Difference between revisions of "1988 AJHSME Problems/Problem 6"

Latest revision as of 16:47, 26 June 2025

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 13: / Line 13: @@
 ~sakshamsethi
+==Solution 3==
+<math>\frac{(0.2)^3}{(0.02)^2}</math> is like equal to <math>\frac{(2)^3  \times (.1)^3}{(2)^2 \times (.1)^4}</math>. Dividing both the numerator and the denominator by <math>(.1)^3</math> we get <math>\frac{(2)^3}{(2)^2 \times (.1)}</math>, and then we divide both sides by <math>(2)^2</math> so we get <math>\frac{2}{.1}</math> which is equal to <math>20</math>.
+~doraluvs2yap
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