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Difference between revisions of "2002 AMC 8 Problems/Problem 18"

(Undo revision 80777 by Soccerpro101 (talk))
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==Problem==
 
==Problem==
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Gage skated 1 hr 15 min each day for 5 days and 1 hr 30 min each day for 3 days. How long would he have to skate the ninth day in order to average 85 minutes of skating each day for the entire time?
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Gage skated <math>1</math> hr <math>15</math> min each day for <math>5</math> days and <math>1</math> hr <math>30</math> min each day for <math>3</math> days. How long would he have to skate the ninth day in order to average <math>85</math> minutes of skating each day for the entire time?
  
 
<math> \text{(A)}\ \text{1 hr}\qquad\text{(B)}\ \text{1 hr 10 min}\qquad\text{(C)}\ \text{1 hr 20 min}\qquad\text{(D)}\ \text{1 hr 40 min}\qquad\text{(E)}\ \text{2 hr} </math>
 
<math> \text{(A)}\ \text{1 hr}\qquad\text{(B)}\ \text{1 hr 10 min}\qquad\text{(C)}\ \text{1 hr 20 min}\qquad\text{(D)}\ \text{1 hr 40 min}\qquad\text{(E)}\ \text{2 hr} </math>
  
 
==Solution==
 
==Solution==
−
 
+
Converting into minutes and adding, we get that she skated <math>75*5+90*3+x = 375+270+x = 645+x</math> minutes total, where <math>x</math> is the amount she skated on day <math>9</math>. Dividing by <math>9</math> to get the average, we get <math>\frac{645+x}{9}=85</math>. Solving for <math>x</math>, <cmath>645+x=765</cmath> <cmath>x=120</cmath> Now we convert back into hours and minutes to get <math>\boxed{\text{(E)}\ 2\ \text{hr}}</math>.
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Converting into minutes and adding, we get that she skated <math>75*5 +90*3 +x=375 +270 +x=645 +x</math> minutes total, where x is the amount she skated on day 9. Dividing by 9 to get the average, we get <math>\frac{645+x}{9}=85</math>. Solving for x, <cmath>645+x=765</cmath> <cmath>x=120</cmath> Now we convert back into hours and minutes to get <math>\boxed{\text{(E)}\ 2\ \text{hr}}</math>.
 
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2002|num-b=17|num-a=19}}
 
{{AMC8 box|year=2002|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:09, 11 February 2020

Problem

Gage skated $1$ hr $15$ min each day for $5$ days and $1$ hr $30$ min each day for $3$ days. How long would he have to skate the ninth day in order to average $85$ minutes of skating each day for the entire time?

$\text{(A)}\ \text{1 hr}\qquad\text{(B)}\ \text{1 hr 10 min}\qquad\text{(C)}\ \text{1 hr 20 min}\qquad\text{(D)}\ \text{1 hr 40 min}\qquad\text{(E)}\ \text{2 hr}$

Solution

Converting into minutes and adding, we get that she skated $75*5+90*3+x = 375+270+x = 645+x$ minutes total, where $x$ is the amount she skated on day $9$. Dividing by $9$ to get the average, we get $\frac{645+x}{9}=85$. Solving for $x$, \[645+x=765\] \[x=120\] Now we convert back into hours and minutes to get $\boxed{\text{(E)}\ 2\ \text{hr}}$.

See Also

2002 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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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