SAT Tip: Direct and Inverse Variation
This tip on improving your SAT score was provided by Vivian Kerr at Veritas Prep.
On the math section of the SAT, direct variation means that two variables change in the same way relative to each other: When one goes up, the other goes up, and when one goes down, the other goes down, too. This is described by the formula y = kx, where x and y are the two variables, and k is a constant. However when variable A increases and variable B decreases, there is an indirect (or inverse) variation between the two variables. 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, like this one, a simple proportion is sufficient to solve the problem:
The number of students on a field trip is directly proportional to the number of buses required to transport them. If 2 buses are required to transport 66 students and both buses are completely full, how many buses are needed to transport 117 students?
This can be restated as: 2 buses / 66 students = x buses / 117 students
Cross-multiply to solve: (2)(117) = (66)(x)
234 = 66x
3.54 = x
Since you can’t have 0.54 of a bus, rounding up gives you the answer: 4.
More challenging test questions will require the use of the y = kx and y = k/x formulas.
If y is inversely proportional to x and y = 20 when x = 5, what is the value of y when x = 25?
Using the inverse variation formula, y = k/ x, this can be restated as 20 = k/5, yielding a value for k of 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
As a double-check, we can see that as we increased our x from 5 to 25, y went down from 20 to 4. This makes sense because we know that y and x are inversely proportional.
The harder proportion questions will involve long word problems and larger values, such as, this stumper:
The weight W of an object varies inversely with the square of the distance d from the center of the earth. At sea level (3,978 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” language tells us we can use our same formula, y = k/x, but substituting the new variables:
W = k/d2
Since when d = 3,978, W = 220, let’s plug these in to solve for 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. Since d = total distance from the center of the earth. We’ll need to add the 3,978 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 / (3,978 + 200)2
W ≈ 200 lbs.
Make sure to memorize both the direct and indirect variation formulas before test day, and look out for those harder-level problems in which you have to solve for the constant first.
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