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Difference between revisions of "1967 AHSME Problems/Problem 30"

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== Solution ==
 
== Solution ==
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<math>\fbox{D}</math>
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The first <math>2</math> radios are sold for <math>\frac{d}{2}</math> dollars each, for income of <math>2 \cdot \frac{d}{2}</math>, or simply <math>d</math>.
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The remaining <math>n-2</math> radios are sold for <math>d+8</math> dollars each, for income of <math>(n-2)(d+8)</math>, which expands to <math>nd +8n -2d - 16</math>.
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The amount invested in buying <math>n</math> radios for <math>d</math> dollars is <math>nd</math>.
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Thus, the total profit is <math>(d) + (nd - 2d + 8n - 16) - nd</math>, which simplifies to <math>8n - d - 16</math>.
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Setting this equal to <math>72</math> gives <math>8n - d - 16 = 72</math>, or <math>n = 11 + \frac{d}{8}</math>.
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Since <math>n</math> must be an integer, <math>d</math> must be a multiple of <math>8</math>.  If we want to minimize <math>n</math>, we want to minimize <math>d</math>.  Since <math>d>0</math>, setting <math>d=8</math> yields <math>n=12</math>, giving the answer of <math>\fbox{D}</math>
  
 
== See also ==
 
== See also ==
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{{AHSME box|year=1967|num-b=29|num-a=31}}   
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{{AHSME 40p box|year=1967|num-b=29|num-a=31}}   
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:40, 16 August 2023

Problem

A dealer bought $n$ radios for $d$ dollars, $d$ a positive integer. He contributed two radios to a community bazaar at half their cost. The rest he sold at a profit of $8 on each radio sold. If the overall profit was $72, then the least possible value of $n$ for the given information is:

$\textbf{(A)}\ 18\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 15\qquad \textbf{(D)}\ 12\qquad \textbf{(E)}\ 11$

Solution

The first $2$ radios are sold for $\frac{d}{2}$ dollars each, for income of $2 \cdot \frac{d}{2}$, or simply $d$.

The remaining $n-2$ radios are sold for $d+8$ dollars each, for income of $(n-2)(d+8)$, which expands to $nd +8n -2d - 16$.

The amount invested in buying $n$ radios for $d$ dollars is $nd$.

Thus, the total profit is $(d) + (nd - 2d + 8n - 16) - nd$, which simplifies to $8n - d - 16$.

Setting this equal to $72$ gives $8n - d - 16 = 72$, or $n = 11 + \frac{d}{8}$.

Since $n$ must be an integer, $d$ must be a multiple of $8$. If we want to minimize $n$, we want to minimize $d$. Since $d>0$, setting $d=8$ yields $n=12$, giving the answer of $\fbox{D}$

See also

1967 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Problem 31
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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