Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
18 Sep 2015, 19:04
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
62% (01:17) correct
38%
(01:36)
wrong
based on 1894
sessions
History
Date
Time
Result
Not Attempted Yet
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
a. 1/9a
b. 1/9a^2
c. 1/3a
d. 9a
e. 9a^2
22 Aug 2018, 21:02
Kudos
Bookmarks
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".
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".
pmklings
Here is a video on inverse variation: https://youtu.be/AT86tjxJ-f0
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)
