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

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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>
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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> we can divide both sides by <math>10</math>.
 
<cmath>n=25+10=35</cmath>
 
<cmath>n=25+10=35</cmath>
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So, our answer is <math>\boxed{\text{D}}</math>
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So, our answer is <math>\boxed{\text{D}}</math>[[User:Stjwyl|Stjwyl]] ([[User talk:Stjwyl|talk]]) 17:13, 29 April 2021 (EDT)stjwyl
  
 
==See Also==
 
==See Also==

Revision as of 16:13, 29 April 2021

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*}



The $n$ dimes' values need to sum to $10$ quarters and $10$ dimes.\[10n=10\cdot25 + 10\cdot 10\] we can divide both sides by $10$. \[n=25+10=35\] So, our answer is $\boxed{\text{D}}$Stjwyl (talk) 17:13, 29 April 2021 (EDT)stjwyl

See Also

1989 AJHSME (Problems • Answer Key • Resources)
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

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