Difference between revisions of "1985 AJHSME Problems/Problem 3"

Revision as of 16:01, 3 July 2013

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 $\boxed{\text{D}}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Revision as of 07:26, 6 May 2009 (view source) 5849206328x (talk \| contribs) m ← Older edit		Revision as of 16:01, 3 July 2013 (view source) MSTang (talk \| contribs) (→‎See Also) Newer edit →
Line 25:		Line 25:

	[[Category:Introductory Algebra Problems]]		[[Category:Introductory Algebra Problems]]
		+
		+
		+	{{MAA Notice}}

Page

Toolbox

Search

Difference between revisions of "1985 AJHSME Problems/Problem 3"

Revision as of 16:01, 3 July 2013

Problem

Solution

See Also