Difference between revisions of "1998 AJHSME Problems/Problem 8"
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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. |
| − | |||
| − | |||
| + | The remaining water is <math>200-15=185=\boxed{C}</math> | ||
== See also == | == See also == | ||
Revision as of 10:13, 31 July 2011
Problem 8
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 | ||
