# 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.

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