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The variable x is inversely proportional to the square of the variable

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The variable x is inversely proportional to the square of the variable  [#permalink]

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New post 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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New post 22 Aug 2018, 21:02
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This is a classic question type that demonstrates a pattern common to many questions on the GMAT. Several of the approaches in this forum focus blindly on the math, but remember: the GMAT is a critical-thinking test. The tactics I will show you here will be useful for numerous questions, not just this one. For those of you preparing for the GMAT, my solution is going to walk through not just what the answer is, but how to strategically think about it. As a result, I will probably include some steps that I would normally just do in my head if it were the actual test, but I want to be as thorough as possible so you can see each step of the process. Ready? Here is the full "GMAT Jujitsu" for this question:

First, we need to make sure we understand the difference between the phrases "directly proportional" and "inversely proportional." Both of these terms describe relative relationships between two variables (or, potentially, relationships between entire chunks of equations.) For purposes of this discussion we will relate \(x\) to \(y\).

If \(x\) and \(y\) are "directly proportional", it means that the relationship between \(x\) and \(y\) can be represented by the equation \(x = Ky\), where \(K\) is a constant called the "coefficient of proportionality" or the "constant of variation." For example, the circumference of a circle is directly proportional to the circle's diameter (\(C = \pi d\)), with \(\pi\) being the "coefficient of proportionality." Some people oversimplify this rule by saying, "when \(x\) increases, so does \(y\)." However, that isn't always accurate. It is possible for \(K\) to be negative, meaning that as \(x\) increases, \(y\) would actually decrease. (Picture a line drawn in coordinate space with a negative slope and you can visualize this quickly. Negative linear slopes are still directly proportional.)

If \(x\) and \(y\) are "inversely proportional", it means that the relationship between \(x\) and \(y\) can be represented by \(xy = K\), with \(K\) still serving as the "coefficient of proportionality." Notice that in this case, in order for the product \(xy\) to always equal the constant, \(K\), as \(x\) increased, \(y\) would have to decrease proportionally, in effect "cancelling out" the change in \(x\). Another way to write this relationship would be \(x = K(\frac{1}{y})\). With the equation in this form, you should be able to see why they call it "inversely proportional" instead of "negatively proportional." \(x\) is proportional to the inverse of \(y\) (in other words, \(\frac{1}{y}\)), not the negative of \(y\). This relationship isn't linear, but is actually curved in coordinate space.

Now that we have the basics, let's take a look at this specific question. It states, "The variable \(x\) is inversely proportional to the square of the variable \(y\)." This means that:

\(x(y^2)=K\). Alternately, we can also visualize it as \(x = K(\frac{1}{y^2})\).

The problem then states that we will be manipulating \(y\) by dividing it by \(3a\). This will make \(y\) (and thus \(y^2\)) proportionally smaller. (Of course, we don't know what "\(a\)" is, but this is a good way to visualize what is happening.) Since \(x\) and \(y^2\) are inversely proportional, this means that anything we do to \(y^2\), we must adjust \(x\) in the proportionately opposite way to cancel out the change and keep \(x(y^2)\) equal to the constant, \(K\).

Thus, when we divide \(y\) by \(3a\), we change \(y^2\) to:

\((\frac{y}{3a})^2=\frac{y^2}{9a^2}\)

The "new" \(y^2\) is now divided by \(9a^2\). In order to reverse this change out, this means that \(x\) would need to be multiplied by something so that \(x(y^2)\) still equals the constant, \(K\). For purposes of visualization, I will call that something "?". Here is what it would look like mathematically:

\((x*?)*(\frac{y^2}{9a^2}) = x(y^2)\)

Solving for "\(?\)" allows us to cancel out \(x\) and \(y^2\) from both sides of the equation, and moves the \(9a^2\) in the denominator over to the other side.

\(?=9a^2\)

Thus, the factor by which we must multiply \(x\) by to maintain the "inversely proportional" relationship is \(9a^2\). The answer is "E".
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Re: The variable x is inversely proportional to the square of the variable  [#permalink]

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New post 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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New post 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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New post 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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New post 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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New post 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: the-variable-x-is-inversely-proportional-to-the-square-of-the-variable-205761.html

Now 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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New post 18 Jan 2017, 15:42
VeritasPrepKarishma wrote:
pmklings wrote:

Here is a post on inverse variation: the-variable-x-is-inversely-proportional-to-the-square-of-the-variable-205761.html

Now 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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New post 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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New post 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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New post 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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New post 19 Jan 2017, 01:00
kanusha wrote:
VeritasPrepKarishma wrote:
pmklings wrote:

Here is a post on inverse variation: the-variable-x-is-inversely-proportional-to-the-square-of-the-variable-205761.html

Now 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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New post 19 Jan 2017, 02:21
kanusha wrote:
VeritasPrepKarishma wrote:
pmklings wrote:

Here is a post on inverse variation: the-variable-x-is-inversely-proportional-to-the-square-of-the-variable-205761.html

Now 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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The variable x is inversely proportional to the square of the variable  [#permalink]

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New post 24 Aug 2018, 04:22
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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


x is inversely proportional to the square of y.
\(xy^2 = k\), where \(k\) is a constant.

Original values:
Let x=1 and y=6, with the result that \(k = 1*6^2 = 36\).

New values:
Let a=2.
Dividing y by 3a, we get:
New \(y = \frac{6}{(3*2)} = 1\).
Substituting y=1 and k=36 into \(xy^2 = k\), we get:
\(x*1^2 = 36\)
New \(x = 36\).

Since old x = 1 and new x = 36, the value of x is multiplied by 36.
Thus, the correct answer must yield 36 when a=2.
Only E works:
\(9a^2 = 9*2^2 = 36\)


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The variable x is inversely proportional to the square of the variable &nbs [#permalink] 24 Aug 2018, 04:22
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