Difference between revisions of "1953 AHSME Problems/Problem 6"

(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>
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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>
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==See Also==
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{{AHSME 50p box|year=1953|num-b=5|num-a=7}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 20:34, 1 April 2017

Problem 6

Charles has $5q + 1$ quarters and Richard has $q + 5$ quarters. The difference in their money in dimes is:

$\textbf{(A)}\ 10(q - 1) \qquad \textbf{(B)}\ \frac {2}{5}(4q - 4) \qquad \textbf{(C)}\ \frac {2}{5}(q - 1) \\  \textbf{(D)}\ \frac{5}{2}(q-1)\qquad \textbf{(E)}\ \text{none of these}$

Solution

$5q+1-q-5\implies 4p-4\implies\frac{4(p-1)}{10}\implies\frac{2(p-1)}{5}$, $\fbox{D}$ If Charles has $5q +1$ quarters and Richard has $q+5$ quarters, then the difference between them is $4q-4$ quarters. Converting quarters to dimes, the difference between them is $\frac{5}{2}\cdot (4)(q-1)$, or $10(q-1)$ dimes, $\implies \fbox{A}$

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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