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

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New post 31 Dec 2017, 12:10
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A
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E

Difficulty:

  25% (medium)

Question Stats:

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

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Re: Is rst = 1 ? (1) r p t = s p t s (2) prt t = p t st  [#permalink]

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New post 31 Dec 2017, 14:23
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!

(B) is our answer.
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Re: Is rst = 1 ? (1) r p t = s p t s (2) prt t = p t st  [#permalink]

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New post 29 Jan 2018, 14:22
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!

(B) is our answer.



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.
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Re: Is rst = 1 ? (1) r p t = s p t s (2) prt t = p t st  [#permalink]

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New post 29 Jan 2018, 21:13
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!

(B) is our answer.



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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Re: Is rst = 1 ? (1) r p t = s p t s (2) prt t = p t st  [#permalink]

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New post 01 Feb 2018, 01:24
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, 01:24
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