SAT Direct and Inverse Variation

Correct Answer: D

Solution:

By the definition of direct variation, there is some constant $k_1$ such that $a = k_1 \cdot b$ .
By the definition of indirect variation, there is some constant $k_2$ such that $b = k_2 \cdot \frac{1}{c}$ .

As such, this gives

$a = k_1 \cdot b = k_1 \cdot k_2 \cdot \frac{1}{c}.$

This tells us that $a$ varies indirectly with $c$ .

Since we also have $c = k_1 \cdot k_2 \cdot \frac{1}{a}$ , thus $c$ varies indirectly with $a$ .

Hence the answer is (D).

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Incorrect Choices:

(A)
If you set $a$ varying directly with $b$ and $b$ varying directly with $c,$ you will get this wrong answer.

Also, if you set $a$ varying inversely with $b$ and $b$ varying inversely with $c$ , you will get this wrong answer

(B) and (C)
By definition, if $y$ varies directly with $x$ , then there is a nonzero constant $k$ for which $y=k\cdot x.$ Rearranging this we get $x= \frac{1}{k} y$ and hence $x$ varies directly with $y.$

Similarly, by definition, if $y$ varies inversely with $x$ , then there is a nonzero constant $k$ , such that $y=k\cdot\frac{1}{x}.$ Rearranging this we obtain $x=k\cdot \frac{1}{y},$ and hence $x$ varies indirectly with $y.$

It is implied then that if $a$ varies directly with $c,$ $c$ must vary directly with $a,$ and if $a$ varies indirectly with $c,$ $c$ must vary indirectly with $a.$

We conclude that these choices are wrong.

(E)
Answer choice (D) is correct.