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

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?

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.

Answer: E.

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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clean E here.
b < 0 but a can be < > 0

|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.

Answer is E.

|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

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.

Answer: E

\(Is |a| + |b| > |a + b| ?\)

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

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

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 ?

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.

Answer: E

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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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)

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

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
Cheers
J

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.

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.

Answer E

|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

E is the answer
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