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# Is rst = 1 ? (1) r p t = s p t s (2) prt t = p t st

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Math Expert
Joined: 02 Sep 2009
Posts: 52296
Is rst = 1 ? (1) r p t = s p t s (2) prt t = p t st  [#permalink]

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31 Dec 2017, 11:10
00:00

Difficulty:

25% (medium)

Question Stats:

73% (01:14) correct 27% (01:48) wrong based on 70 sessions

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Is rst = 1 ?

(1) $$r\sqrt{t} =\frac{s\sqrt{t}}{s}$$

(2) $$\frac{rt}{\sqrt{t}}=\frac{\sqrt{t}}{st}$$

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Joined: 07 Dec 2017
Posts: 868
Re: Is rst = 1 ? (1) r p t = s p t s (2) prt t = p t st  [#permalink]

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31 Dec 2017, 13:23
Bunuel wrote:
Is rst = 1 ?

(1) $$r\sqrt{t} =\frac{s\sqrt{t}}{s}$$

(2) $$\frac{rt}{\sqrt{t}}=\frac{\sqrt{t}}{st}$$

As this looks to be a very technical question (with many equations), we'll just solve it.
This is a Precise approach.

(1) We can cancel out $$s$$ from the right-hand side giving $$r\sqrt{t} =\sqrt{t}$$
As we don't know if $$t = 0$$ or not, we can't cancel it out but we can rewrite as $$r\sqrt{t} -\sqrt{t}=0$$
This simplifies to $$\sqrt{t}(r-1)=0$$ meaning that $$t = 0$$ or $$r = 1$$.
Insufficient!

(2) Since both $$s$$ and $$t$$ are in the denominator, they can't be equal to 0 so we can multiply by them.
Multiplying by $$s$$ and by $$\sqrt{t}$$ gives $$rst = \frac{t}{t}=1$$
Sufficient!

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Joined: 30 Oct 2017
Posts: 9
Re: Is rst = 1 ? (1) r p t = s p t s (2) prt t = p t st  [#permalink]

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29 Jan 2018, 13:22
DavidTutorexamPAL wrote:
Bunuel wrote:
Is rst = 1 ?

(1) $$r\sqrt{t} =\frac{s\sqrt{t}}{s}$$

(2) $$\frac{rt}{\sqrt{t}}=\frac{\sqrt{t}}{st}$$

As this looks to be a very technical question (with many equations), we'll just solve it.
This is a Precise approach.

(1) We can cancel out $$s$$ from the right-hand side giving $$r\sqrt{t} =\sqrt{t}$$
As we don't know if $$t = 0$$ or not, we can't cancel it out but we can rewrite as $$r\sqrt{t} -\sqrt{t}=0$$
This simplifies to $$\sqrt{t}(r-1)=0$$ meaning that $$t = 0$$ or $$r = 1$$.
Insufficient!

(2) Since both $$s$$ and $$t$$ are in the denominator, they can't be equal to 0 so we can multiply by them.
Multiplying by $$s$$ and by $$\sqrt{t}$$ gives $$rst = \frac{t}{t}=1$$
Sufficient!

Can you please explain how you multiplied (2)? Cross multiply?

"2) Since both ss and tt are in the denominator, they can't be equal to 0 so we can multiply by them.
Multiplying by ss and by t√t gives rst=tt=1rst=tt=1
Sufficient!" I'm not seeing it, thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 52296
Re: Is rst = 1 ? (1) r p t = s p t s (2) prt t = p t st  [#permalink]

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29 Jan 2018, 20:13
mekoner wrote:
DavidTutorexamPAL wrote:
Bunuel wrote:
Is rst = 1 ?

(1) $$r\sqrt{t} =\frac{s\sqrt{t}}{s}$$

(2) $$\frac{rt}{\sqrt{t}}=\frac{\sqrt{t}}{st}$$

As this looks to be a very technical question (with many equations), we'll just solve it.
This is a Precise approach.

(1) We can cancel out $$s$$ from the right-hand side giving $$r\sqrt{t} =\sqrt{t}$$
As we don't know if $$t = 0$$ or not, we can't cancel it out but we can rewrite as $$r\sqrt{t} -\sqrt{t}=0$$
This simplifies to $$\sqrt{t}(r-1)=0$$ meaning that $$t = 0$$ or $$r = 1$$.
Insufficient!

(2) Since both $$s$$ and $$t$$ are in the denominator, they can't be equal to 0 so we can multiply by them.
Multiplying by $$s$$ and by $$\sqrt{t}$$ gives $$rst = \frac{t}{t}=1$$
Sufficient!

Can you please explain how you multiplied (2)? Cross multiply?

"2) Since both ss and tt are in the denominator, they can't be equal to 0 so we can multiply by them.
Multiplying by ss and by t√t gives rst=tt=1rst=tt=1
Sufficient!" I'm not seeing it, thanks.

Yes, you can cross-multiply to get: rst^2 = t. Then reduce by t to get rst = 1.
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Joined: 31 Jul 2017
Posts: 518
Location: Malaysia
GMAT 1: 700 Q50 V33
GPA: 3.95
WE: Consulting (Energy and Utilities)
Re: Is rst = 1 ? (1) r p t = s p t s (2) prt t = p t st  [#permalink]

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01 Feb 2018, 00:24
Bunuel wrote:
Is rst = 1 ?

(1) $$r\sqrt{t} =\frac{s\sqrt{t}}{s}$$

(2) $$\frac{rt}{\sqrt{t}}=\frac{\sqrt{t}}{st}$$

Statement I:

$$\sqrt{t} (r-1) = 0$$ ---> $$t = 0 , r = 1$$

Statement II:

$$r\sqrt{t} - 1/s\sqrt{t} = 0$$
$$rst = 1.$$

So, B.
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Re: Is rst = 1 ? (1) r p t = s p t s (2) prt t = p t st &nbs [#permalink] 01 Feb 2018, 00:24
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