Difference between revisions of "1989 AJHSME Problems/Problem 7"
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20(25)+10(10) &= 10(25)+n(10) \\ | 20(25)+10(10) &= 10(25)+n(10) \\ | ||
600 &= 250+10n \\ | 600 &= 250+10n \\ | ||
| − | 35 &= n \ | + | 35 &= n \implies \boxed{\text{D}} |
\end{align*}</cmath> | \end{align*}</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> | ||
| + | So, our answer is <math>\boxed{\text{D}}</math>----stjwyl | ||
==See Also== | ==See Also== | ||
Latest revision as of 16:14, 29 April 2021
Problem
If the value of 
quarters and 
dimes equals the value of quarters and 
dimes, then 
Solution
We have
The dimes' values need to sum to quarters and dimes.![\[10n=10\cdot25 + 10\cdot 10\]](http://latex.artofproblemsolving.com/6/3/3/633298601e689b7dd44aa8d93e7abfc1b7325408.png)
we can divide both sides by .
![\[n=25+10=35\]](http://latex.artofproblemsolving.com/7/3/7/73752d8c2ceb28be89e98ca157981e27b1be202a.png)
So, our answer is 
----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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
