Difference between revisions of "2001 AMC 8 Problems/Problem 5"
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==Solution== | ==Solution== | ||
During the <math> 10 </math> seconds, the sound traveled <math> 1088\times10=10880 </math> feet from the lightning to Snoopy. This is equivalent to <math> \frac{10880}{5280}\approx2 </math> miles, <math> \boxed{\text{C}} </math>. | During the <math> 10 </math> seconds, the sound traveled <math> 1088\times10=10880 </math> feet from the lightning to Snoopy. This is equivalent to <math> \frac{10880}{5280}\approx2 </math> miles, <math> \boxed{\text{C}} </math>. | ||
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| + | ==Solution 2== | ||
| + | Since we are estimating, we can make <math>1088</math> become <math>1000</math>, and <math>5280</math> will become <math>5000</math>. Then do <math>5000÷1000</math> and we get \boxed{\text{C}} $. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2001|num-b=4|num-a=6}} | {{AMC8 box|year=2001|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:12, 2 January 2023
Contents
Problem
On a dark and stormy night Snoopy suddenly saw a flash of lightning. Ten seconds later he heard the sound of thunder. The speed of sound is 1088 feet per second and one mile is 5280 feet. Estimate, to the nearest half-mile, how far Snoopy was from the flash of lightning.
Solution
During the 
seconds, the sound traveled 
feet from the lightning to Snoopy. This is equivalent to 
miles, 
.
Solution 2
Since we are estimating, we can make 
become 
, and 
will become 
. Then do 
and we get \boxed{\text{C}} $.
See Also
| 2001 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
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