Difference between revisions of "1987 AJHSME Problems/Problem 2"

Revision as of 12:01, 25 May 2009

$\frac{2}{25}=$

$\text{(A)}\ .008 \qquad \text{(B)}\ .08 \qquad \text{(C)}\ .8 \qquad \text{(D)}\ 1.25 \qquad \text{(E)}\ 12.5$

$\frac{2}{25}=\frac{2\cdot 4}{25\cdot 4} = \frac{8}{100} = 0.08\rightarrow \boxed{\text{B}}$

1987 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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

@@ Line 3: / Line 3: @@
 <math>\frac{2}{25}=</math>
-<math>\text{(A)}\ .008 \qquad \text{(B)}\ .08 \qquad \text{(C)}\ .8 \qquad \text{(D)} 1.25 \qquad \text{(E)}\ 12.5</math>
+<math>\text{(A)}\ .008 \qquad \text{(B)}\ .08 \qquad \text{(C)}\ .8 \qquad \text{(D)}\ 1.25 \qquad \text{(E)}\ 12.5</math>
 ==Solution==
-<math>\frac{2}{25}=\frac{2\cdot 4}{25\cdot 4} = \frac{8}{100} = \boxed{0.08}</math>
+<math>\frac{2}{25}=\frac{2\cdot 4}{25\cdot 4} = \frac{8}{100} = 0.08\rightarrow \boxed{\text{B}}</math>
 ==See Also==
-[[1987 AJHSME Problems]]
+{{AJHSME box|year=1987|num-b=1|num-a=3}}
+[[Category:Introductory Algebra Problems]]