Difference between revisions of "1998 AJHSME Problems/Problem 8"

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==Problem 8==
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==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?
 
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.
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<math>30</math> days multiplied by <math>0.5</math> gallons a day results in <math>15</math> gallons of water loss.
 
 
<math>200-15=185=\boxed{C}</math>
 
  
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The remaining water is <math>200-15=185=\boxed{C}</math>
  
 
== See also ==
 
== See also ==
{{AJHSME box|year=1998|before=[[1997 AJHSME Problems|1997 AJHSME]]|after=[[1999 AMC 8 Problems|1999 AMC 8]]}}
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{{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]]
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{{MAA Notice}}

Latest revision as of 00:28, 5 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?

$\text{(A)}\ 140 \qquad \text{(B)}\ 170 \qquad \text{(C)}\ 185 \qquad \text{(D)}\ 198.5 \qquad \text{(E)}\ 199.85$

Solution

$30$ days multiplied by $0.5$ gallons a day results in $15$ gallons of water loss.

The remaining water is $200-15=185=\boxed{C}$

See also

1998 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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All AJHSME/AMC 8 Problems and Solutions

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