Difference between revisions of "1953 AHSME Problems/Problem 6"
Katzrockso (talk | contribs) (Created page with "==Problem 6== Charles has <math>5q + 1</math> quarters and Richard has <math>q + 5</math> quarters. The difference in their money in dimes is: <math>\textbf{(A)}\ 10(q - 1)...") |
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== Solution == | == Solution == | ||
<math>5q+1-q-5\implies 4p-4\implies\frac{4(p-1)}{10}\implies\frac{2(p-1)}{5}</math>, <math>\fbox{D}</math> | <math>5q+1-q-5\implies 4p-4\implies\frac{4(p-1)}{10}\implies\frac{2(p-1)}{5}</math>, <math>\fbox{D}</math> | ||
+ | If Charles has <math>5q +1</math> quarters and Richard has <math>q+5</math> quarters, then the difference between them is <math>4q-4</math> quarters. Converting quarters to dimes, the difference between them is <math>\frac{5}{2}\cdot (4)(q-1)</math>, or <math>10(q-1)</math> dimes, <math>\implies \fbox{A}</math> | ||
+ | ==See Also== | ||
+ | |||
+ | {{AHSME 50p box|year=1953|num-b=5|num-a=7}} | ||
+ | |||
+ | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 19:34, 1 April 2017
Problem 6
Charles has quarters and Richard has quarters. The difference in their money in dimes is:
Solution
, If Charles has quarters and Richard has quarters, then the difference between them is quarters. Converting quarters to dimes, the difference between them is , or dimes,
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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