1989 AJHSME Problems/Problem 7: Difference between revisions

Revision as of 16:12, 29 April 2021

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

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

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

The $n$ dimes' values need to sum to $10$ quarters and $10$ dimes. $10n=10\cdot25 + 10\cdot 10$ $n=25+10=35$ So, our answer is $\boxed{\text{D}}$

1989 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AJHSME/AMC 8 Problems and Solutions

@@ Line 18: / Line 18: @@
-The  <math>n</math> dimes' values need to sum to <math>10</math> quarters and <math>10</math> dimes.<cmath>10n=10\cdot25 + 10\cdot 10 \\
+The  <math>n</math> dimes' values need to sum to <math>10</math> quarters and <math>10</math> dimes.<cmath>10n=10\cdot25 + 10\cdot 10</cmath>
-n=25+10=35</cmath>
+<cmath>n=25+10=35</cmath>
+So, our answer is <math>\boxed{\text{D}}</math>
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