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# rt > 0

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rt > 0 [#permalink]  26 Aug 2018, 02:58
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Question Stats:

76% (01:05) correct 23% (00:40) wrong based on 80 sessions
$$rt > 0$$

 Quantity A Quantity B $$\frac{3}{r} + \frac{4}{t}$$ $$\frac{3t + 4r}{r + t}$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

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[Reveal] Spoiler: OA

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Re: rt > 0 [#permalink]  26 Aug 2018, 05:01
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Carcass wrote:
$$rt > 0$$

 Quantity A Quantity B $$\frac{3}{r} + \frac{4}{t}$$ $$\frac{3t + 4r}{r + t}$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E

Here,

r and t could be either both positive , both negative , fraction

if r and t are considered negative integer then QTY A > QTY B

if r and t are considered positive integer then QTY A < QTY B

Hence option D
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Re: rt > 0 [#permalink]  17 Jul 2020, 00:46
Hello. Thnx for solution. But could you please explain, why if r and t considered negative A is bigger than B? Isn't A will be negative (because in nominator we have -(3t+4r) and in denominator it s (-t)*(-r) which is just rt. And in B we have negative numerator and negative denominator, which means the fraction is positive.
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Re: rt > 0 [#permalink]  25 Jul 2020, 10:04
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Carcass wrote:
$$rt > 0$$

 Quantity A Quantity B $$\frac{3}{r} + \frac{4}{t}$$ $$\frac{3t + 4r}{r + t}$$

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Kudos for R.A.E

We are given that the product of r and t is greater than zero. This means that either both values are positive or both are negative.

I solved using test values. If r=t=1, then Quantity A = 7 and Quantity B = 7/2. Quantity A is greater. However, if r=t=-1, then Quantity A = -7 and Quantity B = -7/2. In this case, Quantity B is greater.

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Re: rt > 0 [#permalink]  18 Aug 2020, 05:29
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$$rt > 0$$

 Quantity A Quantity B $$\frac{3}{r} + \frac{4}{t}$$ $$\frac{3t + 4r}{r + t}$$

 Quantity A Quantity B $$\frac{3t}{rt} + \frac{4r}{rt}$$ $$\frac{3t + 4r}{r + t}$$

 Quantity A Quantity B $$\frac{3t + 4r}{rt}$$ $$\frac{3t + 4r}{r + t}$$

 Quantity A Quantity B $$3t + 4r$$ $$3t + 4r\frac{rt}{r + t}$$

Now you have on both sides an known quantity with the same sign $$3t+4r$$. So on the left sign we'll call the sign $$s$$ and on the right hand side you also $$s$$ multiplied by a postive $$rt$$ which is also $$s$$, but unfortunately it is divided by an another unknown sign and we'll call it $$u$$.

You can not guarantee the signs for both sides, and thus the greater quantity cannot be determined.
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Re: rt > 0 [#permalink]  18 Aug 2020, 11:56
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Simple method, always works - assume (r,t) = (1,1) and (r,t) = (-1,-1)

since the ans from both the cases are disparate, we can conclude that D is the ans.
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Re: rt > 0 [#permalink]  18 Aug 2020, 12:58
1
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If r and t are taken as positive values then Quantity A > Quantity B
If r and t are taken as negative values then Quantity B > Quantity A
Hence, the relationship cannot be determined
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Re: rt > 0 [#permalink]  18 Aug 2020, 14:14
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nishita84 wrote:
Simple method, always works - assume (r,t) = (1,1) and (r,t) = (-1,-1)

since the ans from both the cases are disparate, we can conclude that D is the ans.

Yes plugging in numbers is usually simpler. I always try to display an alternative algebraic solution as another tool in your toolset. You’ll come up with some QC problems where plugging in numbers will actually take longer or actually get you the wrong answer by not putting in all the cases (this especially happens with inequality or absolute value QC problems)

One good thing about algebraic solutions that it applies for all cases, some QC problems are simply just set up for them, and sometimes they are faster approach.

This problem is definitely is not suited for this type of solution for many, but if you work this part of your brain and try to find an algebraic solution for QC problems with variables, it will come in as a handy resource when you’re strapped for time and you can look at the problem algebraically without doing much work and solve the problem.
Re: rt > 0   [#permalink] 18 Aug 2020, 14:14
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