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In the equation 3r^{4}=(2/s), if r is doubled, then s must be

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In the equation 3r^{4}=(2/s), if r is doubled, then s must be  [#permalink]

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New post Updated on: 14 Jun 2015, 08:35
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E

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In the equation \(3r^{4}\)=\(\frac{2}{s}\), if r is doubled, the s must be multiplied by:

A.\(\frac{1}{16}\)

B. \(\frac{1}{8}\)

C.\(\frac{1}{4}\)

D. 4

E. 16

Originally posted by petu on 14 Jun 2015, 08:26.
Last edited by petu on 14 Jun 2015, 08:35, edited 1 time in total.
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Re: In the equation 3r^{4}=(2/s), if r is doubled, then s must be  [#permalink]

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New post 15 Jun 2015, 23:22
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irus wrote:
why 1/16 can someone explain pls


\(3*r^4 = \frac{2}{s}\)
\(s= \frac{2}{3*r^4}\) (multiply by s and divide by 3r^4)

What happens when r becomes 2r.

You get \(\frac{2}{3(2*r)^4} = \frac{2}{3*16*r^4}\)

How is this new expression different from the old one? The new one = Old one * (1/16)

Hence, s gets multiplied by 1/16 when r becomes twice.
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Re: In the equation 3r^{4}=(2/s), if r is doubled, then s must be  [#permalink]

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New post 14 Jun 2015, 11:34
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Re: In the equation 3r^{4}=(2/s), if r is doubled, then s must be  [#permalink]

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New post 01 Nov 2015, 21:36
1
just plug in a number for r , like 1

Then solve for s , once you have s, all you have to do is double the value of r (which will now be: 2) , then solve for the new value of y
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Re: In the equation 3r^{4}=(2/s), if r is doubled, then s must be  [#permalink]

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New post 23 Feb 2018, 03:47
VeritasPrepKarishma wrote:
irus wrote:
why 1/16 can someone explain pls


\(3*r^4 = \frac{2}{s}\)
\(s= \frac{2}{3*r^4}\) (multiply by s and divide by 3r^4)

What happens when r becomes 2r.

You get \(\frac{2}{3(2*r)^4} = \frac{2}{3*16*r^4}\)

How is this new expression different from the old one? The new one = Old one * (1/16)

Hence, s gets multiplied by 1/16 when r becomes twice.


Hi Karishma,

I have a small doubt here , I kept r to the left and s to the right and worked at it that way , below is the working :

Given \(3 r^4 = \frac {2}{s}\)
now r is doubled

\((2r)^4 =\frac{2}{3s}\)
\(16r^4= \frac{2}{3s}\)

Now the above can be rearranged as \(3r^4= \frac{2}{16s}\)
on comparing this with the original it seemed that s is being multiplied by 16 when r is doubled , hence I chose E. What is the error in the logic here?
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Re: In the equation 3r^{4}=(2/s), if r is doubled, then s must be  [#permalink]

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New post 24 Feb 2018, 05:17
stne wrote:
VeritasPrepKarishma wrote:
irus wrote:
why 1/16 can someone explain pls


\(3*r^4 = \frac{2}{s}\)
\(s= \frac{2}{3*r^4}\) (multiply by s and divide by 3r^4)

What happens when r becomes 2r.

You get \(\frac{2}{3(2*r)^4} = \frac{2}{3*16*r^4}\)

How is this new expression different from the old one? The new one = Old one * (1/16)

Hence, s gets multiplied by 1/16 when r becomes twice.


Hi Karishma,

I have a small doubt here , I kept r to the left and s to the right and worked at it that way , below is the working :

Given \(3 r^4 = \frac {2}{s}\)
now r is doubled

\((2r)^4 =\frac{2}{3s}\)
\(16r^4= \frac{2}{3s}\)

Now the above can be rearranged as \(3r^4= \frac{2}{16s}\)
on comparing this with the original it seemed that s is being multiplied by 16 when r is doubled , hence I chose E. What is the error in the logic here?


s is in the denominator. You cannot say what it multiplies by.
To know "s is multiplied by ...", you need to separate out the s.
s = Old expression
New s = m * Old expression
You need the value of m.
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Re: In the equation 3r^{4}=(2/s), if r is doubled, then s must be   [#permalink] 24 Feb 2018, 05:17
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