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The variable x is inversely proportional to the square of the variable [#permalink]
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18 Sep 2015, 19:04
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The variable x is inversely proportional to the square of the variable y. If y is divided by 3a, then x is multiplied by which of the following? a. 1/9a b. 1/9a^2 c. 1/3a d. 9a e. 9a^2
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Re: The variable x is inversely proportional to the square of the variable [#permalink]
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06 Nov 2015, 00:23
Tried this question
not sure how to set it up when the say inversely proportional
initially  did x= 1/y^2
But this yielded the wrong answer,
I then re examined the question and tried y^2/x =1
Which I then got y = SQR(X)
subbing into y/3a I got answer E)
not sure what is the correct way to do this. thanks!



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Re: The variable x is inversely proportional to the square of the variable [#permalink]
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07 Nov 2015, 07:11
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Let , \(x = k *\frac{1}{y^2}\) , Where K is constant. \(x = k * \frac{1}{(3a)^2}\) = \(k * \frac{1}{9a^2}\) => \(k = (9a^2)x\) Ans E.
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Re: The variable x is inversely proportional to the square of the variable [#permalink]
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13 Dec 2015, 09:32
GMATDemiGod wrote: Tried this question
not sure how to set it up when the say inversely proportional
initially  did x= 1/y^2
But this yielded the wrong answer,
I then re examined the question and tried y^2/x =1
Which I then got y = SQR(X)
subbing into y/3a I got answer E)
not sure what is the correct way to do this. thanks! hi, u missed a point, y is divided by 3a, i.e y^2 is divided by 9a^2.....
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Re: The variable x is inversely proportional to the square of the variable [#permalink]
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13 Dec 2015, 09:32
GMATDemiGod wrote: Tried this question
not sure how to set it up when the say inversely proportional
initially  did x= 1/y^2
But this yielded the wrong answer,
I then re examined the question and tried y^2/x =1
Which I then got y = SQR(X)
subbing into y/3a I got answer E)
not sure what is the correct way to do this. thanks! hi, u missed a point, y is divided by 3a, i.e y^2 is divided by 9a^2.....
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Re: The variable x is inversely proportional to the square of the variable [#permalink]
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04 Jan 2017, 06:32
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pmklings wrote: The variable x is inversely proportional to the square of the variable y. If y is divided by 3a, then x is multiplied by which of the following?
a. 1/9a
b. 1/9a^2
c. 1/3a
d. 9a
e. 9a^2 Here is a post on inverse variation: thevariablexisinverselyproportionaltothesquareofthevariable205761.htmlNow read the following: x * y^2 = k x' * (y/3a)^2 = k = x * y^2 x' * (y^2)/9a^2 = x * y^2 x' = x * 9a^2 Answer (E)
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Re: The variable x is inversely proportional to the square of the variable [#permalink]
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18 Jan 2017, 15:42
VeritasPrepKarishma wrote: pmklings wrote: Here is a post on inverse variation: thevariablexisinverselyproportionaltothesquareofthevariable205761.htmlNow read the following: x * y^2 = k x' * (y/3a)^2 = k = x * y^2 x' * (y^2)/9a^2 = x * y^2 x' = x * 9a^2 Answer (E) Sorry I don't understand, Can anyone explain. Thanks The variable x is inversely proportional to the square of the variable y ??? Why x*y^2 ?? I wrote as \(\frac{1}{x}\) = y^2
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Re: The variable x is inversely proportional to the square of the variable [#permalink]
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18 Jan 2017, 17:36
VeritasPrepKarishma wrote: VeritasPrepKarishma Did you man to post another link? This is the problem being discussed.



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Re: The variable x is inversely proportional to the square of the variable [#permalink]
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18 Jan 2017, 18:09
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Even i selected B as the answer. When the question says "x is multiplied by" it seems to ask for the constant that is multiplies X to keep the inverse proportion valid, but it really is asking for by what factor does the new value of X differ from the previous one which is Option E. xy^2 = K or X = K/y^2 Xy^2/9a = K => X(new) = 9a (K/y^2) from above = 9a(X(old)) = X(old) * 9a i.e X (new) = X(old) * 9a => Option E I have this problem of not understanding/misinterpreting the question sometimes :'( :'(
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Re: The variable x is inversely proportional to the square of the variable [#permalink]
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19 Jan 2017, 00:56
raeann105 wrote: VeritasPrepKarishma wrote: VeritasPrepKarishma Did you man to post another link? This is the problem being discussed. Oops! Yes, it is this one: https://www.veritasprep.com/blog/2013/0 ... inversely/
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Re: The variable x is inversely proportional to the square of the variable [#permalink]
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19 Jan 2017, 01:00
kanusha wrote: VeritasPrepKarishma wrote: pmklings wrote: Here is a post on inverse variation: thevariablexisinverselyproportionaltothesquareofthevariable205761.htmlNow read the following: x * y^2 = k x' * (y/3a)^2 = k = x * y^2 x' * (y^2)/9a^2 = x * y^2 x' = x * 9a^2 Answer (E) Sorry I don't understand, Can anyone explain. Thanks The variable x is inversely proportional to the square of the variable y ??? Why x*y^2 ?? I wrote as \(\frac{1}{x}\) = y^2 You can write it as k/x = y^2 which is the same as k = x * y^2 Note that it is not necessary that constant is 1. k could take any value. So when you equate, you need to use k, not 1. Check out the post for which the link is given above.
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Re: The variable x is inversely proportional to the square of the variable [#permalink]
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19 Jan 2017, 02:21
kanusha wrote: VeritasPrepKarishma wrote: pmklings wrote: Here is a post on inverse variation: thevariablexisinverselyproportionaltothesquareofthevariable205761.htmlNow read the following: x * y^2 = k x' * (y/3a)^2 = k = x * y^2 x' * (y^2)/9a^2 = x * y^2 x' = x * 9a^2 Answer (E) Sorry I don't understand, Can anyone explain. Thanks The variable x is inversely proportional to the square of the variable y ??? Why x*y^2 ?? I wrote as \(\frac{1}{x}\) = y^2 Hi , The variable x is inversely proportional to the square of the variable y ??? The above statement can be written as follows: \(x \propto \frac{1}{y^{2}} = \frac{k}{y^{2}}\) , where \(k\) is a proportionality constant. or you can wtite it as \(x y^{2} = k\). Hope this helps.



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