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In the formula V = 1/(2r)^3, if r is halved, then V is multi

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In the formula V = 1/(2r)^3, if r is halved, then V is multi  [#permalink]

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New post Updated on: 21 Feb 2014, 04:10
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In the formula V = 1/(2r)^3, if r is halved, then V is multiplied by

A. 64
B. 8
C. 1
D. 1/8
E. 1/64

Originally posted by vksunder on 15 Jul 2008, 11:39.
Last edited by Bunuel on 21 Feb 2014, 04:10, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: In the formula V= 1/(2r)^3, if r is halved, then V is  [#permalink]

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New post 21 Feb 2014, 04:28
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benjamin09 wrote:
Can someone solve this question with an explanation?

Thanks


In the formula V = 1/(2r)^3, if r is halved, then V is multiplied by

A. 64
B. 8
C. 1
D. 1/8
E. 1/64

Given: \(V = \frac{1}{(2r)^3}=\frac{1}{8r^3}\)

Now, halved r is r/2, substitute r/2 instead of r: \(V' = \frac{1}{(2*\frac{r}{2})^3}=\frac{1}{r^3}\).

\(V =\frac{1}{8r^3}\) should be multiplied by 8 to get \(V'=\frac{1}{r^3}\).

Answer: B.

Or:
When r=2, then \(V = \frac{1}{(2r)^3}=\frac{1}{64}\).
When r=1, then \(V' = \frac{1}{(2r)^3}=\frac{1}{8}\).

1/64 should be multiplied by 8 to get 1/8.

Answer: B.

Similar question to practice: if-the-function-q-is-defined-by-the-formula-q-5w-4x-z-131051.html

Hope it helps.
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Re: PS: formula V= 1/(2r)^3  [#permalink]

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New post 15 Jul 2008, 11:42
B - about had that backwards :( :shock:

Make r = 1/2, V = 1

Make r = 1, v = 1/8
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Re: PS: formula V= 1/(2r)^3  [#permalink]

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New post 15 Jul 2008, 12:00
I always mess up these kinds...
Say r=2 => V1 = 1/64
when r=1; V2 = 1/8
V2 = 8*V1.
B?
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Re: PS: formula V= 1/(2r)^3  [#permalink]

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New post 15 Jul 2008, 12:05
I totally agree with you. I mess them up to and about did this one. I saw that if I make 2r = 1, then r=1/2 so to the power of 3 doesn't matter, it's still 1/1. Then I did r =1 (double it) so then V = 1/8. So I was thinking oh, go from 1 to 1/8 multiply by 1/8...but I had worked the problem backwards. There are tricks the authors throw at us, and then there are tricks we bring to the test ourselves. We have to eliminate our own tricks. Easier said than done!

mbawaters wrote:
I always mess up these kinds...
Say r=2 => V1 = 1/64
when r=1; V2 = 1/8
V2 = 8*V1.
B?

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Re: PS: formula V= 1/(2r)^3  [#permalink]

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New post 15 Jul 2008, 12:09
So isnt the qtn essentially testing us on: if r is halved then what should V be multiplied by to retain the intial formula???
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Re: PS: formula V= 1/(2r)^3  [#permalink]

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New post 15 Jul 2008, 14:30
vksunder wrote:
So isnt the qtn essentially testing us on: if r is halved then what should V be multiplied by to retain the intial formula???

I think its asking us to reflect the changes in the value of V if r is halved!
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Re: PS: formula V= 1/(2r)^3  [#permalink]

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New post 15 Jul 2008, 22:22
vksunder wrote:
So isnt the qtn essentially testing us on: if r is halved then what should V be multiplied by to retain the intial formula???


Not really, it's asking what happens to V when r is halved. In this type of problems, just pick an easy value for r and solve both equations.
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Re: In the formula V= 1/(2r)^3, if r is halved, then V is  [#permalink]

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New post 21 Feb 2014, 02:14
Can someone solve this question with an explanation?

Thanks
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Re: In the formula V = 1/(2r)^3, if r is halved, then V is multi  [#permalink]

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New post 21 Feb 2014, 06:34
Thanks a lot Bunuel!
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Re: In the formula V = 1/(2r)^3, if r is halved, then V is multi  [#permalink]

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New post 19 Mar 2018, 09:54
Hi All,

This question is perfect for TESTing Values.

IF... R=2
V = (1/(2x2)^3 = 1/64

Halving R gives us R=1...
V = (1/2x1)^3 = 1/8

1/8 = 8/64...
so going from 1/64 to 8/64 means that V increased 8 times.

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Re: In the formula V = 1/(2r)^3, if r is halved, then V is multi  [#permalink]

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New post 19 Mar 2018, 11:49
Bunuel wrote:
benjamin09 wrote:
Can someone solve this question with an explanation?

Thanks


In the formula V = 1/(2r)^3, if r is halved, then V is multiplied by

A. 64
B. 8
C. 1
D. 1/8
E. 1/64

Given: \(V = \frac{1}{(2r)^3}=\frac{1}{8r^3}\)

Now, halved r is r/2, substitute r/2 instead of r: \(V' = \frac{1}{(2*\frac{r}{2})^3}=\frac{1}{r^3}\).

\(V =\frac{1}{8r^3}\) should be multiplied by 8 to get \(V'=\frac{1}{r^3}\).

Answer: B.

Or:
When r=2, then \(V = \frac{1}{(2r)^3}=\frac{1}{64}\).
When r=1, then \(V' = \frac{1}{(2r)^3}=\frac{1}{8}\).

1/64 should be multiplied by 8 to get 1/8.

Answer: B.

Similar question to practice: http://gmatclub.com/forum/if-the-functi ... 31051.html

Hope it helps.


why answer is B amd not D ? :?
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Re: In the formula V = 1/(2r)^3, if r is halved, then V is multi  [#permalink]

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New post 05 Apr 2018, 10:11
vksunder wrote:
In the formula V = 1/(2r)^3, if r is halved, then V is multiplied by

A. 64
B. 8
C. 1
D. 1/8
E. 1/64


Since (2r)^3 = 8r^3, we can rewrite the formula as V = 1/(8r^3).

Let’s substitute r/2 instead of r in the first equation: V = 1/(2(r/2))^3 = 1/r^3 = 8*1/(8r^3).

As we can see, when we halve r, V gets multiplied by 8.

Alternate Solution:

Let’s let the original value of r = 2. Substituting 2 for r into the original equation yields:

Original V = 1/[(2)(2)]^3,

Original V = 1/(4)^3

Original V = 1/64

Now we halve r, so it is now equal to 1. Substituting 1 for r gives us:

New V = 1/[(2)(1)]^3

New V = 1/ 8

We see that the original value of V (1/64), when multiplied by 8, gives us the new value of V:

(1/64) x 8 = 8/64 = 1/8

Answer: B
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Re: In the formula V = 1/(2r)^3, if r is halved, then V is multi &nbs [#permalink] 05 Apr 2018, 10:11
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