Difference between revisions of "1989 AJHSME Problems/Problem 7"

(Solution)
Line 11: Line 11:
 
20(25)+10(10) &= 10(25)+n(10) \\
 
20(25)+10(10) &= 10(25)+n(10) \\
 
600 &= 250+10n \\
 
600 &= 250+10n \\
35 &= n \rightarrow \boxed{\text{D}}
+
35 &= n \implies \boxed{\text{D}}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  

Revision as of 04:46, 31 August 2015

Problem

If the value of $20$ quarters and $10$ dimes equals the value of $10$ quarters and $n$ dimes, then $n=$

$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 45$

Solution

We have \begin{align*} 20(25)+10(10) &= 10(25)+n(10) \\ 600 &= 250+10n \\ 35 &= n \implies \boxed{\text{D}} \end{align*}

See Also

1989 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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. AMC logo.png