MOD+Inequality

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MOD+Inequality

by kalpita123 » Mon Jul 30, 2012 3:32 pm
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
Source: — Data Sufficiency |

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by eagleeye » Mon Jul 30, 2012 5:28 pm
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 :)

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by mmeital » Mon Jul 30, 2012 6:18 pm
Can you please explain why we read mod a > mod b as |a|>|b|?

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by eagleeye » Mon Jul 30, 2012 6:27 pm
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.

:)

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by kalpita123 » Tue Jul 31, 2012 1:32 pm
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 :)

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by eagleeye » Tue Jul 31, 2012 5:25 pm
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.