Is a < b ? (1) |a - b| < |b| (2) a < 0

E
for b=-2 a =-3 yes
for b=-2 a=-1 no

Is a<b ?

(1) |a−b|<|b| - Insufficient
CounterExample:
a = 3, b = 2. LHS = 1, RHS = 2. a > b
a = 2, b = 3, LHS = 1, RHS = 3, a < b

(2) a<0 - Insufficient
We don't know anything about b. Hence we can't say a < b or a > b or a = b

(1) + (2) - Insufficient
Counterexample:
a = -2, b = -3, putting in expression (1) gives LHS = 1, RHS = 3 Hencce LHS < RHS , a > b
a = -3, b = -2, putting in expression (1) gives LHS = 1, RHS = 2 Hencce LHS < RHS , a < b

Answer - E

Is a <b?

(Statement1): |a−b|<|b|
--> Square both sides:
\((a-b)^{2} < b^{2}\)

\(a^{2} -2ab+ b^{2} < b^{2}\)

\(a^{2} -2ab <0\)
\(a*(a-2b) <0\)

There are two cases to satisfy this inequality. Let's see on of them.
--> if a <0 , then a >2b
a=-3, --> b could be b= -2 or b= -4. (-3> -4 and -3 >-8)
--> but a <b (-3 >-4(NO) and -3< -2 (YES))
Insufficient

(Statement2): a<0
Clearly insufficient.

Taken together 1&2,
The same thing as Statement1,
Insufficient

The answer is E.

Archit3110 Could you please tell how to choose numbers to prove/disprove the given absolute value equations.
I have trouble finding the numbers usually and end up taking lots of time on the DS questions.

On questions such as these, I usually prefer not to put in any values at all.

Statement 1 tells me that a and b must be the same sign, but nothing about if a<>b.
Statement 2 tells me that a <0, again insufficient.
Combining these two statements, all I know is that both a and b are negative. Thus it isn't possible to infer anything.

I had the same approach. Now statement 1 gives us 2 cases-->
Case 1) if a <0 , then a >2b
Case 2 but a > 0, a-2b < 0, a<2b.

Statement 2 Tells us a<0. Hence Case 1 is true. If a >2b, then obviously a has to be greater than b.

Example if b=3, if a>2b--> that means a>6, then a>3.

Then answer is C.

Bunuel Please correct me here.

Yes. I've landed on the same platform. Need experts' opinions. Bunuel VeritasKarishma plz help

Stmnt 1: \(a*(a-2b) <0\)

If a < 0, (a - 2b) > 0 so a > 2b
Now since a is negative, 2b must be negative too (because a is greater than 2b)
a = - 4, b = -3
or a = -4, b = -5
Both will satisfy.

So using both statements, you know that a is negative. But still a > b and a < b both are possible.
Hence answer will be (E)

I would do this question orally within a few secs by just thinking about the number line and the distance concept of absolute values.

Is a < b?
means 'Is a to the left of b on the number line?'

(1) \(|a - b| < |b|\)
Distance of a from b is less than the distance of b from 0. So a is closer to b than b is to 0. Then this is what I imagine.

---------------(a) ---(b)--- (a) ----------- 0 --------- (a) ---(b) ---(a) ------------

b could be to left or right of 0.
a would be nearer to it than 0 would be.

(2) \(a < 0\)
Clearly not sufficient since no info on b.

Using both statements,
So only left side is possible.
---------------(a) ---(b)--- (a) ----------- 0 -------
Still a could be to the left or right of b.
Answer (E)

There are two ways to go about it.

First way is by opening and solving the modulus.

(1) \(|a - b| < |b|\)

\(1]\) \(a - b < b\) = \(a < 2b\) => This means that either (a is negative and b is positive) or (both a and b are positive) - plugin any value and check.
\(2]\) \(a - b < -b\) = \(a < 0\) => a is positive
\(3]\) \(-a + b < b\) = \(-a < 0\) = \(a>0\) => a is negative
\(4]\) \(-a + b < -b\) = \(a > 2b\) => This means that either (a is positive and b is negative) or (both a and b are negative) - plugin any value and check.

From here we know that a and b can anything +ve or -ve. Statement 1 is insufficient.

(2) \(a < 0\)

We don't know anything about b.

(1) + (2) tells us that it's either one of the above highlighted cases. Again, insufficient.


Second way to go about it is by squaring both sides.

(1) \(|a - b| < |b|\)

\((a - b)^2 < b^2\) = \(a^2 + b^2 - 2ab < b^2\) = \(a^2 - 2ab < 0\) = \(a(a - 2b) < 0\)

Now, let's think about this
If \(a(a - 2b) < 0\), then either \(a<0\) or \(a-2b<0\)
a<0 doesn't tell anything about b
\(a<2b\) tells us that either a is negative and b is positive or both a and b are positive or both can be negative as well (Eg: a = -10, b = -4). Only then \(a<2b\) will hold true. It is very easy to miss out on the fact that both a and b can be negative when following this approach.

Hence, Insufficient.

(2) \(a < 0\)
Doesn't tell us anything about b. Insufficient.

(1) and (2)
Combining them still doesn't tell us anything about b - it can be both +ve and -ve as in the highlighted portion above.


Therefore, E should be the correct answer choice.

|x-y| = the distance between x and y
|x| = |x-0| = the distance between x and 0

Statement 1: |a-b| < |b-0|
The distance between a and b is less than the distance between b and 0.
In other words, b is closer to a than to 0.
Statement 2: a < 0

Case 1:
.........b..a..........0..................
Here, b is closer to a than to 0, and a is negative.
In this case, a > b, so the answer to the question stem is NO.

Case 2:
.........a..b..........0..................
Here, b is closer to a than to 0, and a is negative.
In this case, a < b, so the answer to the question stem is YES.

Since the answer is NO in Case 1 but YES in Case 2, the two statements combined are INSUFFICIENT.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.