Brent@GMATPrepNow wrote:

amanjena wrote:A certain musical scale 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√2
(B) 440√(2^7)
(C) 440√(2^12)
(D) 440*12√(2^7)
(E) 440*7√(2^12)

Let's 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 440
We can write: 440(k^12) = (2)440
Divide both sides by 440 to get: k^12 = 2

NOTE: 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 = 2
This means that k^6 = √2

So, the frequency of the 7th note = 440(k^6) = [spoiler]440√2 = A[/spoiler]

Cheers,
Brent

Amrabdelnaby wrote:Brent, I solved it using a very different approach. Please correct me if I am wrong.

I did 880-440 = 440 --> the difference between the 1st and last note

then to determine the 7th note: (440/12)x 6 = 660

660 is very close to 440 root 2, since root 2 is around 1.4

was i thinking right or did i just solve it by luck?

Brent@GMATPrepNow wrote:

amanjena wrote:A certain musical scale 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√2
(B) 440√(2^7)
(C) 440√(2^12)
(D) 440*12√(2^7)
(E) 440*7√(2^12)

Let's 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 440
We can write: 440(k^12) = (2)440
Divide both sides by 440 to get: k^12 = 2

NOTE: 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 = 2
This means that k^6 = √2

So, the frequency of the 7th note = 440(k^6) = [spoiler]440√2 = A[/spoiler]

Cheers,
Brent

Brent@GMATPrepNow wrote:

amanjena wrote:A certain musical scale 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√2
(B) 440√(2^7)
(C) 440√(2^12)
(D) 440*12√(2^7)
(E) 440*7√(2^12)

Let's 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 440
We can write: 440(k^12) = (2)440
Divide both sides by 440 to get: k^12 = 2

NOTE: 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 = 2
This means that k^6 = √2

So, the frequency of the 7th note = 440(k^6) = [spoiler]440√2 = A[/spoiler]

Cheers,
Brent

I am unable to solve as in the question they have specified as "For each of the 12 lower frequencies", in that case how can we take the same constant and multiply for 13th note. Shouldn't we stop at 12th note=440K^11.

amanjena wrote:A certain musical scale 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√2
(B) 440√(2^7)
(C) 440√(2^12)
(D) 440*12√(2^7)
(E) 440*7√(2^12)