Difference between revisions of "1987 AJHSME Problems/Problem 5"

Latest revision as of 22:52, 4 July 2013

Problem

The area of the rectangular region is

$[asy] draw((0,0)--(4,0)--(4,2.2)--(0,2.2)--cycle,linewidth(.5 mm)); label(".22 m",(4,1.1),E); label(".4 m",(2,0),S); [/asy]$

$\text{(A)}\ \text{.088 m}^2 \qquad \text{(B)}\ \text{.62 m}^2 \qquad \text{(C)}\ \text{.88 m}^2 \qquad \text{(D)}\ \text{1.24 m}^2 \qquad \text{(E)}\ \text{4.22 m}^2$

Solution

$(.4) \cdot (.22) = \frac{4}{10} \cdot \frac{22}{100} = \frac{4\cdot 22}{10\cdot 100} = \frac{88}{1000} = 0.088\rightarrow \boxed{\text{A}}$

1987 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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.

@@ Line 13: / Line 13: @@
 ==Solution==
-<math>(.4) \cdot (.22) = \frac{4}{10} \cdot \frac{22}{100} = \frac{4\cdot 22}{10\cdot 100} = \frac{88}{1000} = \boxed{0.088}.</math>
+<math>(.4) \cdot (.22) = \frac{4}{10} \cdot \frac{22}{100} = \frac{4\cdot 22}{10\cdot 100} = \frac{88}{1000} = 0.088\rightarrow \boxed{\text{A}}</math>
 ==See Also==
-[[1987 AJHSME Problems]]
+{{AJHSME box|year=1987|num-b=4|num-a=6}}
+[[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

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Difference between revisions of "1987 AJHSME Problems/Problem 5"

Latest revision as of 22:52, 4 July 2013

Problem

Solution

See Also