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

(New page: ==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> <math>\tex...)
 
Line 13: Line 13:
 
==Solution==
 
==Solution==
  
−
{{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>
  
 
==See Also==
 
==See Also==
  
 
[[1987 AJHSME Problems]]
 
[[1987 AJHSME Problems]]

Revision as of 13:41, 29 January 2009

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

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

See Also

1987 AJHSME Problems

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