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 17:14, 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.
we can divide both sides by
.
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.