Difference between revisions of "1998 AJHSME Problems/Problem 8"
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A child's wading pool contains 200 gallons of water. If water evaporates at the rate of 0.5 gallons per day and no other water is added or removed, how many gallons of water will be in the pool after 30 days? | A child's wading pool contains 200 gallons of water. If water evaporates at the rate of 0.5 gallons per day and no other water is added or removed, how many gallons of water will be in the pool after 30 days? | ||
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==Solution== | ==Solution== | ||
| − | <math>30</math> days multiplied by <math>0.5</math> gallons a day results in <math>15</math> gallons. | + | <math>30</math> days multiplied by <math>0.5</math> gallons a day results in <math>15</math> gallons of water loss. |
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| + | The remaining water is <math>200-15=185=\boxed{C}</math> | ||
== See also == | == See also == | ||
| − | {{AJHSME box|year=1998| | + | {{AJHSME box|year=1998|num-b=7|num-a=9}} |
* [[AJHSME]] | * [[AJHSME]] | ||
* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:28, 4 July 2013
Problem
A child's wading pool contains 200 gallons of water. If water evaporates at the rate of 0.5 gallons per day and no other water is added or removed, how many gallons of water will be in the pool after 30 days?
Solution
days multiplied by 
gallons a day results in 
gallons of water loss.
The remaining water is 
See also
| 1998 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
