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

Final Answer:

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Re: In the formula V = 1/(2r)^3, if r is halved, then V is multi [#permalink]
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: https://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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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 [#permalink]
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In this question, the main point to note is that L.H.S = R.H.S

Original eq : V=1/(2r)^3

Now, when r is halved : 1/(2*r/2)^3 => 8/(2r)^3

Now, to make L.H.S = R.H.S we will have to multiply L.H.S with 8 as well to obtain the original equation.

Please correct me if I'm wrong.
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Re: In the formula V = 1/(2r)^3, if r is halved, then V is multi [#permalink]
We can say that V = k/r^3, whereas k=1/8 or k is some constant
or simplify V*r^3=constant

The key point to note in this type of equation is r is inversely proportional to V, so when r is halved, V is increased or when r is doubled then V will decrease. Why because we need to maintain a constant value.
So we can reject C,D,E before getting into calculation.

dave13 wrote:
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: https://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]
I tend to solve such questions without getting into calculations.

Here's my approach:

If r is halved aka divided by 2, then r^3 will need to be divided by 2^3 i.e. it will need be halved three times.

Because V is inversely proportional, V will also have to be multiplied by the same factor (2^3) to maintain the original value.
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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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