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funke
 
 

Quest for 750 Quant Problem

by funke Sun Sep 21, 2008 4:54 pm

This question comes from the Quest for 750 Quant Homework Set. I have a question about the answer explanation and if anyone has any suggestions on how to solve this problem quickly (possibly by plugging in values?). (The answer is C)

If a and b are distinct integers and their product is not equal to zero, is a > b?

(1) (a(^3)b - b(^3)a)/(a(^3)b + b(^3)a - 2a(^2)b(^2)) < 0

(2) b < 0

----

(1) Rephrased: (a + b)/(a - b) < 0

The answer explanation then instructs the following computations:
So statement (1) simplifies to (a + b)/(a - b) < 0. If a - b is positive (in other words, if a > b), we can simplify even further:

IF a > b:
a + b < a - b (MY MAIN CONCERN: HOW IS IT POSSIBLE TO MULTIPLY (a-b) by 0?)
b < -b
2b < 0
b < 0

This tells us that b < 0 whenever a > b. The converse of this statement is also true. Whenever b < 0, a must be greater than b for the inequality to hold (try plugging sample values into (a + b)/(a - b) < 0 to verify that when b is negative, an a value smaller than b fails to satisfy the inequality).

However, if a - b is negative (in other words, if a < b), we must flip the sign when multiplying through by a - b:
IF a < b:
a + b > a - b (MY MAIN CONCERN: HOW IS IT POSSIBLE TO MULTIPLY (a-b) by 0?)
b > -b
2b > 0
b > 0

This tells us that b > 0 whenever a < b. The converse of this statement is also true. Whenever b > 0, a must be less than b for the inequality to hold (try plugging sample values into (a + b)/(a - b) < 0 to verify that when b is positive, a values greater than b fail to satisfy the inequality).
RonPurewal
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by RonPurewal Tue Oct 14, 2008 4:40 am

um yeah, wow, i don't so much follow that one either.

here's how i'd do it. it's essentially the same work, but approximately in reverse of the order presented there.

--

statement (1):

(a + b)/(a - b) < 0
therefore, a + b and a - b have opposite signs. we can divide this statement into 2 cases.

CASE 1: a + b is positive and a - b is negative
a - b is negative --> immediately know a < b
also, in this case, a + b > a - b, so therefore b > -b, so therefore b is positive.
that's all we know, though; we know nothing about the sign of a. (note that this case works for (a, b) = (2, 4) but also (-2, 4))

CASE 2: a + b is negative and a - b is positive
a - b is positive --> immediately know a > b
in this case, a + b < a - b, so therefore b < -b, so therefore b is negative.
again, that's all we know. (this case works for (a, b) = (2, -4) but also (-2, -4))

this is insufficient, because there's a case in which a < b and a case in which a > b.

statement 2:
obviously insufficient

together:
this has to be CASE 2 above, so therefore a > b.
sufficient.

hth.
jp.jprasanna
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Re: Quest for 750 Quant Problem

by jp.jprasanna Thu Sep 06, 2012 2:04 am

Hi ron - I understand this sol.

If this question was slightly different.

IS (a+b)/(a-b) <0 ?

1/ a<b
2/ b<0

we can rephrase the question to

Case 1/ if a-b > 0

then a+b < a-b so 0>b i.e B should be negative

Case 2/ if a - b < 0

then b>0 i.e B should be +ve

Statement 1 says a-b <0 now all we need to know whether B is negative to answer the questions (YES) But just becasue a-b<0 we cannot say b is -ve; it could be +ve too... -10-1 <0 or -10 - (-1)<0

Statement 2 - Obviously not sufficient

Combined -> From stat1 and stat2 a-b< 0 and b is -ve so can definitely say (a+b)/(a-b) IS NOT LESS THAN 0 (had B been +ve then we can definitely say Yes; either case C is the answer)

Is this sol ok?

Cheers
tim
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Re: Quest for 750 Quant Problem

by tim Thu Sep 06, 2012 2:24 am

this solution is not okay. to be honest, i stopped reading after your first mistake, which was to rephrase the question into a statement. you CANNOT do this! this is one of the biggest mistakes people make in DS. any rephrasing of the question must still have a question mark in it; if you do not still have a QUESTION, i can guarantee you have rephrased the question incorrectly. go back and try again, and i'll be happy to review what you've done..
Tim Sanders
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Follow this link for some important tips to get the most out of your forum experience:
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jp.jprasanna
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Re: Quest for 750 Quant Problem

by jp.jprasanna Thu Sep 06, 2012 2:59 am

tim Wrote:this solution is not okay. to be honest, i stopped reading after your first mistake, which was to rephrase the question into a statement. you CANNOT do this! this is one of the biggest mistakes people make in DS. any rephrasing of the question must still have a question mark in it; if you do not still have a QUESTION, i can guarantee you have rephrased the question incorrectly. go back and try again, and i'll be happy to review what you've done..


Hi tim- WOW thanks for you quick response....

This sol that i have wriiten here is parallel to one in done by Rachel in this post - if-x-does-not-equal-y-t2997-15.html

And apologies for not adding the question mark.....(I did make some mistakes in past leaving out the question mark - thanks for the advise.)


IS (a+b)/(a-b) <0 ?

1/ a<b
2/ b<0

we can rephrase the question to

Case 1/ if a-b > 0

then a+b < a-b so 0>b i.e B should be negative

if either of the statements give us a-b>0 and if we can prove b is negative then we can answer the question with a def YES

if a-b>0 then IS b NEGATIVE?


Case 2/ if a - b < 0

then b>0 i.e B should be +ve

Similar rephrase as done for CASE 1.

Am I ok now?

Cheers
tim
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Re: Quest for 750 Quant Problem

by tim Thu Sep 06, 2012 1:09 pm

this looks acceptable, although i would recommend not splitting a question into cases because there is a high chance of creating confusion regarding what is actually being asked about..
Tim Sanders
Manhattan GMAT Instructor

Follow this link for some important tips to get the most out of your forum experience:
https://www.manhattanprep.com/gmat/forums/a-few-tips-t31405.html