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Difference between revisions of "1952 AHSME Problems/Problem 38"

(Created page with "== Problem == The area of a trapezoidal field is <math>1400</math> square yards. Its altitude is <math>50</math> yards. Find the two bases, if the number of yards in each base i...")
 
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== Solution ==
 
== Solution ==
−
<math>\fbox{}</math>
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Let us denote <math>8m</math> and <math>8n</math> to be our bases. Without loss of generality, let <math>m \le n</math>.
 +
 
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Thus,
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<cmath>50 * \frac{8m + 8n}{2} = 1400</cmath>
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<cmath>4m + 4n = 28</cmath>
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<cmath>m + n = 7</cmath>
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Since <math>m</math> & <math>n</math> are integers, we see that the only solutions to this equation are <math>(1,6)</math>, <math>(2,5)</math>, and <math>(3,4)</math>.
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Therefore, the answer is <math>\fbox{(D) three}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 00:19, 22 December 2015

Problem

The area of a trapezoidal field is $1400$ square yards. Its altitude is $50$ yards. Find the two bases, if the number of yards in each base is an integer divisible by $8$. The number of solutions to this problem is:

$\textbf{(A)}\ \text{none} \qquad \textbf{(B)}\ \text{one} \qquad \textbf{(C)}\ \text{two} \qquad \textbf{(D)}\ \text{three} \qquad \textbf{(E)}\ \text{more than three}$

Solution

Let us denote $8m$ and $8n$ to be our bases. Without loss of generality, let $m \le n$.

Thus, \[50 * \frac{8m + 8n}{2} = 1400\] \[4m + 4n = 28\] \[m + n = 7\]

Since $m$ & $n$ are integers, we see that the only solutions to this equation are $(1,6)$, $(2,5)$, and $(3,4)$. Therefore, the answer is $\fbox{(D) three}$

See also

1952 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 37 		Followed by
Problem 39
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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