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Math Revolution GMAT Instructor
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If (r+t)/(rt)>0, is r>t?
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06 Jun 2016, 22:34
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43% (01:57) correct 57% (01:25) wrong based on 101 sessions
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If (r+t)/(rt)>0, is r>t? 1) t>0 2) r>0 *An answer will be posted in 2 days.
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Re: If (r+t)/(rt)>0, is r>t?
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07 Jun 2016, 06:35
MathRevolution wrote: If (r+t)/(rt)>0, is r>t? 1) t>0 2) r>0
*An answer will be posted in 2 days. The question can be rewritten as: \(\frac{(r+t)}{(rt)}\)>0 Now this is possible when either both numerator and denominator are positive. or both numerator and denominator are negative Case 1: r+t > 0 rt > 0 Add these two: 2r> 0; r > 0 > This is my new question. Show that r> 0 Case 1: r+t < 0 rt < 0 Add these two: 2r < 0; r < 0 > This is my new question. Show that r < 0 So if there is any option that shows that r > 0 or less than 0 > I can get answer. Statement 1Gives no idea about r. Not sufficient. Statement 2We got our answer. r > 0 This means I can say that r+t > 0 (r > t) rt > 0 (r > t) > This is what we wanted to show. Unique answer obtained. B is the answer.



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Re: If (r+t)/(rt)>0, is r>t?
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08 Jun 2016, 16:59
We can modify the original condition and the question. From (r+t)/(rt)>0, we get rt>0? and r+t>0?. If we add two questions together, the question becomes 2r>0?. Hence, r>0?. Then, the condition 2) becomes ‘yes’ and it is sufficient. Hence, the correct answer is B.  Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
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Re: If (r+t)/(rt)>0, is r>t?
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30 Jul 2016, 18:25
Hi! A nice one on how to further solve the question itself...
kindly share some brief notes on equality ...and absolute values please. thanks



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Re: If (r+t)/(rt)>0, is r>t?
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03 Jul 2018, 17:31
I took the longer route and I think the answer is D Taking the cases whether r>0 or t>0, we see that if r and t are on opposite sides of 0, it is already determined whether r>t. We have to determine in cases when both r and t are positive or negative.
Case 1: r>0, t>0 => r+t > 0 => rt has to be more than zero as well. => r > t. Determined
Case 2: r<0, t<0 => r+t < 0 => rt has to be less than zero as well. => r<t. Determined.
I therefore find that Either statement is sufficient for determination of whether r>t or not. Hence answer is D.



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Re: If (r+t)/(rt)>0, is r>t?
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03 Jul 2018, 21:49
slayer1983 wrote: I took the longer route and I think the answer is D Taking the cases whether r>0 or t>0, we see that if r and t are on opposite sides of 0, it is already determined whether r>t. We have to determine in cases when both r and t are positive or negative.
Case 1: r>0, t>0 => r+t > 0 => rt has to be more than zero as well. => r > t. Determined
Case 2: r<0, t<0 => r+t < 0 => rt has to be less than zero as well. => r<t. Determined.
I therefore find that Either statement is sufficient for determination of whether r>t or not. Hence answer is D. Hello I think in first statement we are ignoring one thing here. If say t > 0, then we CAN have a case where r < 0 and still (r+t)/(rt) > 0. Eg, take t = 2 and r = 3. So r+t = 23 which is negative, and rt = 32 which is also negative. So (r+t)/(rt) = negative/negative which is positive. And here r < t. Thats why statement 1 is NOT sufficient to answer the question.



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Re: If (r+t)/(rt)>0, is r>t?
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03 Jul 2018, 21:54
amanvermagmat wrote: slayer1983 wrote: I took the longer route and I think the answer is D Taking the cases whether r>0 or t>0, we see that if r and t are on opposite sides of 0, it is already determined whether r>t. We have to determine in cases when both r and t are positive or negative.
Case 1: r>0, t>0 => r+t > 0 => rt has to be more than zero as well. => r > t. Determined
Case 2: r<0, t<0 => r+t < 0 => rt has to be less than zero as well. => r<t. Determined.
I therefore find that Either statement is sufficient for determination of whether r>t or not. Hence answer is D. Hello I think in first statement we are ignoring one thing here. If say t > 0, then we CAN have a case where r < 0 and still (r+t)/(rt) > 0. Eg, take t = 2 and r = 3. So r+t = 23 which is negative, and rt = 32 which is also negative. So (r+t)/(rt) = negative/negative which is positive. And here r < t. Thats why statement 1 is NOT sufficient to answer the question. Hi, If we are considering this case when t>0 and r<0, then we are already establishing that r<t.



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Re: If (r+t)/(rt)>0, is r>t?
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03 Jul 2018, 22:04
slayer1983 wrote: amanvermagmat wrote: slayer1983 wrote: I took the longer route and I think the answer is D Taking the cases whether r>0 or t>0, we see that if r and t are on opposite sides of 0, it is already determined whether r>t. We have to determine in cases when both r and t are positive or negative.
Case 1: r>0, t>0 => r+t > 0 => rt has to be more than zero as well. => r > t. Determined
Case 2: r<0, t<0 => r+t < 0 => rt has to be less than zero as well. => r<t. Determined.
I therefore find that Either statement is sufficient for determination of whether r>t or not. Hence answer is D. Hello I think in first statement we are ignoring one thing here. If say t > 0, then we CAN have a case where r < 0 and still (r+t)/(rt) > 0. Eg, take t = 2 and r = 3. So r+t = 23 which is negative, and rt = 32 which is also negative. So (r+t)/(rt) = negative/negative which is positive. And here r < t. Thats why statement 1 is NOT sufficient to answer the question. Hi, If we are considering this case when t>0 and r<0, then we are already establishing that r<t. Hello Yes, but from first statement, there is another possibility: t>0 and r>0. Eg, if we take t = 4, r = 5: then r+t = 9 and rt = 1, and 9/1 is also positive. Here t > 0 and r > t. So in my previous example (previous comment) t>0 while r<0 (so r is less than t) And in my current example above, t>0 and r>0 (and r is greater than t) Thats why this statement 1 is NOT sufficient to answer the question.




Re: If (r+t)/(rt)>0, is r>t? &nbs
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03 Jul 2018, 22:04






