Is |a| |b| ? (1) a − b a^2 − b^2 (2) a b

Question

Is |a| > |b| ?

(1) a − b > a^2 − b^2

(2) a > b

chetan2u · Accepted Answer

Is |a| > |b| ? (1) a − b > a^2 − b^2 a − b > a^2 − b^2..........a − b -( a^2 − b^2)>0...............(a-b)(1-(a+b)>0 two cases # a-b>0 or a>b, 1-(a+b)>0 or 1>a+b, a=-2 and b=-3, here |a|<|b| OR a=1 and b=-0.5, here |a|>|b|...... # a-b<0 or a b 2>-3 or 3>2 give different answers clearly insuff combined.. one case still remains.. # a-b>0 or a>b, 1-(a+b)>0 or 1>a+b, a=-2 and b=-3, here |a|<|b| OR a=1 and b=-0.5, here |a|>|b|. insufficient E

KanishkM · Answer

Is |a|>|b|?
(1) a−b>a^2−b^2
(2) a>b

Statement (1)
when you solve this inequality,
a-b > a^2 - b^2
will give you
=> 0 > a^2 - b^2 - (a-b)
=> a-b (a+b-1) < 0

Now there can be 2 cases that a-b > 0 and a+b-1 < 0 or a-b < 0 and a+b-1 > 0

Satisfying the above will not give you a definite answer, as if you take Case 1, a > b and a+b < 1

the question |a|>|b| can be No when a = -1 and b = -2 and Yes when a = 1 and b =-7 -> %

Statement (2), just mentions a > b, which can take both the cases as mentioned in %

Combining both the statements will give the mentioned scenario in %

Making the answer as E.

DavidTutorexamPAL · Answer

The answer is E. 
 
we can solve this question by using numbers. 
1) 1) a−b>a^2−b^2: 
this can be true for a=2, b=1 - for which |2|>|1| - yes!
but it can also be true for a=1, b=-2, for which |1|<|-2| - no! 
two contradicting answers - insufficient!

2) a>b:
same example as before - this is true for both (2,1), for which the answer is yes, and (1,-2) - for which it is no - contradicting answers - insufficient!

Combined: since both examples ((2,1) & (1,-2)) hold for both statements - combining them adds nothing. we can still get either a "yes" or a "no" . insufficient!

kunalbean · Answer

Hi!  Bunuel 
Can you please let me know what is wrong with this approach :

Squaring the question i.e a^2 > b^2 = a^2 - b^2 > 0. 
Hence , a>b or a>-b.

Therefore, stmnt (2) will be sufficient.

Sayon · Answer

We need to find whether |a| > |b|

(1) a − b > a^2 − b^2
(a – b) > (a + b) (a – b)
1 > a + b

Let a = -5, b = -2
|-5| > |-2| so |a| > |b|

Let a = 2, b = -5
|2| < |-5| so |a| < |b|

Not Sufficient

(2) a > b

Let a = 2, b = -5
|2| < |-5| so |a| < |b|

Let a = 1, b= -0.5
|1| > |-0.5| so |a| > |b|

Not Sufficient

(1) + (2) -> [1 > a + b] and [a > b]
Let a = 2, b = -5
|2| < |-5| so |a| < |b|

Let a = 1, b= -0.5
|1| > |-0.5| so |a| > |b|

Not Sufficient

Answer E

kunalbean · Answer

Hello! Can we square the question? We will get (a+b)(a-b) > 0

MahmoudFawzy · Answer

yes we can because we are sure that both sides of the inequality are positive already.

so the question can be :
is  a^2>a^2 ?
is  a^2-b^2 > 0  ?

from statement (1), we don't know the value of a-b to decide -->  insufficient

from statement (2), a-b>0, so (a-b) is an unknown positive number.
(a,b) can be (2,-3) or (2,1)
but we don't know any thing about  a^2-b^2 
(is (a+b) positive ?? we can't know; (a,b) can be (2,-3) or (2,1) so produce different results) -->  insufficient

by combining (1),(2)
 a^2-b^2  can be a positive number smaller than the positive number (a-b),
or  a^2-b^2  can be a negative number ... we don't know -->  insufficient

kunalbean · Answer

Can you please let me know what is wrong with this approach :

Squaring the question i.e a^2 > b^2 = a^2 - b^2 > 0. = (a-b)(a+b) > 0
Hence , a>b or a>-b.

Therefore, stmnt (2) should be sufficient right?

MahmoudFawzy · Answer

if a>b, this never mean that a>-b

if a = 2 and b = -3, so a >b BUT a<-b

kunalbean · Answer

Correct.

So whenever, (a+b)(a-b) > 0 , how will we determine the values of a & b?

MahmoudFawzy · Answer

you don't have to.
the question is not asking you to determine the value,
the question is asking whether the info given is sufficient.
because the info is insufficient, we can't know the value. that's why we choose E.

kunalbean · Answer

Correct. Not in this question. But my objective here was to simplify the question so how can we simplify it further.

Jatin108 · Answer

hi sir, for statement A=2, B =1 is not a valid solution right ? then how can we disprove using that 
a-b>a^2-b^2 = 2-1> 4-1 -- this is not true

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Is |a| > |b| ? (1) a − b > a^2 − b^2 (2) a > b

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