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 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.

x1*z1/y1 = x2*z2/y2 = k (In any two instances, xz/y must remain the same)

x1*z1/y1 = x2*(1/2)z1/2*y1

x2 = 4*x1

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.

x/y = k

x/z = k

Joint variation: x/yz = k

2. x varies directly with y and y varies inversely with z.

x/y = k

yz = k

Joint variation: 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.

x/y^2 = k

y/z = k which implies that y^2/z^2 = k

Joint variation: 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.

x/y^2 = k

yz = k which implies y^2*z^2 = k

z/p^3 = k which implies z^2/p^6 = k

Joint variation: (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:

Rate/M^2 = k

Rate*N = k

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)

Simple enough?