# SAT Math: How to Solve Direct & Inverse Variation Problems

*Direct variation*: This is when two things change in the same way over time. If Column A increases and Column B increases at the same time, we can say that the two columns vary directly.

This is described by the formula **y = kx**, where x and y are the two variables, and k is the constant.

*Indirect variation*: If when Column A increases, Column B decreases, there is an indirect (also called inverse) variation between the two columns.

This is described by the formula **y = k/x **where x and y are the two variables and k is the constant. For easier variation questions, a simple proportion is sufficient to solve.

1. The number of kids on a field trip is directly proportional to the number of buses required to transport them. If 2 buses are required to transport 66 kids and both buses are completely full, how many buses are needed to transport 117 kids?

2 buses / 66 kids = x buses / 117 kids

Cross-multiply to solve.

(2)(117) = (66)(x)

234 = 66x

3.54 = x

More challenging test questions will require the use of the y = kx and y = k/x formulas. Let’s look at a Grockit question:

2. If y is inversely proportional to x and y = 20 when x = 5, what is the value of y when x = 25?

y = k/ x, so 20 = k/5, and k = 100.

Now our equation for this question looks like y = 100/x. All we need to do is plug in our new x, 25, and solve for y.

y = 100/25

y = 4

Try this SAT Math question for more practice today!

The harder proportion questions will involve long word problems and larger values, like this stumper:

3. The weight W of an object varies inversely as the square of the distance d from the center of the earth. At sea level (3978 mi from the center of the earth), an astronaut weighs 220lbs. What is his weight when he is 200 mi above the surface of the earth and the spacecraft is not in motion?

This question replaces x and y with W and d, but the “varies inversely” tells us we can use our same formula: y = k/x.

W = k/d²

d must be squared because of the phrase “as the square of the distance d” in the problem.

We are told that when d = 3978, W = 220. Let’s plug these in to solve for our constant, k.

220 = k / (3978)²

220 = k / 15,824,484

3,481,386,480 = k

Now we can rewrite our formula as:

W = 3,481,386,480 / d²

The question asks what will W be when d = 200. Consider that d = total distance from the center of the earth. We’ll need to add the 3978 from sea level to the center, and the 200 from sea level to space to find the total d this time. Again, all we do is plug in to solve:

W = 3,481,386,480 / (3978 + 200)²

W ≈ 200 lbs.