 on 02 Oct 2017, 04:54.
Last edited by niks18 on 14 Oct 2017, 20:30, edited 3 times in total.
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Bunuel
If rs ≠ 0, does 1/r + 1/s = 5 ?

(1) rs > 1
(2) s < –r
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\(\frac{1}{r}+\frac{1}{s}=5\) or \(\frac{(r+s)}{rs}=5\). we need to find the values of r & s to verify the equation.

Statement 1: from this we know \(rs>1\) but we cannot find the value of r & s. Hence Insufficient.

Statement 2: \(s<-r\) or \(r+s<0\). again value of \(r\) & \(s\) cannot be calculated. Insufficient

Combining 1 & 2 we know that \(rs>0\) and \(r+s<0\) hence \(\frac{(r+s)}{rs}\) will be negative so it cannot be equal to 5. Hence sufficient

Option C
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If rs ≠ 0, does 1/r + 1/s = 5 ? 02 Oct 2017, 22:39
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niks18
Bunuel
If rs ≠ 0, does 1/r + 1/s = 5 ?

(1) rs > 1
(2) s < –r

\(\frac{1}{r}+\frac{1}{s}=5\) is only possible when both \(r\) and \(s\) are fractions for eg \(r=1\) and \(s=\frac{1}{4}\) or \(r=\frac{1}{2}\) and \(s=\frac{1}{3}\), if \(r\) & \(s\) are integers then LHS will has a value less than \(1\).

Statement 1: from this we know \(rs>1\) this implies that r and s are not a proper fraction because the multiplication is greater than 1. Hence LHS cannot be equal to RHS. Sufficient.

Statement 2: \(s<-r\) or \(r+s<0\). again value of \(r\) & \(s\) cannot be calculated. Insufficient

Option A
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both r and s not necessarily be fractions or integers.
eg r=5 and s= 5/24
then rs= 25/24 ----> rs>1
and 1/5 +1/(5/24) = 1/5 + 24/5 = 5

Hence Statement 1 is not sufficient.

stmt 2 says s< -r
that means s+r <0
stmt 2 also gives nothing hence insufficient.

Combining stmt 1 and 2 we get rs>1 means rs is positive, and r+s<0 means r+s is negative
1/r + 1/s
=(r+s)/rs
=negative / positive
so in any case 1/r + 1/s will not be equal to 5.
Hence Both statements together are sufficient.

Answer is Option C.


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If rs ≠ 0, does 1/r + 1/s = 5 ? 03 Oct 2017, 02:07
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pushkarajnjadhav
niks18
Bunuel
If rs ≠ 0, does 1/r + 1/s = 5 ?

(1) rs > 1
(2) s < –r

\(\frac{1}{r}+\frac{1}{s}=5\) is only possible when both \(r\) and \(s\) are fractions for eg \(r=1\) and \(s=\frac{1}{4}\) or \(r=\frac{1}{2}\) and \(s=\frac{1}{3}\), if \(r\) & \(s\) are integers then LHS will has a value less than \(1\).

Statement 1: from this we know \(rs>1\) this implies that r and s are not a proper fraction because the multiplication is greater than 1. Hence LHS cannot be equal to RHS. Sufficient.

Statement 2: \(s<-r\) or \(r+s<0\). again value of \(r\) & \(s\) cannot be calculated. Insufficient

Option A


both r and s not necessarily be fractions or integers.
eg r=5 and s= 5/24
then rs= 25/24 ----> rs>1
and 1/5 +1/(5/24) = 1/5 + 24/5 = 5

Hence Statement 1 is not sufficient.

stmt 2 says s< -r
that means s+r <0
stmt 2 also gives nothing hence insufficient.

Combining stmt 1 and 2 we get rs>1 means rs is positive, and r+s<0 means r+s is negative
1/r + 1/s
=(r+s)/rs
=negative / positive
so in any case 1/r + 1/s will not be equal to 5.
Hence Both statements together are sufficient.

Answer is Option C.


Kudos if it helps.
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Hi pushkarajnjadhav,

thanks for highlighting the mistake :-) . initially i had solved correct but then got swayed by the fraction :? Great work in identifying that unique fraction 5/24 :thumbup:
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If rs ≠ 0, does 1/r + 1/s = 5 ? 15 Jul 2022, 08:04
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