A certain musical scale has has 13 notes, each having a different freq

A quick gut check on this problem is that every choice but A is way more than 880. Logically the 7th note must have a smaller value(the problem also states 'lower' frequency) than than the 13th. If crunched for time, good guess would be A.

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

Say the notes are:

\(n_1, \ n_2, \ n_3, \ ..., \ n_{12}, n_{13}\)

We are told that "for each of the 12 lower frequencies, the ratio of a frequency to the next higher frequency is a fixed constant". So the ratio of 12th lowest frequency (\(n_{12}\)) to the next higher frequency (which would be \(n_{13}\)) is constant. Basically we are told that:

\(\frac{n_1}{n_2}= \frac{n_2}{n_3}= \frac{n_3}{n_4}=...=\frac{n_{12}}{n_{13}}=constant\) (notice that this perfectly follows info given in the stem: "for each of the 12 lower frequencies, the ratio of a frequency to the next higher frequency is a fixed constant").

We have a geometric progression with the first, the lowest term (\(n_1\)) equal to 440:

\(n_1=440\);
\(n_2=440k\);
\(n_3=440k^2\);
...
\(n_7=440k^6\);
...
\(n_{13}=440k^{12}\)

The stem also says that "the highest frequency is twice the lowest":

\(n_{13}=2*n_1\)

\(440k^{12}=2*440\)

\(k^{12}=2\);

\(k=\sqrt[12]{2}\).

The question asks to find the value of \(n_7\):

\(n_7=440k^6=440(\sqrt[12]{2})^6=440\sqrt{2}\).

Answer: A.

Hope it's clear.
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let the constant be k

F1 = 440

F2 = 440k

F3 = 440 k * k = 440 * k^2

F13= 440 * k^12

we know F13 = 2 *F1 = 2 * 440 = 880

880/440 = k^12

k = twelfth root of 2

for F7...

F7 = 440 * k^6 ( as we wrote for F2 and F3)
F7 = 440 * (twelfth root of 2) ^ 6

F7 = 440 * sqrt (2)


Answer A

Let k be the constant.

Hence, f2/f1=k. given, f1=440, f13=highest frequency=2*440=880.

Also, f13=440*k^12

therefore, 440*k^12=880 , or k^12=2, or (k^6)^2=2 or k^6=root 2.

f7=440k^6= 440*root2.

The question stem says that : the ratio of frequency to next higer frequency is fixed constant ...

Doesnt that mean F1/F2 = k

F2/F3 = k^2 and something like that ..... ??? and not F2/F1 = k which is F2 = kF1???

Siddhans,

It says the ratio of a frequency to the next higher frequency is a constant. f2/f1=f3/f2=f4/f3=.....=fixed number. This is an example of a geometric progression.

Here is a video explanation:

https://www.gmatquantum.com/og10-journal ... ition.html

Dabral

Responding to a pm:

The statement, "the ratio of a frequency to the next higher frequency is a fixed constant." means that the ratio of two consecutive frequencies is always the same.
It doesn't matter how you write it.
You can say F1/F2 = F2/F3 = F3/F4 = ... = F12/F13 = k
You can also say F2/F1 = F3/F2 = F4/F3 = ... = F13/F12 = k
The two constants are different. Your k will be reciprocal of each other in the two cases. You can follow any approach. You will get the value of k accordingly.

Mind you, F2/F3 is also k, not k^2. The ratio remains constant. It is a geometric progression. The ratio between any two consecutive values is always the same.

Why do you multiply and not add?

Like
1. 440
2. 440 + k and so on?

Because we are given that the ratio of a frequency to the next higher frequency is a fixed constant: F2/F1=k.
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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.

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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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

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 ?

"For each of the 12 lower frequencies, the ratio of a frequency to the next higher frequency is a fixed constant."

Ratio of each of the 12 lower frequencies to the next higher frequency is fixed.

So ratio of the 1st frequency to the next higher frequency (which is 2nd) is fixed.
So ratio of the 2nd frequency to the next higher frequency (which is 3rd) is fixed.
...
So ratio of the 12th frequency to the next higher frequency (which is 13th) is fixed.

All 13 are in GP

Let's say you get a question like this and have NO CLUE what to do. Or maybe you have an idea, but you know it's going to take you a while and you're not sure you'd make it to the finish line. Or maybe you like getting right answers without having to do all the math because you train yourself to think like a test-writer, not just as a test-taker. Let's use the answer choices and deploy a little logic.

The smallest term is 440. The largest term is 880.
Answer choices B, C, and E are all larger than 880. We've done nothing at all, and in a matter seconds, we have it down to two.

The 7th term, huh? Why'd they pick the 7th? Is it because there's some symmetry in getting from 1 to 7 and then from 7 to 13 (with 7 half way between 1 and 13 and multiplying by \(\sqrt{2}\) each half)? That sure sounds like something a GMAC test-writer would do, and it sure would make answer choice A appealing. Also, answer choice D seems to suggest that if we plug the term number into \(440 * \sqrt[12]{2^7}\) in place of the 7, we'd get the term. But we can quickly tell that won't work for the first term since \(440 * \sqrt[12]{2^1}\) isn't 440, and it won't work for the last term, either, since \(440 * \sqrt[12]{2^{13}}\) isn't 880. I don't have much confidence in D, especially given that I like the potential symmetry of A.

Answer choice A.

@bunnel

in the question it is stated that for 12 of the lower frequencies ,

how can we conclude NOTE13/NOTE12 = K?????

NOTE12/NOTE11 = K is right

Good question.

Say the notes are:

\(n_1, \ n_2, \ n_3, \ ..., \ n_{12}, n_{13}\)

We are told that "for each of the 12 lower frequencies, the ratio of a frequency to the next higher frequency is a fixed constant". So the ratio of 12th lowest frequency (\(n_{12}\)) to the next higher frequency (which would be \(n_{13}\)) is constant. Basically we are told that:

\(\frac{n_1}{n_2}= \frac{n_2}{n_3}= \frac{n_3}{n_4}=...=\frac{n_{12}}{n_{13}}=constant\) (notice that this perfectly follows info given in the stem: "for each of the 12 lower frequencies, the ratio of a frequency to the next higher frequency is a fixed constant").

Hope it's clear.
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1st frequency = 440 c/m
13th frequency = 880 c/m

Now a = 440
Let the constant ratio be r
then 2nd frequency = ar
3rd frequency = ar^2
4th frequency = ar^3
..............
..............
13th frequency = ar^12

Now, ar^12 = 880
=> r^12 = 880/440 = 2
=> r = 12th root of 2

Hence 7th note would be ar^6 = 440 * (12th root of 2)^6
=> 440 * √2

Option A

I still didn't understand that we are given "the ratio of Frequency to the next higher Frequency" => F1, F2.... are frequencies then F1/F2 = k but people have used F2/F1 = k. Shouldn't that mean that "the ratio of Frequency to the previous higher Frequency"?

­
Your doubt is addressed here. 

Hope it helps.
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