Difference between revisions of "1998 AJHSME Problems/Problem 3"

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== See also ==
 
== See also ==
{{AJHSME box|year=1998|before=[[1997 AJHSME Problems|1997 AJHSME]]|after=[[1999 AMC 8 Problems|1999 AMC 8]]}}
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{{AJHSME box|year=1998|num-b=2|num-a=4}}
 
* [[AJHSME]]
 
* [[AJHSME]]
 
* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]

Revision as of 00:05, 10 June 2011

Problem 3

$\dfrac{\dfrac{3}{8} + \dfrac{7}{8}}{\dfrac{4}{5}} =$

$\text{(A)}\ 1 \qquad \text{(B)} \dfrac{25}{16} \qquad \text{(C)}\ 2 \qquad \text{(D)}\ \dfrac{43}{20} \qquad \text{(E)}\ \dfrac{47}{16}$

Solution

$\frac{\frac{3}{8}+\frac{7}{8}}\frac{4}{5}}=\frac{\frac{10}{8}}{\frac{4}{5}=\frac{\frac{5}{4}}{\frac{4}{5}}=\frac{25}{16}$ (Error compiling LaTeX. Unknown error_msg)

$\boxed{B}$

See also

1998 AJHSME (ProblemsAnswer KeyResources)
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