1988 AJHSME Problems/Problem 6

Problem

$\frac{(.2)^3}{(.02)^2} =$

$\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 20$

Solution

Converting the decimals to fractions gives us $\frac{(.2)^3}{(.02)^2} =\frac{\left( \frac{1}{5}\right)^3}{\left(\frac{1}{50}\right)^2}=\frac{50^2}{5^3}=\frac{2500}{125}=20\Rightarrow \mathrm{(E)}$ .

Solution 2

We expand $\frac{(0.2)^3}{(0.02)^2}$ , and get $\frac{(0.2) \times (0.2) \times (0.2)}{(0.02) \times (0.02)}$ . The two $0.02$ 's "cancel" out with the two $0.2$ 's, leaving the fraction as: $(10) \times (10) \times (0.2)$ . Using basic calculations, we compute this expression to get $20\Rightarrow \mathrm{(E)}$ .

~sakshamsethi

Solution 3

$\frac{(0.2)^3}{(0.02)^2}$ is like equal to $\frac{(2)^3 \times (.1)^3}{(2)^2 \times (.1)^4}$ . Dividing both the numerator and the denominator by $(.1)^3$ we get $\frac{(2)^3}{(2)^2 \times (.1)}$ , and then we divide both sides by $(2)^2$ so we get $\frac{2}{.1}$ which is equal to $20$ .

~doraluvs2yap

1988 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AJHSME/AMC 8 Problems and Solutions

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1988 AJHSME Problems/Problem 6

Contents

Problem

Solution

Solution 2

Solution 3

See Also