vinni.k wrote:Is 1 / a - b < b - a ?

(1) a < b
(2) 1 < |a - b|

Answer is A

Hi All,
I got the above question correct; but I am not sure whether my method to solve this question is correct or not. O.G explanation is little different. Please check my method.

1 / a - b < b - a
a - b > b - a [ This doesn't logically follow from the above]

subtracting b from both sides
a - b - b > b - b - a
a - 2b > - a

subtracting a from both sides
a - a - 2b > -a - a
-2b > -2a

Dividing both sides by -2
b < a (flipping the inequality sign).

So, Is b < a ?

Option (1) says a < b that means b > a. (No but sufficient)[ It's to be 'YES!']

Option (2) 1 < | a - b |
Means, |a - b| > 1
Now, a -b > 1 and a - b < -1

For example, if a = 7 and b = 4 then 1/3 > -3, but if a = 4 and b = 7, then -1/3 < 3.
So, not sufficient

Thanks & Regards
Vinni

Hi Vinni, For the most part, your solution is correct, except in the areas bold-faced.

The question is: Is 1/(a-b) < b-a? Here note that a-b is a negative of b-a.
To simplify our calculation, let's represent a-b by 'm'; hence, b-a shall be '-m'.

Now the question is: Is 1/m < -m?
Bring the variable on the right-hand side to the left.
= (1/m) + m <0
= (1 + m^2)/m < 0
We know for sure that the numerator is positive; therefore for the inequality to hold good, m has to be < 0.
So, simply the question is as follows: Is m<0? To make it relevant, let's undo our representation.
we used m to stand for a-b.
Therefore the question is as follows: Is a-b <0?
This means: Is a < b [By transferring -b to the right-hand side]
Statement 1: a < b.....Yes! So sufficient.
Statement 2: Similar to what you've done.

Hence the answer is A.

You can also do it by plugging in numbers:

1/(1-b) < b-a?

(1) a < b

We have 2 variables, hence create 3 versions of their relationship.
1. where they are both negative: a = -4 and b = -1
2. Where they are both positive: a = 1 and b = 2
3. Where one is positive and one is negative. a = -1 and b = 1

Now when you solve, you'll see that 1/(1-b) < b-a? holds true for all of the versions.

How do I know to create the 3 versions, when ever you see inequality signs comparing variables, usually negatives/positives and/or integers/non-integers are whats being tested.

(2) This doesn't tell us anything about each variable or their relationship. Simply gives us a range. range = multiple values, hence insufficient.

(a)

There are two ways this problem can be solved. Here is a video explanation.

https://www.gmatquantum.com/shared-posts ... ds120.html

Dabral

Thanks YAMLAKSIRA, Gurpinder, and dabral for checking my question. I should have plugged in the values to check whether 1/a - b is less than b - a. I was simply solving the question quickly and thought changing the reciprocal will change the inequality sign. I don't know where i got lost.

But thanks for giving the explanations. All the explanation really helped.

Thanks & Regards
Vinni

vinni.k wrote:Thanks YAMLAKSIRA, Gurpinder, and dabral for checking my question. I should have plugged in the values to check whether 1/a - b is less than b - a. I was simply solving the question quickly and thought changing the reciprocal will change the inequality sign. I don't know where i got lost.

But thanks for giving the explanations. All the explanation really helped.

Thanks & Regards
Vinni

Taking the reciprocal of both sides usually does simply flip the inequality, which is what you did. However, it doesn't always work, most notably when one side is negative and the others side is positive (which, incidentally, is the case here):

-4 < 3 (clearly true)
-1/4 > 1/3 (clearly false)

The easiest pure-number-logic way to solve this problem is to realize that b-a and a-b are negatives of each other, and thus that if one is positive, the other is negative, depending simply on which is bigger, a or b. This is actually a really useful fact because something like "a-b" shows up quite often on the GMAT, and it's nice to know that a-b is ALWAYS positive for a>b and ALWAYS negative for b>a, regardless of the signs of the variables. Thus:

1 / a-b < b-a? If a>b, it's + < - (answer: no); if b>a, it's - < + (answer: yes).

(1) tells us straight off a<b, which we already know is Sufficient (answer: yes).
(2) has an absolute value around |a-b|, which we should quickly realize ignores which one is bigger; that is, since |a-b| = |b-a| (again, they're just negatives of one another), either case a>b or b>a could work equally here. Thus it's Insufficient.

This is really just a non-algebraic version of Yamlaksira's approach above, but because you're likely to see a-b and b-a in some form or another on test day, I think it's worth it to learn how they work so you can get directly and confidently to the answer on questions like this without risking algebraic mishap.

I would with smart numbers instead of long equations.
St 1 says- a<b. So just pick a= 2 , b = 3.. So the equation says, 1/a-b < b-a so, 1/(2-3)<(3-2).So, --1<1 (Yes). Take any negative number, a=-3. b = -2. So,now 1/(-3+2)< -2+3. So, -1<1 Yes..
Both YES..SO A is sufficient.

St 2 says- 1 < |a - b| or |a-b| >1. So, a> 1+b OR -a+b>1 or b>a+1..Mixed answers...NOT SUF...

Ans-A