Is 1/(a-b) < b-a ?
Stat1) a<b
Stat2) 1<!a-b!
Absolute value
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Guys I am having some other opinion at all. The Qs is saying that -
Is 1/(b - a) < (a - b) - But this we can reformat to 1/-(a - b) <a> -1. So we have to show - Is (a - b)^2 > -1?
This we can conclude when option - 1 and option - 2 independently holds true. So IMO D. What's the OA?
Is 1/(b - a) < (a - b) - But this we can reformat to 1/-(a - b) <a> -1. So we have to show - Is (a - b)^2 > -1?
This we can conclude when option - 1 and option - 2 independently holds true. So IMO D. What's the OA?
Correct me If I am wrong
Regards,
Amitava
Regards,
Amitava
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(1) if a is less than b, then (a-b) is negative. So, 1/(a-b) is also negative.Bronson wrote:Is 1/(a-b) < b-a ?
Stat1) a<b
Stat2) 1<!a-b!
Also, if a - b is negative, then b-a is positive.
So, 1/(a-b) is negative and b-a is positive.. therefore, 1/(a-b) is definitely less than b-a. Statement (1) is sufficient.
(2) if 1 is less than the abs val of (a-b), then a-b is greater than 1 or a-b is less than -1. However, this doesn't tell us the relationship between a and b, so statement 2 is insufficient.
For example, we could choose the values:
a=10 b=5, since 1 is less than 5. We then ask the original question:
is 1/(10-5) greater than (5-10)? Or is 1/5 greater than -5. Yes it is!
We could also choose the values a= -10 and b = -5 (since 1 is less than 5). We then ask the original question:
Is 1/(-10 --5) greater than (-5 --10)?
In other words, is 1/-5 greater than +5 ? No, it isn't!
Since we can get both a yes and a no, statement (2) is insufficient.
1 is sufficient, 2 isn't.. choice (A) is correct.
ps: I had to edit this post 8 times to finally get it to format correctly - any time I used a greater than or less than sign it messed everythign up.. very frustrating!
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Hey Stuart, if I interpret the Qs
1/-(b - a) < (b - a)
so -1 < (b - a)^2
So the Qs is coming like -
in the following way -Is 1/(a-b) < b-a ?
1/-(b - a) < (b - a)
so -1 < (b - a)^2
So the Qs is coming like -
what wrong I am doing? Am I missing something or doing something wrong? Stuart, looking for ur suggestion/ concern...Is -1 < (b-a)^2 ?
Correct me If I am wrong
Regards,
Amitava
Regards,
Amitava
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Answer should be A .. a<b means a-b is negative which makes 1/(a-b) negative and also a<b means (b-a) is positive so 1/(a-b)<(b-a).Bronson wrote:Is 1/(a-b) < b-a ?
Stat1) a<b
Stat2) 1<!a-b!
the second statement could mean a-b is positive or negative so it is insufficient.
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You need to be very careful when manipulating inequalities.
In this question, to get to your final statement, you're multiplying both sides by (b-a). However, if b-1 turns out to be negative, you have to change the direction of the inequality, so you'd actually end up with:
Is -1 greater than (b-a)^2
to which the answer is "no".
So, if b-a is positive, then the answer will be yes. However, if b-a is negative, the answer will be no. That's another way to deal with statement (2) and decide that it's insufficient.
In this question, to get to your final statement, you're multiplying both sides by (b-a). However, if b-1 turns out to be negative, you have to change the direction of the inequality, so you'd actually end up with:
Is -1 greater than (b-a)^2
to which the answer is "no".
So, if b-a is positive, then the answer will be yes. However, if b-a is negative, the answer will be no. That's another way to deal with statement (2) and decide that it's insufficient.
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Hi Stuart:
gabriel just let me know about your issues with posting. My apologies for your frustration, I'm investigating right now what the problem is with the greater then, less then symbols...
gabriel just let me know about your issues with posting. My apologies for your frustration, I'm investigating right now what the problem is with the greater then, less then symbols...
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Quick update. I ran some tests with the symbols and didn't encounter any issues. I'll keep investigating, but this might have been a fluke incident.
Again, my apologies!
Again, my apologies!
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I disabled HTML, I think that should fix it (since HTML uses < and > for tags).beatthegmat wrote:Quick update. I ran some tests with the symbols and didn't encounter any issues. I'll keep investigating, but this might have been a fluke incident.
Again, my apologies!
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That did it. Thanks for the QA work, Stuart.Stuart Kovinsky wrote:I disabled HTML, I think that should fix it (since HTML uses <and> for tags).beatthegmat wrote:Quick update. I ran some tests with the symbols and didn't encounter any issues. I'll keep investigating, but this might have been a fluke incident.
Again, my apologies!
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