kalpita123 wrote:I am able to solve the below problem using "number plugging" method,however, looking for some other ways to approach the problem:

If a is not equal to b, is [ 1/(a-b)]> ab ?
1) mod a > mod b
2) a<b

ans- E

Here's a way to do it using mathematical reasoning:
First, let's rephrase the expression:

We are asked whether:
1/(a-b) <ab
=> (a-b)^2*1/(a-b) <ab(a-b)^2 (multiplying both sides by (a-b)^2, since it is positive as a!=b)
=> (a-b)<ab(a-b)^2
=> (a-b)(1-ab(a-b)) <0.
So the question rephrases as Do (a-b) and (1-ab(a-b)) have opposite signs?

With that in mind, let's look at the statements:
1) |a| > |b|.

If a is positive, (a-b)>0, and (1-ab(positive)) may be positive or negative depending on sign of b. Insufficient.

2) a<b. If a<b, (a-b) <0, again using the same argument as above, depending on the sign of b, (1-ab(a-b)) can be positive or negative. Insufficient.

Combining the two, |a|>|b| but a<b. This means that a<0. But b may still be positive or negative, which again, may give us positive or negative values for (1-ab(a-b)). Still insufficient.
Hence E is correct.

Although we didn't pick numbers, you can check the above reasoning using the following 2 cases:
1. a=-2, b=-1
2. a=-2, b=1

Let me know if this helps

Can you please explain why we read mod a > mod b as |a|>|b|?

mmeital wrote:Can you please explain why we read mod a > mod b as |a|>|b|?

"|a|" is called either:
1) absolute value of a, or
2) modulus of a

Some people shorten "modulus" and call it "mod". (although strictly speaking, calling modulus "mod" is incorrect, because mod refers to "modulo" in modular arithmetic which refers to divisibility). I knew what the poster meant and hence interpreted it accordingly.

eagleeye wrote:

kalpita123 wrote:I am able to solve the below problem using "number plugging" method,however, looking for some other ways to approach the problem:

If a is not equal to b, is [ 1/(a-b)]> ab ?
1) mod a > mod b
2) a<b

ans- E

Here's a way to do it using mathematical reasoning:
First, let's rephrase the expression:

We are asked whether:
1/(a-b) <ab
=> (a-b)^2*1/(a-b) <ab(a-b)^2 (multiplying both sides by (a-b)^2, since it is positive as a!=b)
=> (a-b)<ab(a-b)^2
=> (a-b)(1-ab(a-b)) <0.
So the question rephrases as Do (a-b) and (1-ab(a-b)) have opposite signs?

With that in mind, let's look at the statements:
1) |a| > |b|.

If a is positive, (a-b)>0, and (1-ab(positive)) may be positive or negative depending on sign of b. Insufficient.

2) a<b. If a<b, (a-b) <0, again using the same argument as above, depending on the sign of b, (1-ab(a-b)) can be positive or negative. Insufficient.

Combining the two, |a|>|b| but a<b. This means that a<0. But b may still be positive or negative, which again, may give us positive or negative values for (1-ab(a-b)). Still insufficient.
Hence E is correct.

Although we didn't pick numbers, you can check the above reasoning using the following 2 cases:
1. a=-2, b=-1
2. a=-2, b=1

Let me know if this helps

eagleeye,

Just a rectification!
Here, we are asked to check whether 1/(a-b)> ab and not whether 1/(a-b)]< ab
Hence, we will rephrase the question as, Do (a-b) and (1-ab(a-b)) have same signs?

Your process still fits the bill, as we have no evidence for the sign of "b" in either case.

Thanks for your response! that helped

kalpita123 wrote:

eagleeye wrote:

kalpita123 wrote:I am able to solve the below problem using "number plugging" method,however, looking for some other ways to approach the problem:

If a is not equal to b, is [ 1/(a-b)]> ab ?
1) mod a > mod b
2) a<b

ans- E

Here's a way to do it using mathematical reasoning:
First, let's rephrase the expression:

We are asked whether:
1/(a-b) <ab
=> (a-b)^2*1/(a-b) <ab(a-b)^2 (multiplying both sides by (a-b)^2, since it is positive as a!=b)
=> (a-b)<ab(a-b)^2
=> (a-b)(1-ab(a-b)) <0.
So the question rephrases as Do (a-b) and (1-ab(a-b)) have opposite signs?

With that in mind, let's look at the statements:
1) |a| > |b|.

If a is positive, (a-b)>0, and (1-ab(positive)) may be positive or negative depending on sign of b. Insufficient.

2) a<b. If a<b, (a-b) <0, again using the same argument as above, depending on the sign of b, (1-ab(a-b)) can be positive or negative. Insufficient.

Combining the two, |a|>|b| but a<b. This means that a<0. But b may still be positive or negative, which again, may give us positive or negative values for (1-ab(a-b)). Still insufficient.
Hence E is correct.

Although we didn't pick numbers, you can check the above reasoning using the following 2 cases:
1. a=-2, b=-1
2. a=-2, b=1

Let me know if this helps

eagleeye,

Just a rectification!
Here, we are asked to check whether 1/(a-b)> ab and not whether 1/(a-b)]< ab
Hence, we will rephrase the question as, Do (a-b) and (1-ab(a-b)) have same signs?

Your process still fits the bill, as we have no evidence for the sign of "b" in either case.

Thanks for your response! that helped

You're welcome. The one good thing about reducing an equality to 0 on one side is that whatever the answer is for A>0 for DS will be the same as A<0. At the same time, I need to be more careful while typing.