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If a and b are real numbers, is a < b?

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If a and b are real numbers, is a < b?  [#permalink]

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New post 17 Jul 2018, 19:35
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A
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E

Difficulty:

  95% (hard)

Question Stats:

31% (02:50) correct 69% (02:01) wrong based on 45 sessions

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If a and b are real numbers, is a < b?


(1) \(a^b < b^a\)

(2) \(\frac{a}{b} > 1\)
Math Expert
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Joined: 02 Aug 2009
Posts: 7764
If a and b are real numbers, is a < b?  [#permalink]

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New post 17 Jul 2018, 21:08
1
If a and b are real numbers, is a < b?


(1) \(a^b < b^a\)..
a) let a be 4 and b be 3......a>b
\(a^b<b^a........4^3<3^4........64<81\)
b) let a be -2 and b be -1......a<b
\((-2)^{-1}<(-1)^{-2}...........-1/2<1\)
So insufficient

(2) \(\frac{a}{b} > 1\)
This means |a|>|b|...
a=4, and b=3..........4/3>1 and a>b
a=-4, and b=-1..........(-4)/(-1)=4>1 and a<b
So insufficient

Combined
Both cases remain..
a=4, and b=3......a>b
a=-2, and b =-1.....a<b
Insufficient

E
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Re: If a and b are real numbers, is a < b?  [#permalink]

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New post 17 Jul 2018, 21:45
Karthik200 wrote:
If a and b are real numbers, is a < b?


(1) \(a^b < b^a\)

(2) \(\frac{a}{b} > 1\)



(1) \(a^b < b^a\)
\(a=2\), \(b=3\) \(a<b\)
\(8<9\) True

\(a=-3\), \(b=-2\) \(a<b\)
\(9<\frac{-1}{8}\) False

Insufficient.

(2) \(\frac{a}{b} > 1\)

\(\frac{a}{b} > 1\)
\(a>b\)
However
\(\frac{-a}{-b} >1\)
\(-a<-b\)
Eg: \(a=+-5\), \(b=+-4\)

Insufficient.

(1) and (2) Insufficient.

Hence, E
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Status: MBA Student
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If a and b are real numbers, is a < b?  [#permalink]

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New post 18 Jul 2018, 02:17
chetan2u wrote:
If a and b are real numbers, is a < b?


(1) \(a^b < b^a\)..
a) let a be 4 and b be 3......a>b
\(a^b<b^a........4^3<3^4........64<81\)
b) let a be -2 and b be -1......a<b
\((-2)^{-1}<(-1)^{-2}...........-1/2<1\)
So insufficient

(2) \(\frac{a}{b} > 1\)
This means |a|>|b|...
a=4, and b=3..........4/3>1 and a>b
a=-4, and b=-1..........(-4)/(-1)=4>1 and a<b
So insufficient

Combined
Both cases remain..
a=4, and b=3......a>b
a=-2, and b =-1.....a<b
Insufficient

E


Hi chetan2u,

In these kinds of problems where plugging in of numbers required, I am finding it difficult to try out all the possibilities to disprove the given statements under two minutes. Is there any strategy to plug in the smart numbers that work best, especially for number property DS questions? I hope it will benefit many others as well.

Thanks & Regards,
Karthik
Math Expert
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Joined: 02 Aug 2009
Posts: 7764
Re: If a and b are real numbers, is a < b?  [#permalink]

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New post 18 Jul 2018, 02:26
Karthik200 wrote:
chetan2u wrote:
If a and b are real numbers, is a < b?


(1) \(a^b < b^a\)..
a) let a be 4 and b be 3......a>b
\(a^b<b^a........4^3<3^4........64<81\)
b) let a be -2 and b be -1......a<b
\((-2)^{-1}<(-1)^{-2}...........-1/2<1\)
So insufficient

(2) \(\frac{a}{b} > 1\)
This means |a|>|b|...
a=4, and b=3..........4/3>1 and a>b
a=-4, and b=-1..........(-4)/(-1)=4>1 and a<b
So insufficient

Combined
Both cases remain..
a=4, and b=3......a>b
a=-2, and b =-1.....a<b
Insufficient

E


Hi chetan2u,

In these kinds of problems where plugging in of numbers required, I am finding it difficult to try out all the possibilities to disprove the given statements under two minutes. Is there any strategy to plug in the smart numbers that work best, especially for number property DS questions? I hope it will benefit many others as well.

Thanks & Regards,
Karthik


If I were to do this problem, I'll pick up the statement II as it is more friendly...
a/b>1 means a and b are of same sign and numeric value of a is more than that of b.. hence |a|>|b|
And |a|>|b| means a>b if both positive and a<b if both negative

So check for the same cases in statement I too..
Both negative and a<b.....-4<-3....(-4)^{-3}<(-3)^{-4}.....-1/64<1/81
Both positive and a>b......4>3.....4^3<3^4.........64<81
So insufficient
_________________
Intern
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Status: MBA Student
Joined: 20 Nov 2017
Posts: 30
Location: India
Concentration: Strategy, Marketing
GPA: 3.9
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Re: If a and b are real numbers, is a < b?  [#permalink]

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New post 18 Jul 2018, 02:42
chetan2u wrote:
Karthik200 wrote:
chetan2u wrote:
If a and b are real numbers, is a < b?


(1) \(a^b < b^a\)..
a) let a be 4 and b be 3......a>b
\(a^b<b^a........4^3<3^4........64<81\)
b) let a be -2 and b be -1......a<b
\((-2)^{-1}<(-1)^{-2}...........-1/2<1\)
So insufficient

(2) \(\frac{a}{b} > 1\)
This means |a|>|b|...
a=4, and b=3..........4/3>1 and a>b
a=-4, and b=-1..........(-4)/(-1)=4>1 and a<b
So insufficient

Combined
Both cases remain..
a=4, and b=3......a>b
a=-2, and b =-1.....a<b
Insufficient

E


Hi chetan2u,

In these kinds of problems where plugging in of numbers required, I am finding it difficult to try out all the possibilities to disprove the given statements under two minutes. Is there any strategy to plug in the smart numbers that work best, especially for number property DS questions? I hope it will benefit many others as well.

Thanks & Regards,
Karthik


If I were to do this problem, I'll pick up the statement II as it is more friendly...
a/b>1 means a and b are of same sign and numeric value of a is more than that of b.. hence |a|>|b|
And |a|>|b| means a>b if both positive and a<b if both negative

So check for the same cases in statement I too..
Both negative and a<b.....-4<-3....(-4)^{-3}<(-3)^{-4}.....-1/64<1/81
Both positive and a>b......4>3.....4^3<3^4.........64<81
So insufficient


Thanks chetan2u. I will try to apply this from my next DS problem onwards.
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Re: If a and b are real numbers, is a < b?   [#permalink] 18 Jul 2018, 02:42
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