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505-555 (Easy)|   Fractions and Ratios|               
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vksunder
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In the formula V = 1/(2r)^3, if r is halved, then V is multi Updated on: 21 Feb 2014, 05:10
Originally posted by vksunder on 15 Jul 2008, 12:39.
Last edited by Bunuel on 21 Feb 2014, 05:10, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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B
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75% (01:05) correct 25% (01:23) wrong based on 3126 sessions

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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
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B
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In the formula V = 1/(2r)^3, if r is halved, then V is multi 21 Feb 2014, 05:28
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benjamin09
Can someone solve this question with an explanation?

Thanks
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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

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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jallenmorris
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In the formula V = 1/(2r)^3, if r is halved, then V is multi 15 Jul 2008, 12:42
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B - about had that backwards :( :shock:

Make r = 1/2, V = 1

Make r = 1, v = 1/8
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mbawaters
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In the formula V = 1/(2r)^3, if r is halved, then V is multi 15 Jul 2008, 13:00
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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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In the formula V = 1/(2r)^3, if r is halved, then V is multi 15 Jul 2008, 13:05
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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
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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In the formula V = 1/(2r)^3, if r is halved, then V is multi 15 Jul 2008, 13:09
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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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In the formula V = 1/(2r)^3, if r is halved, then V is multi 15 Jul 2008, 15:30
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vksunder
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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I think its asking us to reflect the changes in the value of V if r is halved!
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In the formula V = 1/(2r)^3, if r is halved, then V is multi 15 Jul 2008, 23:22
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vksunder
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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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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In the formula V = 1/(2r)^3, if r is halved, then V is multi 21 Feb 2014, 03:14
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Can someone solve this question with an explanation?

Thanks
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In the formula V = 1/(2r)^3, if r is halved, then V is multi 21 Feb 2014, 07:34
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Thanks a lot Bunuel!
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In the formula V = 1/(2r)^3, if r is halved, then V is multi 19 Mar 2018, 10:54
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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:
Show Spoiler
B


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In the formula V = 1/(2r)^3, if r is halved, then V is multi 19 Mar 2018, 12:49
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benjamin09
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.
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why answer is B amd not D ? :?
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In the formula V = 1/(2r)^3, if r is halved, then V is multi 05 Apr 2018, 11:11
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vksunder
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
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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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In the formula V = 1/(2r)^3, if r is halved, then V is multi 29 Jul 2019, 21:51
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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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In the formula V = 1/(2r)^3, if r is halved, then V is multi 08 Oct 2021, 10:21
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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
Bunuel
benjamin09
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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In the formula V = 1/(2r)^3, if r is halved, then V is multi 05 Jul 2023, 07:31
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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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In the formula V = 1/(2r)^3, if r is halved, then V is multi 23 Oct 2024, 02:21
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Given the formula:
V = 1 / (2r)^3
1. If r is halved, substitute r/2 for r:
V' = 1 / (2 * (r/2))^3 = 1/r^3
= 8 * 1 / (2r)^3 = 8V
2. Therefore, V is multiplied by 8.
Final Answer:
(B) 8
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In the formula V = 1/(2r)^3, if r is halved, then V is multi 09 Nov 2025, 10:11
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