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Intern  Joined: 17 Jun 2013
Posts: 4
Schools: ISB '14
Re: A certain musical scale has has 13 notes, each having a different freq  [#permalink]

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Pushpinder wrote:
Lowest frequency = 440
Highest frequency = 880
Lowest frequency (n) ^12 = Highest frequency
N^12 = 2 ---------------- 1
7th note = Lowest Frequency x (n)^6
7th note = 440 x (2)^6/12

Hence the answer is A

Pushpinder Ji I couldn't understand from here. Can u tell me please

Originally posted by dasikasuneel on 12 Jul 2013, 22:40.
Last edited by Bunuel on 13 Jul 2013, 00:28, edited 2 times in total.
Edited.
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Re: A certain musical scale has has 13 notes, each having a different freq  [#permalink]

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Bunuel wrote:
dasikasuneel wrote:
Pushpinder 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 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)

Lowest frequency = 440
Highest frequency = 880
Lowest frequency (n) ^12 = Highest frequency
N^12 = 2 ---------------- 1
7th note = Lowest Frequency x (n)^6
7th note = 440 x (2)^6/12

Hence the answer is A

Pushpinder Ji I couldn't understand from here. Can u tell me please

1st = $$440$$
2nd = $$440k$$
3rd = $$440k^2$$
...
7th = $$440k^6$$
...
13th = $$440k^{12}=2*440=880$$ --> $$440k^{12}=880$$ --> $$k^{12}=2$$ --> $$k=\sqrt{2}$$.

Thus, 7th = $$440k^6=440(\sqrt{2})^6=440\sqrt{2}$$.

Answer: A.

Hope it's clear.

Hi Bunuel ,

It says that the ratio of ratio of a frequency to the next higher frequency is a fixed constant.
Doesnt that mean f1/f2 = k ??

Just a little lost.

Cheers
HeirApparent.
Math Expert V
Joined: 02 Sep 2009
Posts: 57026
Re: A certain musical scale has has 13 notes, each having a different freq  [#permalink]

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heirapparent wrote:
Bunuel wrote:
dasikasuneel wrote:
Pushpinder Ji I couldn't understand from here. Can u tell me please

1st = $$440$$
2nd = $$440k$$
3rd = $$440k^2$$
...
7th = $$440k^6$$
...
13th = $$440k^{12}=2*440=880$$ --> $$440k^{12}=880$$ --> $$k^{12}=2$$ --> $$k=\sqrt{2}$$.

Thus, 7th = $$440k^6=440(\sqrt{2})^6=440\sqrt{2}$$.

Answer: A.

Hope it's clear.

Hi Bunuel ,

It says that the ratio of ratio of a frequency to the next higher frequency is a fixed constant.
Doesnt that mean f1/f2 = k ??

Just a little lost.

Cheers
HeirApparent.

Does not matter how you write: f1/f2=constant --> f2/f1=1/constant.
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Posts: 19
Re: A certain musical scale has has 13 notes, each having a different freq  [#permalink]

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In a geometric progression, the median is the geometric mean given by SQRT (First * Last).
Here, First is 440, Last is 2*440 = 880 and 7th Note is the median, so it's value = SQRT (440*880) = SQRT (440*440*2) = 440*SQRT(2)
A is correct.
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Re: A certain musical scale has has 13 notes, each having a different freq  [#permalink]

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Hello ,

This is the solution if we take f1/f2=k

Given: f1= 440, f 13 =2(440)=880

Also f1/f2 =f2/f3 =f3/f4 =f4/f5 =f5/f6 =f6/f7 =f7/f8 =f8/f9 =f9/f10 =f10/f11 =f11/f12 =f12/f13 = K ( Some constant)

need to find f7?

we can write f2/f1= 1/k
Acc to GP fn= f1(1/k)^n-1
Then f7= 440(1/k)^6

and f13= 440(1/k)^12

880=440 (1/k)^12
(1/k)^12= 2

{(1/K)^6}^2= 2 or
(1/k)^6= √ ( 2)

Substitute the value of (1/k)^6 in equation

So f7= 440√ 2

@Banuel is this also a correct approach.

Choice A
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Abhimanyu
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Re: A certain musical scale has has 13 notes, each having a different freq  [#permalink]

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The sequence is in GP a= 440
Ar^n-1 =term of GP
Now 2ar^1-1 = ar^13-1
2= r^12--------------------1
7th term = ar^6
440*root(2)
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Re: A certain musical scale has has 13 notes, each having a different freq  [#permalink]

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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 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)

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.

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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Re: A certain musical scale has has 13 notes, each having a different freq  [#permalink]

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Top Contributor
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. $$440 * \sqrt{2^7}$$

E. $$440 * \sqrt{2^{12}}$$

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 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) = 440√2

Answer = A

Cheers,
Brent
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A certain musical scale has has 13 notes, each having a different freq  [#permalink]

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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. $$440 * \sqrt{2^7}$$

E. $$440 * \sqrt{2^{12}}$$

My reasoning:

$$a_1 = 440$$ and $$a_13 = 880$$

Since we know the frequencies are increasing with a constant ratio we know this is a geometric sequence.

Thus $$a_{13} = a_1 * r^{13-1}$$

We can solve for r which is $$2^{\frac{1}{12}}$$

With r we can solve for the 7th term

$$a_7 = 440 * 2^{\frac{1}{12} * 6}$$

$$a_7 = 440 * \sqrt 2$$
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Re: A certain musical scale has has 13 notes, each having a different freq  [#permalink]

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Bunuel VeritasKarishma isn't the gp for the first 12 numbers of the sequence? how/why are we assuming that 880 is the 13th and the last term of the gp when the question says that the gp is of the 1st 12 scale notes ? Re: A certain musical scale has has 13 notes, each having a different freq   [#permalink] 30 Mar 2019, 14:27

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