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Difference between revisions of "1995 AHSME Problems/Problem 5"

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==Solution==
 
==Solution==
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The rectangular field is 3600 inches wide and 4800 inches long, so it has an area of 17280000 square inches. We multiply by three to account for the ants: 51840000, which is closes to 50 million <math>\Rightarrow \mathrm{(C)}</math>
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The rectangular field is <math>300 \text{ feet} \cdot \frac{12 \text{ inches}}{1 \text{ foot}} =  3600 \text{ inches}</math> wide and <math>400 \text{ feet} \cdot \frac{12 \text{ inches}}{1 \text{ foot}} =  4800 \text{ inches}</math> inches long.
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It has an area of <math>3.6\cdot 10^3 \cdot 4.8 \cdot 10^3 = 17.28 \cdot 10^6</math> square inches.  
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We multiply by <math>3</math> to account for the ants to get approximately <math>50 \cdot 10^6</math>, which is <math>50</math> million <math>\Rightarrow \mathrm{(C)}</math>
  
 
==See also==
 
==See also==
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{{Old AMC12 box|year=1995|num-b=4|num-a=6}}
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{{AHSME box|year=1995|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 12:58, 5 July 2013

Problem

A rectangular field is 300 feet wide and 400 feet long. Random sampling indicates that there are, on the average, three ants per square inch through out the field. [12 inches = 1 foot.] Of the following, the number that most closely approximates the number of ants in the field is


$\mathrm{(A) \ \text{500 thousand} } \qquad \mathrm{(B) \ \text{5 million} } \qquad \mathrm{(C) \ \text{50 million} } \qquad \mathrm{(D) \ \text{500 million} } \qquad \mathrm{(E) \ \text{5 billion} }$

Solution

The rectangular field is $300 \text{ feet} \cdot \frac{12 \text{ inches}}{1 \text{ foot}} = 3600 \text{ inches}$ wide and $400 \text{ feet} \cdot \frac{12 \text{ inches}}{1 \text{ foot}} = 4800 \text{ inches}$ inches long.

It has an area of $3.6\cdot 10^3 \cdot 4.8 \cdot 10^3 = 17.28 \cdot 10^6$ square inches.

We multiply by $3$ to account for the ants to get approximately $50 \cdot 10^6$, which is $50$ million $\Rightarrow \mathrm{(C)}$

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AHSME Problems and Solutions

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