# SAT Direct and Inverse Variation

To solve problems about direct and inverse variation on the SAT, you need to know:

- how to work with fractions
- the rules of exponents
- how to translate words into math
- the definition of direct variation

If $y$ varies directly with $x$, then there is a constant $k$ for which

$y=k\cdot x, \quad k\neq0$

$k$ is called the constant of proportionality, or the constant of variation.

Notice that as $x$ increases (or decreases) by some factor, $y$ increases (or decreases) by the same factor.

Sometimes the definition is written as follows to demonstrate that the ratio of $y$ and $x$ is constant:

$\frac{y}{x}=k$

- the definition of inverse variation

If $y$ varies inversely with $x$, then there is a constant $k$, such that:

$y=k\cdot\frac{1}{x}, \quad k\neq0$

$k$ is called the constant of proportionality, or the constant of variation.

Notice that as $x$ increases (or decreases) by some factor, $y$ (decreases or increases) by the same factor.

Sometimes the definition is written as follows to demonstrate that the product of $x$ and $y$ is constant:

$x\cdot y =k$

#### Contents

## Examples

$p$ varies directly with $n$. Which of the following equations could represent the relationship between $p$ and $n$?

(A) $\ \ p=\frac{1}{2n}$

(B) $\ \ n=\frac{2}{p}$

(C) $\ \ p\cdot n=\frac{1}{2}$

(D) $\ \ p\cdot n=2$

(E) $\ \ p=\frac{1}{2}\cdot n$

Correct Answer: E

Solution 1:By the definition of direct variation, for some constant $k$, $p=kn$. Here, $k=\frac{1}{2}.$ You should be able to tell that this answer choice is correct at a glance.

Solution 2:

Tip: Replace variables with numbers.

In a direct variation, as one variable increases by a factor, the other one increases by the same factor; and as one variable decreases by a factor, the other one decreases by the same factor. We analyze each choice to see which one meets this expectation.(A) For $n=1, p=\frac{1}{2\cdot1}=\frac{1}{2}.$

For $n=2, p=\frac{1}{2\cdot2}=\frac{1}{4}.$Contrary to the definition of direct variation, as $n$ doubles, $p$ is divided by $4$. Eliminate (A).

(B) For $p=1, n=\frac{2}{1}=2.$

For $p=2, n=\frac{2}{2}=1.$Contrary to the definition of direct variation, as $p$ doubles, $n$ halves. Eliminate (B).

(C) Dividing both sides of this equation by $n$, we get:

$p \cdot n=\frac{1}{2}$ $p=\frac{1}{2n}$

This is identical to (A), which we already eliminated. Eliminate (C).

(D) Dividing both sides of this equation by $n$, we get:

$p \cdot n=2$ $p=\frac{2}{n}$

For $n=1, p=\frac{2}{1}=2.$

For $n=2, p=\frac{2}{2}=1.$As $n$ doubles, $p$ halves. This contradicts the definition of direct variation. Eliminate (D).

(E) For $n=1, p=\frac{1}{2} \cdot 1 = \frac{1}{2}.$

For $n=2, p=\frac{1}{2} \cdot 2 = \frac{2}{2}=1.$As $n$ doubles, so does $p$. Therefore, this is the correct answer.

Incorrect Choices:

(A),(B),(C), and(D)

See Solution 2 for how to eliminate these answers.

$a$ varies directly with $b$.

$b$ varies indirectly with $c$.

If the above statements are true for nonzero $a, b,$ and $c$, which of the following statements must be true?

(A) $\ \ a$ varies directly with $c$ and $c$ varies directly with $a$.

(B) $\ \ a$ varies directly with $c$ and $c$ varies indirectly with $a$.

(C) $\ \ a$ varies indirectly with $c$ and $c$ varies directly with $a$.

(D) $\ \ a$ varies indirectly with $c$ and $c$ varies indirectly with $a$.

(E) $\ \$ None of the above.

If $y$ is inversely proportional to $x$, and $y=20$ when $x=4$, what is the value of $y$ when $x=10$?

(A) $\ \ 4$

(B) $\ \ 5$

(C) $\ \ 8$

(D) $\ \ 20$

(E) $\ \ 50$

Correct Answer: C

Solution 1:If $y$ is inversely proportional to $x$, then for some constant $k\neq 0, y=k\frac{1}{x}.$

We are told that when $x=4, y=20$. So,

$\begin{array}{l c l l l} y&=&\frac{k}{x} &\quad \text{definition of inverse variation} &(1)\\ 20&=&\frac{k}{4} &\quad \text{replace variables with given values} &(2)\\ 80&=&k &\quad \text{multiply both sides by}\ 4 &(3)\\ \end{array}$

We found the constant $k$. Therefore, $y=\frac{80}{x}$.

When $x=10$, $y=\frac{80}{10}=8$.

Solution 2:When $x=4$, $y=20$. We are looking for $y$ when $x=10$. By definition, when two quantities vary inversely, their product is constant. Here we create two products, set them equal to each other, and solve for the unknown.

$\begin{array}{l c l l l} 20\cdot 4&=&y \cdot 10 &\quad \text{set the two products equal to each other}\\ 80 &=&y \cdot 10 &\quad 20\cdot 4=80\\ 8 &=& y &\quad \text{divide both sides by}\ 10\\ \end{array}$

Incorrect Choices:

(A)

Tip: Just because a number appears in the question doesn’t mean it is the answer.

(B)

This answer is meant to confuse you. It is the quotient of the numbers $20$ and $4$, which appear in the prompt.

(D)

Tip: Just because a number appears in the question doesn’t mean it is the answer.

(E)

If we vary $y$ directly with $x$, instead of inversely, we will start with the equation $y=kx$. Using $y=20$ and $x=4$, we will find that $20=k\cdot 4$. Dividing both sides by $4$, we will get $k=5$. Then, for $x=10, y=5\cdot 10 =50$. But, this is wrong.

## Review

If you thought these examples difficult and you need to review the material, these links will help:

## SAT Tips for Direct Variation

- Know the Rules of Exponents.
- Follow order of operations.
- SAT General Tips

**Cite as:**SAT Direct and Inverse Variation.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/sat-direct-and-inverse-variation/