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A certain musical scale has 13 notes, each having a

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A certain musical scale has 13 notes, each having a [#permalink]  22 Nov 2004, 09:02
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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*root 2
b. 440*root 2^7
c. 440*root 2^12
d. 440*12root 2^7
e. 440*7root2^12

Pls pls explain
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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.
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This is the only PS from the OG I did not understand
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Paul wrote:
This is the only PS from the OG I did not understand

I'm with you, Paul, it took me a while to get it, too...
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Did this type of sum when i was in the 11th standard for physics must say its more of a physics problem.
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It is basically Geometric Progression...
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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.
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Took me a white to get it [#permalink]  23 Nov 2004, 17:38
It took me about 3.5 minutes to get the answer, even though I did get it right. I had to think about it for a few seconds then write down what I knew in order to see the solution. How difficult is this problem in relation to the ones I'm likely to see on the test if I'm aiming for 710-750?
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Re: Took me a white to get it [#permalink]  23 Nov 2004, 17:54
toddmartin wrote:
It took me about 3.5 minutes to get the answer, even though I did get it right. I had to think about it for a few seconds then write down what I knew in order to see the solution. How difficult is this problem in relation to the ones I'm likely to see on the test if I'm aiming for 710-750?

Do not worry too much. I think this type of question does not appear unless you have 51 in quant. You can definitely break the 700 barrier without reaching 51 quant as long as you have a good verbal score also.
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I had no idea how to start it so I just
wrote down what I had and tried reorganizing
it.

440 * k = 2nd tone
440 * k * k = 3rd tone
.
.
440 * k ^6 = 7th tone.

At this point, I was wondering why none of
power, but after I wrote 440 (k^12) = 880
and started simplifying, I stumbled on the answer.

Congratulations to ruhi160184.
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Re: [#permalink]  09 Nov 2011, 03:03
ruhi wrote:
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???
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Re: A certain musical scale has 13 notes, each having a [#permalink]  09 Nov 2011, 19:01
Slightly different way of approaching it...

Given

n_1 = 440

n_{13} = 880

n_i = 440(1+k)^{i-1}

Solve for n_7

Given that: n_7 = 440(1+k)^{6}

n_{13} = 440(1+k)^{12}

880 = (440(1+k)^{6})(1+k)^{6}

440*880 = (440(1+k)^{6})(440(1+k)^{6})

2*440^2 = (n_7)^2

n_7 = \sqrt{2}(440)
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Re: Re: [#permalink]  10 Nov 2011, 21:32
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Expert's post
siddhans wrote:
ruhi wrote:
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???

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.
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Re: Re:   [#permalink] 10 Nov 2011, 21:32
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