stoolfi wrote:

A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. \(440 * \sqrt 2\)

B. \(440 * \sqrt {2^7}\)

C. \(440 * \sqrt {2^{12}}\)

D. \(440 * \sqrt[12]{2^7}\)

E. \(440 * \sqrt[7]{2^{12}}\)

Let k = the multiplier for each successive note. That is, each note is k TIMES the note before it.

Then we'll start

listing each note:

1st note =

440 cycles per second

2nd note = 440(k) cycles per second

3rd note = 440(k)(k) cycles per second

4th note = 440(k)(k)(k) cycles per second

.

.

.

7th note = 440(k)(k)(k)(k)(k)(k) =

440(k^6) cycles per second

.

.

.

13th note = 440(k)(k)(k)(k)(k)(k)(k)(k)(k)(k)(k)(k) =

440(k^12) cycles per second

We're told that

the highest frequency is twice the lowest.

In other words,

440(k^12) is twice as big as

440We can write:

440(k^12) = (2)

440Divide both sides by 440 to get:

k^12 = 2NOTE: Our goal is to find the frequency for the 7th note. In other words, we want to find the value of

440(k^6)Since

k^12 = 2, we can rewrite this as

(k^6)^2 = 2This means that k^6 = √2

So, the frequency of the 7th note =

440(k^6) = 440√2

Answer = A

Cheers,

Brent

_________________

Brent Hanneson – Founder of gmatprepnow.com