Difference between revisions of "1986 AJHSME Problems/Problem 1"
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==Problem== | ==Problem== | ||
| − | In July 1861, <math>366</math> inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month? | + | In July 1861, <math>366</math> inches of rain fell in Cherrapunji, India. What was the [[average]] rainfall in inches per hour during that month? |
<math>\text{(A)}\ \frac{366}{31\times 24}</math> | <math>\text{(A)}\ \frac{366}{31\times 24}</math> | ||
| Line 15: | Line 15: | ||
==Solution== | ==Solution== | ||
| − | {{ | + | We can express the number of inches of rain in a [[ratio]], <math>\frac{366 \text{ inches}}{1\text{ month}}</math>. We need to find to find the number of hours, but first we need to find the number of days. We know there are 31 days in July, so we get <cmath>\frac{366 \text{ inches}}{1 \text{ month}} \times \frac{1 \text{ month}}{31 \text{ days}}</cmath> |
| + | |||
| + | We know there are 24 hours in a day, so we get <cmath>\frac{366 \text{ inches}}{1 \text{ month}} \times \frac{1 \text{ month}}{31 \text{ days}} \times \frac{1 \text{ day}}{24 \text{ hours}}</cmath> | ||
| + | |||
| + | And simplifying gets us <math>\dfrac{366 \text{ inches}}{31 \times 24 \text{ hours}}</math>, which is <math>\boxed{\text{A}}</math>. | ||
| + | |||
| + | Note that in the simplifying step, we noted that the month and day units cancel, leaving us with only inches per hour. | ||
==See Also== | ==See Also== | ||
| − | [[ | + | {{AJHSME box|year=1986|before=First<br>Problem|num-a=2}} |
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:02, 3 July 2013
Problem
In July 1861, 
inches of rain fell in Cherrapunji, India. What was the average rainfall in inches per hour during that month?
Solution
We can express the number of inches of rain in a ratio, 
. We need to find to find the number of hours, but first we need to find the number of days. We know there are 31 days in July, so we get ![\[\frac{366 \text{ inches}}{1 \text{ month}} \times \frac{1 \text{ month}}{31 \text{ days}}\]](http://latex.artofproblemsolving.com/e/4/7/e47c78f8893cf81d05f3720052b63b88cdadd5f0.png)
We know there are 24 hours in a day, so we get ![\[\frac{366 \text{ inches}}{1 \text{ month}} \times \frac{1 \text{ month}}{31 \text{ days}} \times \frac{1 \text{ day}}{24 \text{ hours}}\]](http://latex.artofproblemsolving.com/3/9/d/39d190cbf01f0e5c1301dbd3c0fc11ae31016fb7.png)
And simplifying gets us 
, which is 
.
Note that in the simplifying step, we noted that the month and day units cancel, leaving us with only inches per hour.
See Also
| 1986 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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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