1985 AJHSME Problems/Problem 3

Problem

$\frac{10^7}{5\times 10^4}=$

$\text{(A)}\ .002 \qquad \text{(B)}\ .2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$

Solution

We immediately see some canceling. We see powers of ten on the top and on the bottom of the fraction, and we make quick work of this: $\frac{10^7}{5 \times 10^4} = \frac{10^3}{5}$

We know that $10^3 = 10 \times 10 \times 10$ , so

$\begin{align*} \frac{10^3}{5} &= \frac{10\times 10\times 10}{5} \\ &= 2\times 10\times 10 \\ &= 200 \\ \end{align*}$

So the answer is $\boxed{\textbf{(D)}\ 200}$ ~shunyipanda (Minor edit)

Solution 2 (Brute Force)

$10^7$ is $10000000$

$5 \times 10^4$ is $50000$

Thus the answer is $200$ , or $\boxed{\textbf{(D)}\ 200}$

Solution 3 (Reasoning)

As we can see, the numerator is way larger than the denominator so answer choices $\textbf{(A)}\ .002$ , $\textbf{(B)}\ .2$ , and $\textbf{(C)}\ 20$ are eliminated. $10^7/10^4$ is $1000$ , but in this question the expression, $10^7/5 \times 10^4$ , is smaller than $1000$ since the denominator is larger than $10^4$ . Therefore the answer is $\boxed{\textbf{(D)}\ 200}$ . ~shunyipanda

Video Solution by BoundlessBrain!

https://youtu.be/BfFv227egOg

Video Solution

https://youtu.be/KW5HexBjEHU

~savannahsolver

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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.

Page

Toolbox

Search