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Manhnip
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If ab 0, is ((a - b)/(a^(-1) - b^(-1)))^(-1) > a + b ? Updated on: 30 Sep 2022, 06:22
Originally posted by Manhnip on 06 Jun 2013, 11:21.
Last edited by Bunuel on 30 Sep 2022, 06:22, edited 2 times in total.
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85% (hard)

Question Stats:

51% (02:48) correct 49% (02:46) wrong based on 182 sessions

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If ab ≠ 0, is \((\frac{a - b}{a^{-1} - b^{-1}})^{-1} > a + b\) ?

(1) |a| > |b|

(2) a < b
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Bunuel
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If ab 0, is ((a - b)/(a^(-1) - b^(-1)))^(-1) > a + b ? 10 Jun 2013, 10:54
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WholeLottaLove
How did you simplify the inequality like you did?

Thanks!

arpanpatnaik
Manhnip
If ab≠0, is \(\frac{[ ( a - b )}{( a^{-1} - b^{-1})]}^{-1}> (a+b)\)

(1) |a| > |b|

(2) a < b

What is the answer and fast way to solve ?

Well my answer would be [E] for the question!

My approach would be to further reduce the expression \(\frac{( a - b )}{( a^{-1} - b^{-1})}^{-1}\) in such a manner:

\(\frac{( a - b )}{( a^{-1} - b^{-1})}^{-1} > (a+b)\) =>\(\frac{-1}{( a^{-1} + b^{-1})} > 1\)

For the above to be greater than 1, the denominator needs to be -ve and a fraction i.e between {-1,0}.
[b]Statement 1 gives away the difference in magnitude of a and b but doesn't tell anything about sign. Hence insufficient.
Statement 2[/b] tells us that a < b but then again we can not ascertain the sign of either numbers. Hence insufficient.

Combining then both, we realize that a is -ve, but then again b can be positive or negative. Hence we cannot be sure that 1/a + 1/b will be both negative and a fraction. Hence [E]. Not sure if this is the shortest way or not! Admins plz help out if there is a better solution to the question :)

Regards,
Arpan
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Above simplification is not correct. It should be:

    \([\frac{( a - b )}{( a^{-1} - b^{-1})}]^{-1}\) (notice that the last -1 is the exponent of the whole fraction);

    \(\frac{( a^{-1} - b^{-1})}{(a - b)}\);

    \(\frac{(\frac{1}{a}-\frac{1}{b})}{a-b}\);

    \(\frac{(\frac{b-a}{ab})}{a-b}\);

    \(-\frac{1}{ab}\).

So, the question asks whether \(-\frac{1}{ab}>a+b\).

If ab ≠ 0, is \((\frac{a - b}{a^{-1} - b^{-1}})^{-1} > a + b\) ?

As shown above the question asks:

    Is \(-\frac{1}{ab}>a+b\)?


(1) |a| > |b|:

The above means that a is further from 0 than b.
    If a = 2 and b = 1, then the answer is NO.
    If a = -2 and b = 1, then the answer is NO.

Not sufficient,

(2) a < b

    If a = 1 and b = 2, then the answer is NO.
    If a = -1 and b = 1, then the answer is YES.

Not sufficient,

(1)+(2) From (1) we have that a is further from 0 than b and from (2) we have that a < b. We can have the following cases:


----a--------0---b--
----a---b---0-----

    If a = -2 and b = 1, then the answer is YES.
    If a = -1 and b = -1/2, then the answer is NO.



Not sufficient.

Answer: E.
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If ab 0, is ((a - b)/(a^(-1) - b^(-1)))^(-1) > a + b ? 06 Jun 2013, 13:16
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Manhnip
If ab≠0, is \(\frac{[ ( a - b )}{( a^{-1} - b^{-1})]}^{-1}> (a+b)\)

(1) |a| > |b|

(2) a < b

What is the answer and fast way to solve ?
Show more

Well my answer would be [E] for the question!

My approach would be to further reduce the expression \(\frac{( a - b )}{( a^{-1} - b^{-1})}^{-1}\) in such a manner:

\(\frac{( a - b )}{( a^{-1} - b^{-1})}^{-1} > (a+b)\) =>\(\frac{-1}{( a^{-1} + b^{-1})} > 1\)

For the above to be greater than 1, the denominator needs to be -ve and a fraction i.e between {-1,0}.
[b]Statement 1 gives away the difference in magnitude of a and b but doesn't tell anything about sign. Hence insufficient.
Statement 2[/b] tells us that a < b but then again we can not ascertain the sign of either numbers. Hence insufficient.

Combining then both, we realize that a is -ve, but then again b can be positive or negative. Hence we cannot be sure that 1/a + 1/b will be both negative and a fraction. Hence [E]. Not sure if this is the shortest way or not! Admins plz help out if there is a better solution to the question :)

Regards,
Arpan
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If ab 0, is ((a - b)/(a^(-1) - b^(-1)))^(-1) > a + b ? 10 Jun 2013, 09:57
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How did you simplify the inequality like you did?

Thanks!

arpanpatnaik
Manhnip
If ab≠0, is \(\frac{[ ( a - b )}{( a^{-1} - b^{-1})]}^{-1}> (a+b)\)

(1) |a| > |b|

(2) a < b

What is the answer and fast way to solve ?

Well my answer would be [E] for the question!

My approach would be to further reduce the expression \(\frac{( a - b )}{( a^{-1} - b^{-1})}^{-1}\) in such a manner:

\(\frac{( a - b )}{( a^{-1} - b^{-1})}^{-1} > (a+b)\) =>\(\frac{-1}{( a^{-1} + b^{-1})} > 1\)

For the above to be greater than 1, the denominator needs to be -ve and a fraction i.e between {-1,0}.
[b]Statement 1 gives away the difference in magnitude of a and b but doesn't tell anything about sign. Hence insufficient.
Statement 2[/b] tells us that a < b but then again we can not ascertain the sign of either numbers. Hence insufficient.

Combining then both, we realize that a is -ve, but then again b can be positive or negative. Hence we cannot be sure that 1/a + 1/b will be both negative and a fraction. Hence [E]. Not sure if this is the shortest way or not! Admins plz help out if there is a better solution to the question :)

Regards,
Arpan
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If ab 0, is ((a - b)/(a^(-1) - b^(-1)))^(-1) > a + b ? 21 Jun 2024, 10:51
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