Is IaI > IbI

[log2bala]

Hello Guys

Would anyone kindly explain how to proceed further in this problem?

. Is IaI > IbI ?

a) b < - a
b) a < 0

Answer.

Taking Option B again and eliminating it as it does not have any information on BD is crossed out

Taking Option A now, How do you rephrase this inequality? b < -a

Please clarify on these types how to approach them?

Thank you guys for your help

Bala

[rchitta]

Assuming IaI & IbI are 3 digit numbers with the digit in the units and hundreds place being the same, aren't they actually asking:

If a < b internally ?

example 121 and 131 should be the same as 2 < 3?
also, 151 and 141 should be same as 5 < 4 ?

1) b < -a
=> a + b < 0 (INSUFFICIENT) not sure where a and b line on the integer line.
i.e, is a +ve, b -ve? or
is a -ve & b +ve? or
is a -ve & b -ve?
So, we cannot determine if a < b.

2. a < 0 (same thing), we have no information about b (INSUFFICIENT)


combining 1 & 2

a + b < 0 and a < 0
this again is INSUFFICIENT, I think. b could be -ve or b could be just +ve making a + b still < 0

So, I would say the answer is E

[rchitta]

Oops, I think I mis-read the question. I will try to solve it again assuming IaI means absolute value of a and not Integer A Integer. Sorry about the mis-lead.

[rchitta]

Sorry, I completely forgot about this...

This is what I think...

1. b < -a implies that a + b < 0 (you get this when you add a to either sides of the inequality).

This implies either,
i) both a and b are -ve (cannot say if |a| > |b| here because we are not sure if b is more -ve (far down on the number line from 0) or if a is more -ve. e.g; a = -1 and b = -2, a = -2, b = -1) --> THIS PROVES INSUFFICIENCY.

ii) a is +ve and b -ve: (I'm sure you could come up with some a, b value here)
iii) a is -ve and b is +ve: (I'm sure you could come up with some a, b value here)


2. if a < 0, this doesn't say anything about b and b either could be more -ve or less -ve than a (same thing as above ) INSUFFICIENT.


Combining both: a + b < 0 and a < 0 doesn't tell us much either. either b could be more -ve or less -ve than a. Hence both statements are insufficient together. D

Hope this helps!

[rchitta]

Sorry again! I meant E is the answer choice: Statements 1 and 2 TOGETHER are NOT sufficient

[esledge]

I take pretty much the same pos/neg case approach as rchitta. To save time, make a chart of the scenarios. I actually started with a chart that had 4 possibilities:

a..........b
pos.....pos
pos.....neg
neg.....pos
neg.....neg

Then I add columns for checking which scenarios are allowed by the statement (esp. statement 1: we rule out the pos/pos scenario with that one) and for answering.

At the end, my chart looked something like this:
a...........b......b < -a...........Interpret...............|a| > |b|?
pos......pos.....imposs............(blank)
pos......neg.....OK..............a is closer to 0.............NO
neg.....pos....OK..............b is closer to 0.............YES
neg.....neg....OK..........either a or b closest to ....MAYBE

Bold text indicates what I circled on paper. The chart is for (1), the circle is for (2). Together, we still get a MAYBE result.

[yousuf_azim]

What is the ans?

BR

[JohnHarris]

. Is IaI > IbI ?

a) b < - a
b) a < 0

Try a = -1, b = -100. (a) is satisfied, (b) is satisfied, (a) and (b) are satisfied. The inequality is not true. Neither (a), nor (b), nor (a) & (b) are sufficient.

What (a) says is that b is to the right of -a on the number line but makes no restrictions on the magnitude of b. Case (b) ignores b and thus makes no restrictions on the magnitude of b. That is, choose any negative a [satisfies (b)] and any b < a. Then b is also negative and, since -a is positive, b < -a [satisfies (a)]. However, -b > -a [multiply previous by minus one], i.e. |b| > |a|. Thus (a) & (b) are not sufficient.

[jnelson0612]

The answer is E.