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Varying Inversely

BY VeritasPrepKarishma, VERITAS PREP


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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Linear Relations in GMAT Questions

BY VeritasPrepKarishma, VERITAS PREP


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

48 (answer choice C) is our answer.

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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Variations On The GMAT - All In One Topic [#permalink]
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Bunuel wrote:
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?

y1 in denominator..please correct me if i am wrong
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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]
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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Re: Variations On The GMAT - All In One Topic [#permalink]
Really Good post, which brushed up my school days concept and additional insights.
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Re: Variations On The GMAT - All In One Topic [#permalink]
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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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?


Are you talking about this question: https://gmatclub.com/forum/the-rate-of- ... 90119.html ?

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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Re: Variations On The GMAT - All In One Topic [#permalink]
Bunuel wrote:
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?


Thanks a lot in advance for clarifying the red parts
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Variations On The GMAT - All In One Topic [#permalink]
Bunuel wrote:
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?


Hi VeritasKarishma


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.
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Anuragjn wrote:
Bunuel wrote:
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?


Hi VeritasKarishma


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
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Re: Variations On The GMAT - All In One Topic [#permalink]
GMATestaker wrote:
Bunuel wrote:
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?


Thanks a lot in advance for clarifying the red parts


May someone clarify the red parts? I didn't see it posted.
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Re: Variations On The GMAT - All In One Topic [#permalink]
Bunuel and KarishmaB
This is making no sense to me as i do not get how can two different relations be expresed by same constant we can literally never do that.There is no clear explanation or proof of this

Bunuel wrote:
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)

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: \(\frac{x*z^2}{y^2} = k\)


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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Re: Variations On The GMAT - All In One Topic [#permalink]
Expert Reply
Elite097 wrote:
Bunuel and KarishmaB
This is making no sense to me as i do not get how can two different relations be expresed by same constant we can literally never do that.There is no clear explanation or proof of this

Bunuel wrote:
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)

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: \(\frac{x*z^2}{y^2} = k\)


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?


k stands for 'constant,' and it is not the same constant.

Everywhere you have k, just put 'constant.'
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Re: Variations On The GMAT - All In One Topic [#permalink]
 
CEdward wrote:
GMATestaker wrote:
Bunuel wrote:
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?

Thanks a lot in advance for clarifying the red parts

May someone clarify the red parts? I didn't see it posted.

­Even I have the same doubt man
like I'm not able to comprehend how squaring the variables will still keep the constant as it is
eg 2/3=k then 4/16 also k? maybe I'm having some conceptual gap
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