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# Variations On The GMAT - All In One Topic

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07 Jul 2017, 01:02
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Variations On The GMAT - All In One Topic

Varying Directly

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!).
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07 Jul 2017, 01:38
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Important Caveat on Joint Variation GMAT Questions

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 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.
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07 Jul 2017, 01:17
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Varying Inversely

As promised, we will discuss inverse variation today. The concept of inverse variation is very simple – two quantities x and y vary inversely if increasing one decreases the other proportionally.

If x takes values $$x_1$$, $$x_2$$, $$x_3$$… and y takes values $$y_1$$, $$y_2$$, $$y_3$$ … correspondingly, then $$x_1*y_1 = x_2*y_2 = x_3*y_3 =$$ some constant value

This means that if x doubles, y becomes half; if x becomes 1/3, y becomes 3 times etc. In other words, product of x and y stays the same in different instances.

Notice that $$\frac{x_1}{x_2} = \frac{y_2}{y_1}$$; The ratio of x is inverse of the ratio of y.

The concept will become clearer after working on a few examples.

Question 1: The price of a diamond varies inversely with the square of the percentage of impurities. The cost of a diamond with 0.02% impurities is $2500. What is the cost of a diamond with 0.05% impurities (keeping everything else constant)? (A)$400
(B) $500 (C)$1000
(D) $4000 (E)$8000

Solution:

$$Price_1*(\% \ of \ Impurities1)^2 = Price_2*(\% \ of \ Impurities2)^2$$

$$2500*(.02)^2 = Price_2*(.05)^2$$

Price = \$400

Answer (A) This question is discussed HERE.

The answer is quite intuitive in the sense that if % of impurities in the diamond increases, the price of the diamond decreases.

There is an important question type related to inverse variation. It often uses the formula:

Total Price = Number of units*Price per unit

If, due to budgetary constraints, we need to keep the total money spent on a commodity constant, number of units consumed varies inversely with price per unit. If price per unit increases, we need to reduce the consumption proportionally.

Question 2: The cost of fuel increases by 10%. By what % must the consumption of fuel decrease to keep the overall amount spent on the fuel same?

(A) 5%
(B) 9%
(C) 10%
(D) 11%
(E) 20%

Solution: Do you think the answer is 10%? Think again.

Total Cost = Number of units*Price per unit

If the price per unit increases by 10%, it becomes 11/10 of its original value. To keep the total cost same, you need to multiply number of units by 10/11. i.e. you need to decrease the number of units by 1/11 i.e. 9.09%. In that case,

New Total Cost = (10/11)*Number of units*(11/10)*Price per unit

This new total cost will be the same as the previous total cost.

Answer (B) This question is discussed HERE.

Let’s look at one more example of the same concept but this one is a little trickier.

Question 3: Recently, fuel price has seen a hike of 20%. Mr X is planning to buy a new car with better mileage as compared to his current car. By what % should the new mileage be more than the previous mileage to ensure that Mr X’s total fuel cost stays the same for the month? (assuming the distance traveled every month stays the same)

(A) 10%
(B) 17%
(C) 20%
(D) 21%
(E) 25%

Solution: The problem here is ‘how is mileage related to fuel price?’

Total fuel cost = Fuel price * Quantity of fuel used

Since the ‘total fuel cost’ needs to stay the same, ‘fuel price’ varies inversely with ‘quantity of fuel used’.

Quantity of fuel used = Distance traveled/Mileage

Distance traveled = Quantity of fuel used*Mileage

Since the same distance needs to be traveled, ‘quantity of fuel used’ varies inversely with the ‘mileage’.

We see that ‘fuel price’ varies inversely with ‘quantity of fuel used’ and ‘quantity of fuel used’ varies inversely with ‘mileage’. So, if fuel price increases, quantity of fuel used decreases proportionally and if quantity of fuel used decreases, mileage increases proportionally. Hence, if fuel price increases, mileage increases proportionally or we can say that fuel price varies directly with mileage.

If fuel price becomes 6/5 (20% increase) of previous fuel price, we need the mileage to become 6/5 of the previous mileage too i.e. mileage should increase by 20% too.

Another method is that you can directly plug in the expression for ‘Quantity of fuel used’ in the original equation.

Total fuel cost = Fuel price * Distance traveled/Mileage

Since ‘total fuel cost’ and ‘distance traveled’ need to stay the same, ‘fuel price’ is directly proportional to ‘mileage’.

Answer (C) This question is discussed HERE.

We hope you are comfortable with fundamentals of direct and inverse variation now. More next week!
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07 Jul 2017, 01:29
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Varying Jointly

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?
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07 Jul 2017, 01:44
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Linear Relations in GMAT Questions

We have covered the concepts of direct, inverse and joint variation in previous posts. Today, we will discuss what we mean by “linearly related”. A linear relation is one which, when plotted on a graph, is a straight line. In linear relationships, any given change in an independent variable will produce a corresponding change in the dependent variable, just like a change in the x-coordinate produces a corresponding change in the y-coordinate on a line.

We know the equation of a line: it is y = mx + c, where m is the slope and c is a constant.

Let’s illustrate this concept with a GMAT question. This question may not seem like a geometry question, but using the concept of linear relations can make it easy to find the answer:

A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?

(A) 20
(B) 36
(C) 48
(D) 60
(E) 84

Let’s think of the two scales R and S as x- and y-coordinates. We can get two equations for the line that depicts their relationship:

30 = 6m + c ……. (I)

60 = 24m + c ……(II)

(II) – (I)

30 = 18m

m = 30/18 = 5/3

Plugging m = 5/3 in (I), we get:

30 = 6*(5/3) + c

c = 20

Therefore, the equation is S = (5/3)R + 20. Let’s plug in S = 100 to get the value of R:

100 = (5/3)R + 20

R = 48

Alternatively, we have discussed the concept of slope and how to deal with it without any equations in this post. Think of each corresponding pair of R and S as points lying on a line – (6, 30) and (24, 60) are points on a line, so what will (r, 100) be on the same line?

We see that an increase of 18 in the x-coordinate (from 6 to 24) causes an increase of 30 in the y-coordinate (from 30 to 60).

So, the y-coordinate increases by 30/18 = 5/3 for every 1 point increase in the x-coordinate (this is the concept of slope).

From 60 to 100, the increase in the y-coordinate is 40, so the x-coordinate will also increase from 24 to 24 + 40*(3/5) = 48. Again, C is our answer. This question is discussed HERE.
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07 Jul 2017, 01:53
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Re: Variations On The GMAT - All In One Topic  [#permalink]

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07 Jul 2017, 02:03
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Bunuel wrote:
Varying Jointly

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?

y1 in denominator..please correct me if i am wrong
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07 Jul 2017, 02:07
mynamegoeson wrote:
y1 in denominator..please correct me if i am wrong

Edited. Hope now it's correct. Thank you.
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Re: Variations On The GMAT - All In One Topic  [#permalink]

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07 Jul 2017, 02:12
Bunuel wrote:
mynamegoeson wrote:
y1 in denominator..please correct me if i am wrong

Edited. Hope now it's correct. Thank you.

Thanks a lot..These are really helpful
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07 Jul 2017, 11:03
Really Good post, which brushed up my school days concept and additional insights.
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02 Apr 2018, 22:01
In the "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?"

How come m, k, and n constants were used in this problem whereas just constant k was used in the previous examples?

Shouldn't all the examples use unique constants for each direct/inverse relationship?
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02 Apr 2018, 22:26
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ViciousK wrote:
In the "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?"

How come m, k, and n constants were used in this problem whereas just constant k was used in the previous examples?

Shouldn't all the examples use unique constants for each direct/inverse relationship?

Here "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..." can be expressed as $$RATE=\frac{A^2}{B}*k$$, so you'll need only one variable to describe this relationship.

In case of any farther questions please post in the following topic: https://gmatclub.com/forum/the-rate-of- ... 90119.html
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18 Jul 2019, 18:58
Bunuel wrote:
Varying Jointly

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?

Thanks a lot in advance for clarifying the red parts
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Joined: 28 Jan 2019
Posts: 30
Location: India
Variations On The GMAT - All In One Topic  [#permalink]

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10 Jun 2020, 19:58
Bunuel wrote:
Varying Jointly

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?

In Point 2 " x varies directly with y and y varies inversely with z."

As per my understanding, if Z increase, Y should decrease and if Y decreases, X should decrease as well. So, Z is inversely proportional to X.

But in Join Variation equation you have mentioned

X/yz = K => which means X is directly proportional to Z

Can you please explain this further ? I am a bit confused.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10623
Location: Pune, India
Re: Variations On The GMAT - All In One Topic  [#permalink]

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14 Jun 2020, 19:55
1
Anuragjn wrote:
Bunuel wrote:
Varying Jointly

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?

In Point 2 " x varies directly with y and y varies inversely with z."

As per my understanding, if Z increase, Y should decrease and if Y decreases, X should decrease as well. So, Z is inversely proportional to X.

But in Join Variation equation you have mentioned

X/yz = K => which means X is directly proportional to Z

Can you please explain this further ? I am a bit confused.

The biggest problem people face in joint variation - they forget that the other factors need to be held CONSTANT. If I want the relation between Z and X, I want it when Y is constant.

If Z increases, Y decreases. But I need Y constant. So Y needs to be increased back. When Y is increased, X will increase too. Hence Z and X are directly varying.

x/yz = k
_________________
Karishma
Veritas Prep GMAT Instructor

Re: Variations On The GMAT - All In One Topic   [#permalink] 14 Jun 2020, 19:55