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Difference between revisions of "1953 AHSME Problems/Problem 6"
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== Solution ==
 
== Solution ==
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<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>
 
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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Note that <math>q</math> does not stand for the number of quarters that they have.
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==See Also==
 
==See Also==
  

Latest 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

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, $\implies \fbox{A}$

Note that $q$ does not stand for the number of quarters that they have.

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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All AHSME Problems and Solutions

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