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Preparing for the GMAT? Avoid these 5 major mistakes that can lower your score and hurt your MBA dreams. In this expert panel discussion, 4 top GMAT coaches - GMAT Ninja, Jeff Miller from TTP, Rida Shafiq from eGMAT, and Chris Gentry from Manhattan Prep- Mar 27
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Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
07 Jul 2017, 02:02
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Variations On The GMAT - All In One Topic
CONTENT
Varying Directly
BY VeritasPrepKarishma, VERITAS PREP
So let’s move on to another topic now: Variation.
Basically, variation describes the relation between two or more quantities. e.g. workers and work done, children and noise, entrepreneurs and start ups. More workers means more work done; more children means more noise; more entrepreneurs means more start ups and so on… These are examples of direct variation i.e. if one quantity increases, the other increases proportionally. Then there are quantities that have inverse variation between them e.g. workers and time taken. If there are more workers, time taken to complete a work will be less.
Let’s discuss direct variation today.
Formally, let’s say x varies directly with y. If x takes values \(x_1\), \(x_2\), \(x_3\)… and y takes values \(y_1\), \(y_2\), \(y_3\) … correspondingly, then \(\frac{x_1}{y_1} = \frac{x_2}{y_2} = \frac{x_3}{y_3} =\) Some constant value
In other words, ratio of x and y stays the same in different instances.
(Notice that this is the same as \(\frac{x_1}{x_2} = \frac{y_1}{y_2}\))
It might seem a little cumbersome when put this way but the truth is that direct variation is quite intuitive. A couple of questions will make it clear.
Question 1: 20 workmen can make 35 widgets in 5 days. How many workmen are needed to make 105 widgets in 5 days?
(A) 7
(B) 20
(C) 25
(D) 30
(E) 60
Solution: Notice that the number of days stays the same so we can ignore it. Now think, how are workmen and widgets related? If the number of workmen increases, the number of widget made also increases proportionally. You need to find the new number of workmen required. The number of widgets has become thrice (105/35 = 3) so number of workmen needed will become thrice as well (remember, the number of workmen will increase in the same proportion).
We need 20*3 = 60 workmen
Answer (E) This question is discussed HERE.
The concept of variation is very intuitive. If the number of widgets required doubles, the number of workmen required to make them in the same amount of time will double too. If the number of widgets required becomes one fourth, the number of workmen required to make them in the same amount of time will become one fourth too.
A quantity can directly vary with some power of another quantity. Let’s take an example of this scenario too.
Question 2: If the ratio of the volumes of two right circular cylinders is given by 64/9, what is the ratio of their radii? (Both the cylinders have the same height)
(A) 4/3
(B) 8/3
(C) 16/9
(D) 4/1
(E) 16/3
Solution: This question involves a little bit of geometry too. The volume of a right circular cylinder is given by Area of base * height i.e.
Volume of a right circular cylinder = \(\pi*radius^2 * height\)
So volume varies directly with the square of radius.
\(\frac{Va}{Vb} = \frac{64}{9} = \frac{Ra^2}{Rb^2}\)
\(\frac{Ra}{Rb} = \frac{8}{3}\)
Answer (B) This question is discussed HERE.
We hope this little concept is not hard to understand. We will work on inverse proportion next week and then work on problems involving both (that’s where the good questions are!).
07 Jul 2017, 02:38
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Important Caveat on Joint Variation GMAT Questions
BY VeritasPrepKarishma, VERITAS PREP
Before we start today’s discussion, recall a previous post on joint variation. A question arose some days back on the applicability of this concept. This official question was the case in point:
Question: In a certain business, production index p is directly proportional to efficiency index e, which is in turn directly proportional to investment i. What is p if i = 70?
Statement I: e = 0.5 whenever i = 60
Statement II: p = 2.0 whenever i = 50
This was the issue that was raised:
If one were to follow the method given in the post on joint variation, one would arrive at this solution:
p/e = k (a constant)
e/i = m (another constant)
Hence, p*i/e = n is the joint variation expression
(where k, m and n are constants)
So we get that p is inversely proportional to i, that is, p*i = Constant
Statement II gives us the values of p and i which can help us get the value of the Constant.
2*50 = Constant
The question asks us the value of p given the value of i = 70. If Constant = 100,
p = 100/70.
But actually, this is wrong and the value that you get for p in this question is different.
The question is “why is it wrong?”
Valid question, right? It certainly seems like a joint variation scenario – relation between three variables. Then why does’t it work in this case?
The takeaway from this question is very important and before you proceed, we would like you to think about it on your own for a while and then proceed to the the rest of the discussion.
Here is how this question is actually done:
Taking one statement at a time:
“production index p is directly proportional to efficiency index e,”
implies p = ke (k is the constant of proportionality)
“e is in turn directly proportional to investment i”
implies e = mi (m is the constant of proportionality. Note here that we haven’t taken the constant of proportionality as k since the constant above and this constant could be different)
Then, p = kmi (km is the constant of proportionality here. It doesn’t matter that we depict it using two variables. It is still just a number)
Here, p seems to be directly proportional to i!
So if you have i and need p, you either need this constant directly (as you can find from statement II) or you need both k and m (statement I only gives you m).
So the issue now is that is p inversely proportional to i or is it directly proportional to i?
Review the joint variation post – In it we discussed that joint variation gives you the relation between 2 quantities keeping the third (or more) constant.
p will vary inversely with i if and only if e is kept constant.
Think of it this way: if p increases, e increases. But we need to keep e constant, we will have to decrease i to decrease e back to original value. So an increase in p leads to a decrease in i to keep e constant.
But if we don’t have to keep e constant, an increase in p will lead to an increase in e which will increase i.
It is all about the sequence of increases/decreases
Here, we are not given that e needs to be kept constant. So we will not use the joint variation approach.
Note how the independent question is framed in the joint variation post:
The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical M present and inversely proportional to the concentration of chemical N present. If the concentration of chemical N is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical M required to keep the reaction rate unchanged?
You need relation between N and M when reaction rate is constant.
You are given no such constraint here. So an increase in p leads to an increase in e which in turn, increases i.
So let’s complete the solution to our original question:
\(p = ke\)
\(e = mi\)
\(p = kmi\)
Statement I: \(e = 0.5\) whenever \(i = 60\)
\(0.5 = m * 60\)
\(m = \frac{0.5}{60}\)
We do not know k so we cannot find p given i and m.
This statement alone is not sufficient.
Statement II: \(p = 2.0\) whenever \(i = 50\)
\(2 = km * 50\)
\(km = \frac{1}{25}\)
If \(i = 70\), \(p = (\frac{1}{25})*70 = \frac{14}{5}\)
This statement alone is sufficient.
Answer (B) This question is discussed HERE.
07 Jul 2017, 02:29
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Varying Jointly
BY VeritasPrepKarishma, VERITAS PREP
Now that we have discussed direct and inverse variation, joint variation will be quite intuitive. We use joint variation when a variable varies with (is proportional to) two or more variables.
Say, x varies directly with y and inversely with z. If y doubles and z becomes half, what happens to x?
“x varies directly with y” implies x/y = k (keeping z constant)
If y doubles, x doubles too.
“x varies inversely with z” implies xz = k (now keeping y constant)
If z becomes half, x doubles.
So the overall effect is that x becomes four times of its initial value.
The joint variation expression in this case will be \(\frac{xz}{y} = k\). Notice that when z is constant, x/y = k and when y is constant, xz = k; hence both conditions are being met. Once you get the expression, it’s very simple to solve for any given conditions.
\(x_1*\frac{z_1}{y_1} = x_2*\frac{z_2}{y_2} = k\) (In any two instances, xz/y must remain the same)
\(x_1*\frac{z_1}{y_1} = x_2*(\frac{1}{2})*\frac{z_1}{2*y_1}\)
\(x_2 = 4*x_1\)
Let’s look at some more examples. How will you write the joint variation expression in the following cases?
1. x varies directly with y and directly with z.
2. x varies directly with y and y varies inversely with z.
3. x varies inversely with y^2 and inversely with z^3.
4. x varies directly with y^2 and y varies directly with z.
5. x varies directly with y^2, y varies inversely with z and z varies directly with p^3.
Solution: Note that the expression has to satisfy all the conditions.
1. x varies directly with y and directly with z.
\(\frac{x}{y} = k\)
\(\frac{x}{z} = k\)
Joint variation: \(\frac{x}{yz} = k\)
2. x varies directly with y and y varies inversely with z.
\(\frac{x}{y} = k\)
\(yz = k\)
Joint variation: \(\frac{x}{yz} = k\)
3. x varies inversely with y^2 and inversely with z^3.
\(x*y^2 = k\)
\(x*z^3 = k\)
Joint variation: \(x*y^2*z^3 = k\)
4. x varies directly with y^2 and y varies directly with z.
\(\frac{x}{y^2} = k\)
\(\frac{y}{z} = k\) which implies that \(\frac{y^2}{z^2} = k\)
Joint variation: \(\frac{x*z^2}{y^2} = k\)
5. x varies directly with y^2, y varies inversely with z and z varies directly with p^3.
\(\frac{x}{y^2} = k\)
\(yz = k\) which implies \(y^2*z^2 = k\)
\(\frac{z}{p^3} = k\) which implies \(\frac{z^2}{p^6} = k\)
Joint variation: \(\frac{(x*p^6)}{(y^2*z^2)} = k\)
Let’s take a GMAT prep question now to see these concepts in action:
Question 1: The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical M present and inversely proportional to the concentration of chemical N present. If the concentration of chemical N is increased by 100 percent, which of the following is closest to the percent change in the concentration of chemical M required to keep the reaction rate unchanged?
(A) 100% decrease
(B) 50% decrease
(C) 40% decrease
(D) 40% increase
(E) 50% increase
Solution:
\(\frac{Rate}{M^2} = k\)
\(Rate*N = k\)
\(\frac{Rate*N}{M^2} = k\)
If Rate has to remain constant, N/M^2 must remain the same too.
If N is doubled, M^2 must be doubled too i.e. M must become ?2 times. Since ?2 = 1.4 (approximately),
M must increase by 40%.
Answer (D) This question is discussed HERE.
Simple enough?
