Let me try putting it in words. Broadly - (1) Using example numbers or (2) sticking to abstract algebraic relationships.
PS: I am assuming the question says 1 / (a-b) and not (1/a)-b
1) Using example numbers.
i. Try taking numbers which are unrelated. eg., do not take a and b as 2 and 6. Take fairly unrelated numbers like 2 and 97.
ii. Try taking negative numbers. If they say a is a number, assume that a is -5.
iii. Try taking fractional numbers. If they say a is a number, assume that a is -3.5. More often, we make the mistake of assuming easy numbers and the question setter will try to exploit the pitfall here.
iv. Specifically here, take a = -2 and b = 3.5. The point which one will realize is that a-b is always a -ve number irrespective of the positivity or negativity of a and b. Once u hit this, the rest will fall in place.
v. modulus (a-b) > 1. Assume b to be, say 7. The point is that to satisfy this condition, a has to be < 6 or a has to be > 8. You will realize that this implies that the gap between a and b is greater than 1. There is no suggestion as to whether a is greater or b is greater
2) Sticking to abstract algaebraic relationships.
Using this will lead you to the conclusions of (iv) and (v) highlighted previously. Repeated practice with method (1) will help you reach these conclusions quicker and with more accuracy.
Remember that using algaebraic relationships is more often better.
Examples often come with the following pitfall: A rule may be true for a 1000 different numbers. But, there might be one number for which the rule is not satisfied. You have to be tremendously lucky to use that single number as an example.
I hope this helps. I am not sure if I have hit the nail on its head as the question was a little vague.
Data Sufficiency-Inequalities
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4GMAT_Mumbai
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eaakbari
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IMO A
Stem:
1/-(b-a) > b-a
If we think for what value of b-a can satisfy is when b-a = positive integer like say 3. We shall keep that in mind.
Statement one
If b>a lets get their difference b-a to a positive integer: (-3,-5) serve
See if following b>a can any numbers give you b-a as negative quantity as that wont satisfy.
No numbers like that
Hence Suff
Statement two
Tells us the difference is more than one. Does not help otherwise
Answer A.
Stem:
1/-(b-a) > b-a
If we think for what value of b-a can satisfy is when b-a = positive integer like say 3. We shall keep that in mind.
Statement one
If b>a lets get their difference b-a to a positive integer: (-3,-5) serve
See if following b>a can any numbers give you b-a as negative quantity as that wont satisfy.
No numbers like that
Hence Suff
Statement two
Tells us the difference is more than one. Does not help otherwise
Answer A.
@eaakbari
When I saw statement 2, I was trying to figure out how should I remove the modulus. I was thinking of squaring the inequalities or doing cases for which the inequality exist (+ve or -ve).
When should we do the squaring to remove the modulus or the multiple possibilities?
I got this for st 2:-
|a - b| > 1
a - b > 1 or - (a - b) > 1
Thus insufficient cos the two inequalities can result in a few scenarios for question, right?
When I saw statement 2, I was trying to figure out how should I remove the modulus. I was thinking of squaring the inequalities or doing cases for which the inequality exist (+ve or -ve).
When should we do the squaring to remove the modulus or the multiple possibilities?
I got this for st 2:-
|a - b| > 1
a - b > 1 or - (a - b) > 1
Thus insufficient cos the two inequalities can result in a few scenarios for question, right?
-
eaakbari
- Master | Next Rank: 500 Posts
- Posts: 435
- Joined: Mon Mar 15, 2010 6:15 am
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Your approach is definitely right. In this question, I did not go with squaring the modulus because the R.H.S was one , so no avail. If squaring the modulus gave me some expression which could be simplified then that would be the primary step.dxgamez wrote:@eaakbari
When I saw statement 2, I was trying to figure out how should I remove the modulus. I was thinking of squaring the inequalities or doing cases for which the inequality exist (+ve or -ve).
When should we do the squaring to remove the modulus or the multiple possibilities?
I got this for st 2:-
|a - b| > 1
a - b > 1 or - (a - b) > 1
Thus insufficient cos the two inequalities can result in a few scenarios for question, right?
Picking numbers and checking for statement two is not wrong but definitely time consuming and frankly cause statement one also required picking numbers , saving time is an important deal for this question
HTH
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