chiragatara wrote:
Hi Bunuel,
I am not very much comfortable with Absolute value Problems, e.g.
Is |a| + |b| > |a + b| ?
(1) a2 > b2
(2) |a| × b < 0
Is there any GYAN Material for improvement? Kindly help.
Hi, Don't worry. You will learn the things and GMAT absolute value tricks sooooon.
The be low is my approcah for any modulus qtn in GMAT.
Remember.
The meaning of |x-y| is "On the number line, the distance between X and +Y"
The meaning of |x+y| is "On the number line, the distance between X and -Y"
The meaning of |x| is "On the number line, the distance between X and 0".
On the # line, Left to 0 are all the -ve #s and right to 0 are all +ve #s
Original Qtn:
Is |a| + |b| > |a + b| ?
menas "is the SUM of the distance between a and 0, and the distance bewtween b and 0 > the distance bewtween a+b and 0"
let us take some cases to simplify the qtn.
case1: if a and b both are RIGHT to 0 on the # line then a+b will also be RIGHT to zero and even more far away from 0 than a or/and b is/are.
Then |a| + |b| > |a + b| is FLASE
and infact LHS is always = RHS
e.g: a = 1 and b = 3 then a+b = 4
a is 1 unit away right to 0
b is 3 units away right to
a+b is 4 units away right to 0
==>
the SUM of the distance between a and 0, and the distance between b and 0 = 1+3 = 4
and the distance b/w a+b and 0 is 4
-------------0----------b-----a---------a+b---
case2: if a and b both are LEFT to 0 on the # line then a+b will also be LEFT to zero and even more far away from 0 than a and/or b is.
Then |a| + |b| > |a + b| is FLASE
and infact LHS is always = RHS
e.g: a = -1 and b = -3 then a+b = -4
a is 1 unit away left to 0
b is 3 units away left to
a+b is 4 units away left to 0
==>
the SUM of the distance between a and 0, and the distance between b and 0 = 1+3 = 4
and the distance b/w a+b and 0 is 4
----a+b------a---------b-------0-------------
HENCE,
for |a| + |b| > |a + b| or
|a| + |b| < |a + b| or
|a| + |b| not= |a + b| to be true,
a and b, on the # line, should NOT be on the same side with respect to 0.
So BASICALLY the qtn is ASKING IF a AND b ARE on DEFFERENT SIDES, with respect to 0, on THE # LINE>
stmnt1: a^2 > b^2
from this we can say that the magnitude of a is > the magnitude of b. magnitude means with out sign.
==> it says that |a| > |b|
FROM THIS,
a and b can be on the same side on the # line with respect to 0
-------0------b----------------a , obviosly a is toofar ways from 0 than b is ==> |a| > |b|
OR,
a can be on a different side than b
-----a-------------0---b------ again, a is toofar ways from 0 than b is ==> |a| > |b|
so we have both the cases...hence stmnt 1 is NOT suff.
stmnt2: |a| × b < 0
|a| is always +ve hence b shud be -ve for the product to be -ve.
so all that stmnt 2 says is b is LEFT to 0 on the # line but it does not say whether a is on LEFT or RIGHT to 0. hence NOT suff.
stmnts 1&2 together
stmnt:1 says that a is too far from zero than b is.
stmnt2 says that b is LEFT to 0 on the # line
hence using both too, we can not say if a and b are on different sides to 0 on the #line or on the on the same sides
hence NOT suff.
ANSWER..."E"
Regards,
Murali.
KUDOS?