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# Is |a| + |b| > |a + b| ?

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Is |a| + |b| > |a + b| ? [#permalink]

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28 Nov 2010, 07:05
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Is |a| + |b| > |a + b| ?

(1) a^2 > b^2
(2) |a| * b < 0
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Dec 2012, 01:57, edited 1 time in total.
Renamed the topic and edited the question.

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28 Nov 2010, 07:48
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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.

Regards,
Murali.
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28 Nov 2010, 08:13
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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.

You should notice that inequality $$|a|+|b|>|a+b|$$ holds true if and only $$a$$ and $$b$$ have opposite sign, as only in this case absolute value of positive+negative will be less than |positive|+|negative|. For example |-2|+|3|>|-2+3|. In all other cases $$|a|+|b|=|a+b|$$.

So, basically the question is whether $$a$$ and $$b$$ have opposite sign.

(1) a^2 > b^2 --> can not determine whether $$a$$ and $$b$$ have opposite sign. Not sufficient.
(2) |a|*b<0 --> just tells us that $$b<0$$, but we don't know the sign of $$a$$. Not sufficient.

(1)+(2) $$b<0$$ and $$a^2>b^2$$ --> still can not get the sign of $$a$$. Not sufficient.

For theory check About Value chapter of Math Book created by Walker: math-absolute-value-modulus-86462.html

Absolute value PS questions: search.php?search_id=tag&tag_id=58
Absolute value DS questions: search.php?search_id=tag&tag_id=37

Hard absolute value questions: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.
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16 Jun 2011, 00:32
clean E here.
b < 0 but a can be < > 0
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24 Aug 2011, 20:09
|a| + |b| > |a+b|?

if a and b are of same sign then LHS = RHS. Hence No.
if a and b are of opposite sign then LHS > RHS . Then Yes.

1. Not sufficient

|a|>|b|

we cannot say anything about the sign of a and b.

2. Not sufficient

b<0, but we dont know the sign of a.

together,
|a|>|b| and b<0 => a can be positive or negative
Hence not sufficient.

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30 Aug 2011, 02:08
|a| + |b| > |a + b|

the expression will be true if "a,b have opposite signs e.g. a<0, b>0 etc"

st1) insufficient. a or b could have same or different sign
st2) this proves that b<0. but it is also possible that a<0. so, insufficient.

E is the ans
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04 Dec 2012, 01:54
Is |a| + |b| > |a + b| ?
(1) Both positive |5| + |5| > |5 + 5| NO!
(2) Both negative |-5| + |-5| > |-10| NO!
(3) Oppositve signs |5| + |-5| > |5-5| YES!

(1) a2 > b2
|a| > |b|
This doesn't tell us anything about the sign. INSUFFICIENT.

(2) |a| × b < 0
This tells us b is negative. INSUFFICIENT.

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06 Jun 2013, 08:03
$$Is |a| + |b| > |a + b| ?$$

$$(1) a^2>b^2$$

($$2)|a| * b<0$$

got me confusing for a while and choose E.

used the std rule of | A + B | < |A| + | B | when XY > 0

Since it is difficult to find the sign of A in any case selected E.

Please let me know if the approach to this problem is correct ?
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Re: Ds Modules and inequalities [#permalink]

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06 Jun 2013, 09:43
Normally I solve it by plugging numbers.

1. a^2>b^2
lets take a=-3 and b=-2; so, |a|+|b|=5 & |a+b|=5
Lets take a=-3 and b=2; so |a|+|b| = 5 & |a+b|=1
two different values, hence insufficient.

2. |a|*b<0
This implies b<0.
Lets take a=-3 and b=-2; so |a| + |b| = 5 & |a+b|=5
Lets take a=3 and b=-2; so |a| + |b| = 5 & |a+b|=1
two different values, hence insufficient.

Taking both together:
we can use the same numbers in statement 2. Hence, insufficient.

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Re: Ds Modules and inequalities [#permalink]

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06 Jun 2013, 10:01
Manhnip wrote:
$$Is |a| + |b| > |a + b| ?$$

$$(1) a^2>b^2$$

($$2)|a| * b<0$$

got me confusing for a while and choose E.

used the std rule of | A + B | < |A| + | B | when XY > 0

Since it is difficult to find the sign of A in any case selected E.

Please let me know if the approach to this problem is correct ?

Merging similar topics.

Look at the original post here: is-a-b-a-b-105457.html#p824216 The name of the topics MUST be "Is |a| + |b| > |a + b| ?"
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Re: Is |a| + |b| > |a + b| ? [#permalink]

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30 Jun 2013, 12:46
Is |a| + |b| > |a + b| ?

Is a^2 + b^2 > (a+b)^2?
is a^2 + b^2 > a^2 + 2ab + b^2?
Is 0 > 2ab?
Is ab negative?

(1) a^2 > b^2

This tells us that a^2 is greater than b^2 but it tells us nothing about the signs of a and b. For example, (-3)^2 > (2)^2 but -3 is less than 2. In this case, ab would be negative. On the other hand, (3)^2 > (2)^2 in which case 3 > 2 and ab would be positive.
INSUFFICIENT

(2) |a| * b < 0

This tells us that b must be negative as |a| will always be positive. however, a could be negative in which case ab would be positive or a could be negative in which case ab would be negative.
INSUFFICIENT

1+2) a^2 > b^2 and b is negative: that means the absolute value of |a| > |b|.

|-4| > |-1| ===> 4>1 Valid ===> ab = (-4)*(-1) = 4 (positive)
|4| > |-1| ===> 4>1 Valid ===> ab = (4)*(-1) = -4 (negative)

In other words, ab could be either positive or negative.
INSUFFICIENT

(E)

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Re: Is |a| + |b| > |a + b| ? [#permalink]

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22 Jul 2013, 13:41
Is |a| + |b| > |a + b| ?

|4| + |2| > |4 + 2| 6>6 Invalid

|-4| + |-2| > |-4 + -2| 6>6 Invalid

|-4| + |2| > |-4 + 2| 6>2 Valid

|4| + |-2| > |4 + -2| 6> 2 Valid

(Take note that the only two valid cases are when a and b have opposite signs)

The question then becomes, do a and b have opposite signs?

(1) a^2 > b^2
|a| > |b| but we don't know the signs of either.
INSUFFICIENT

(2) |a| * b < 0
For this to hold true, regardless of what a is (any number but zero) b must be negative. Like #1, this tells us nothing about the sign of a.
INSUFFICIENT

1+2) a^2 > b^2, |a| * b < 0
we know that b is negative but a could be positive or negative and a^2 > b^2 could still hold true. For example:

-6^2 > -5^2
36>25
-a, -b

6^2 > -5^2
36 > 25
a, -b

INSUFFICIENT

(E)

There is a similar problem in which you plug in numbers to solve (is-xy-0-1-x-3-y-5-x-y-2-0-2-x-y-x-y-86780.html) but when I try to solve it, it becomes a confusing mess. Could someone explain to me where I went wrong?

So, valid when:
|-a| > |b|
|a| > |-b|

(1) a^2 > b^2
The absolute value of a must be greater than the absolute value of b. For #1:
|a| > |b|
|-a| > |b|
|-a| > |-b|

Therefore, according to #1, the values of a and b could make the inequality |4| + |2| > |4 + 2| invalid (for example, if |a|>|b| or valid if |-a| > |b|)
INSUFFICIENT

(2) |a| * b < 0
|a|*b < 0
This is saying [positive or zero] * [b] < 0. If a is positive, b is negative. a and b cannot = 0. b is always negative regardless of what a is as |a| will always be positive or zero.

|-a| * (-b) < 0
|a| * (-b) < 0

The problem is valid when |-a| > |-b|. We don't know if |-a| > |b| as is stated in the stem. For example, |-1| * (-4) < 0 but |-1| is not greater than |-4|
INSUFFICIENT

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Re: Is |a| + |b| > |a + b| ? [#permalink]

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07 Apr 2014, 05:55
From question we need only to know if ab<0.

Statement 1 tells us that abs A > abs B, which is clearly insufficient.

Statement 2 tells us that b<0.

From both statements we can't tell whether a is negative.

Therefore E is the correct answer choice

Hope this helps
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Is |a| + |b| > |a + b| ? Manhattan Inquality [#permalink]

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14 Dec 2014, 05:15
Is |a| + |b| > |a + b| ?

(1) a2 > b2

(2) |a| × b < 0

I got this from one of the closed group, thought to share for the advantage of vast global aspirant.

N.B. a2 = a X a = a square
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Re: Is |a| + |b| > |a + b| ? Manhattan Inquality [#permalink]

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14 Dec 2014, 05:30
honchos wrote:
Is |a| + |b| > |a + b| ?

(1) a2 > b2

(2) |a| × b < 0

I got this from one of the closed group, thought to share for the advantage of vast global aspirant.

N.B. a2 = a X a = a square

st.1

a^2>b^2

if a=3, b=1, then answer to the question is no.
if a=-3,b=1, then answer to the question is yes

hence insufficient

st.2
|a| × b < 0
from here we can conclude that b<0 as |a| will always be positive. but here a can be both positive as well as negative.

if a=4, b=-2 then answer to the question is yes.
if a=-4 ,b=-1 then answer to the question is no.

st.1 and st.2

we know for sure than b<0, but a can still take both positive as well as negative values. hence both yes and no. answers are possible. therefore answer must be E.

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Re: Is |a| + |b| > |a + b| ? Manhattan Inquality [#permalink]

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14 Dec 2014, 17:46
The question 'Is |a| + |b| > |a + b| ?' is basically asking if a and b have opposite signs.

(1) Insufficient. Says nothing about the signs of a and b.
(2) Insufficient. Just tells that b is negative, but we still don't know about the sign of a.

(1)+(2) Insufficient. We still don't know about the sign of a.

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Re: Is |a| + |b| > |a + b| ? [#permalink]

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15 Dec 2014, 08:08
honchos wrote:
Is |a| + |b| > |a + b| ?

(1) a2 > b2

(2) |a| × b < 0

I got this from one of the closed group, thought to share for the advantage of vast global aspirant.

N.B. a2 = a X a = a square

Merging topics. Please refer to the discussion above.
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Re: Is |a| + |b| > |a + b| ? [#permalink]

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Re: Is |a| + |b| > |a + b| ? [#permalink]

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02 Jun 2016, 13:07
chiragatara wrote:
Is |a| + |b| > |a + b| ?

(1) a^2 > b^2
(2) |a| * b < 0

|a| + |b| will have both a and b +ve

|a + b| will have positive sum of a and b

But we don't know signs of both a and b

If both the signs are +ve then |a| + |b| = |a + b|
If both the signs are -ve then |a| + |b| = |a + b|
If both have opposite signs then |a| + |b| > |a + b|

(1) a^2 > b^2

It doesn't tell us anything about signs of a or b

(2) |a| * b < 0
|a| will be +ve, hence b is -ve

but a can be +ve or -ve. We don't get the actual sign.

Combining both statements is not giving us answer

(-6)^2> -5^2
6^2> -5^2
Not sufficient

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Re: Is |a| + |b| > |a + b| ? [#permalink]

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12 Jul 2017, 04:09
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Re: Is |a| + |b| > |a + b| ?   [#permalink] 12 Jul 2017, 04:09
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