Can anyone help with a SAT practice problem?

[max]

after the formula V = 4/3*pi*r^3 has been evaluated for some positive value of r, the formula is again evaluated using one half of the original value of r. The new value of V is what fractional part of the original value of V?

I don't understand what I'm suppose to do.
Please, do not just post the answer but explain how to come to the correct idea of what to do. Also, could you point out the key words?

Thanks a bunch!

 

[morson]

V = \(\displaystyle \frac{4}{3}\ \pi\ r^3\) has been evaluated for some positive value of r, the formula is again evaluated using one half of the original value of r. The new value of V is what fractional part of the original value of V?

Key words are in bold.

Using "one half of the original value of r," \(\displaystyle V' = \frac{4}{3}\ \pi\ (1/2 r)^3\) = \(\displaystyle \frac{1}{6}\ \pi\ r^3\), where V' is the new volume.

Now, \(\displaystyle \frac{V}{V'}\ = 8\)

Finish.

 

[soroban]

Hello, max!

After the formula \(\displaystyle V\:=\:\frac{4}{3}\pi r^3\) has been evaluated for some positive value of \(\displaystyle r\),
the formula is again evaluated using one half of the original value of \(\displaystyle r\).
The new value of \(\displaystyle V\) is what fractional part of the original value of \(\displaystyle V\)?

The original value is: \(\displaystyle \:V_o \:=\:\frac{4}{3}\pi r^3\)

We use half the value of \(\displaystyle r:\;\frac{r}{2}\)

The new value is: \(\displaystyle \:V_1 \:=\:\frac{4}{3}\pi\left(\frac{r}{2}\right)^3 \:=\:\frac{1}{6}\pi r^3\)

How does this compare with the original \(\displaystyle V\)?

Make a fraction: \(\displaystyle \L\:\frac{V_1}{V_o} \:=\:\frac{\frac{1}{6}\pi r^3}{\frac{4}{3}\pi r^3} \:=\:\frac{1}{8}\)

 

[max]

Thanks

Thanks a bunch! I understand it.