Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 17 Jun 2013
Posts: 4

Re: A certain musical scale has has 13 notes, each having a different freq
[#permalink]
Show Tags
Updated on: 13 Jul 2013, 00:28
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.



Intern
Joined: 17 Jul 2013
Posts: 6
Location: United States
Concentration: Operations, Strategy
GPA: 3.18
WE: Engineering (Telecommunications)

Re: A certain musical scale has has 13 notes, each having a different freq
[#permalink]
Show Tags
26 Jul 2013, 05:50
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 = \(440k^{12}=2*440=880\) > \(440k^{12}=880\) > \(k^{12}=2\) > \(k=\sqrt[12]{2}\). Thus, 7th = \(440k^6=440(\sqrt[12]{2})^6=440\sqrt{2}\). 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
Joined: 02 Sep 2009
Posts: 60647

Re: A certain musical scale has has 13 notes, each having a different freq
[#permalink]
Show Tags
26 Jul 2013, 06:40
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 = \(440k^{12}=2*440=880\) > \(440k^{12}=880\) > \(k^{12}=2\) > \(k=\sqrt[12]{2}\). Thus, 7th = \(440k^6=440(\sqrt[12]{2})^6=440\sqrt{2}\). 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.
_________________



Intern
Joined: 02 Mar 2010
Posts: 19

Re: A certain musical scale has has 13 notes, each having a different freq
[#permalink]
Show Tags
24 Sep 2013, 20:12
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.



Intern
Joined: 14 Oct 2016
Posts: 35
Location: India
WE: Sales (Energy and Utilities)

Re: A certain musical scale has has 13 notes, each having a different freq
[#permalink]
Show Tags
07 Sep 2017, 08:31
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)^n1 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
_________________



Director
Joined: 14 Nov 2014
Posts: 584
Location: India
GPA: 3.76

Re: A certain musical scale has has 13 notes, each having a different freq
[#permalink]
Show Tags
07 Sep 2017, 10:42
The sequence is in GP a= 440 Ar^n1 =term of GP Now 2ar^11 = ar^131 2= r^121 7th term = ar^6 440*root(2)



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2806

Re: A certain musical scale has has 13 notes, each having a different freq
[#permalink]
Show Tags
11 Dec 2017, 18:17
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
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4226
Location: Canada

Re: A certain musical scale has has 13 notes, each having a different freq
[#permalink]
Show Tags
17 Jan 2018, 17:01
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. \(440 * \sqrt[12]{2^7}\)
E. \(440 * \sqrt[7]{2^{12}}\) 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 440We can write: 440(k^12) = (2) 440Divide both sides by 440 to get: k^12 = 2NOTE: 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 = 2This means that k^6 = √2 So, the frequency of the 7th note = 440(k^6) = 440√2 Answer = A Cheers, Brent
_________________
Test confidently with gmatprepnow.com



Manager
Joined: 22 Sep 2018
Posts: 237

A certain musical scale has has 13 notes, each having a different freq
[#permalink]
Show Tags
06 Feb 2019, 21:01
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. \(440 * \sqrt[12]{2^7}\)
E. \(440 * \sqrt[7]{2^{12}}\) 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 \(a_{13} = a_1 * r^{131}\) We can solve for r which is \(2^{\frac{1}{12}}\) With r we can solve for the 7th term \(a_7 = 440 * 2^{\frac{1}{12} * 6}\) \(a_7 = 440 * \sqrt 2\)



Manager
Joined: 04 Dec 2015
Posts: 147
WE: Operations (Commercial Banking)

Re: A certain musical scale has has 13 notes, each having a different freq
[#permalink]
Show Tags
30 Mar 2019, 14:27
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 ?



Intern
Joined: 13 Nov 2019
Posts: 2

Re: A certain musical scale has has 13 notes, each having a different freq
[#permalink]
Show Tags
05 Dec 2019, 02:41
Praetorian wrote: stoolfi wrote: On another thread, someone with a very good GMAT score said he couldn't really understand this question. It is posted for your perusal:
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)
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 Could you please explain how you got 440 * sqrt (2) from 440 * (twelfth root of 2) ^ 6? Thank you very much! Posted from my mobile device




Re: A certain musical scale has has 13 notes, each having a different freq
[#permalink]
05 Dec 2019, 02:41



Go to page
Previous
1 2
[ 31 posts ]



