Art of Problem Solving

1951 AHSME Problems/Problem 4: Difference between revisions

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New page: The answer is (E) 720. 4[50+65]+2[130]=720
 
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The answer is (E) 720.
==Problem==
A barn with a roof is rectangular in shape, <math>10</math> yd. wide, <math>13</math> yd. long and <math>5</math> yd. high.  It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is:


4[50+65]+2[130]=720
<math> \mathrm{(A) \ } 360 \qquad \mathrm{(B) \ } 460 \qquad \mathrm{(C) \ } 490 \qquad \mathrm{(D) \ } 590 \qquad \mathrm{(E) \ } 720 </math>
 
==Solution==
The walls are <math>13*5=65</math> and <math>10*5=50</math> in area, and the ceiling has an area of <math>10*13=130</math>.
 
<math>((65+50)2)2+130=590 \Rightarrow \boxed{\mathrm{(D) \ }}</math>
 
==See also==

Revision as of 09:01, 5 February 2008

Problem

A barn with a roof is rectangular in shape, $10$ yd. wide, $13$ yd. long and $5$ yd. high. It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is:

$\mathrm{(A) \ } 360 \qquad \mathrm{(B) \ } 460 \qquad \mathrm{(C) \ } 490 \qquad \mathrm{(D) \ } 590 \qquad \mathrm{(E) \ } 720$

Solution

The walls are $13*5=65$ and $10*5=50$ in area, and the ceiling has an area of $10*13=130$.

$((65+50)2)2+130=590 \Rightarrow \boxed{\mathrm{(D) \ }}$

See also