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) 440sqrt(2)
(B) 440sqrt(2^7)
(C) 440sqrt(2^12)
(D) 440*12rt(2^7)
(E) 440*7rt(2^12)
Word problem
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- Brent@GMATPrepNow
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Let's let k = the multiplier for each successive note. That is, each note is k TIMES the note before it.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)
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
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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?
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:Let's let k = the multiplier for each successive note. That is, each note is k TIMES the note before it.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)
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
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Amr, You assumed that the frequency increases linearly but is not true. It increases exponentially. That is, your solution assumes that the difference between any frequency and the next one is constant whereas the question mentions that the ratio between a frequency and the next one is constant. So, Brent's answer is not equivalent to yours.
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:Let's let k = the multiplier for each successive note. That is, each note is k TIMES the note before it.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)
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
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That said, given the significant differences in magnitude between the answers and the fact that k is significantly less than 2, Amrab's approximation (using the arithmetic mean on the base instead of the exponent) wasn't that bad. Seems like a decent way to make an educated guess on test day.
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.
Brent@GMATPrepNow wrote:Let's let k = the multiplier for each successive note. That is, each note is k TIMES the note before it.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)
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
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Look at the wording again: For each of the 12 lower frequencies, the ratio of a frequency to the next higher frequency is a fixed constant.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.
So each frequency is defined in terms of its relationship to the next higher frequency. If the 13th is the highest, it can't have a relationship to a 14th frequency that doesn't exist. And if the 12th note is 440k^11, and the ratio of each note to the next higher frequency is 'k' then 13th/440k^11 = k; Solving for the 13th, we get 13th = 440k^12
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We are given that a certain musical scale has 13 notes, ordered from least to greatest. We also know that each next higher frequency is equal to the preceding frequency multiplied by some constant. Since the first frequency is 440 cycles per second, the second frequency is 440k, the third is 440k^2, the fourth is 440k^3...the seventh frequency is 440k^6, and the thirteenth frequency is 440k^12.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)
Since the highest frequency is twice the lowest, we can create the following equation:
440 x 2 = 440k^12
2 = k^12
(^12)√2 = k
Thus, the seventh frequency is 440((^12)√2)^6 = 440√2.
Answer: A
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