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

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<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>
  
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==Solution==
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==Solution 1==
 
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>.
 
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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==Solution 2==
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For the first five days, each day you are 10 minutes short of 85 minutes. And for the nest three days, you are 5 minutes above 85 minutes. So in total you are missing 3*5-5*10 which equals to negative 35. So on the ninth day, to have an average of 85 minutes, you need to skate for 85+35 minutes, which is 120 minutes, or <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 20:29, 22 December 2021

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 1

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 $\boxed{\text{(E)}\ 2\ \text{hr}}$.

Solution 2

For the first five days, each day you are 10 minutes short of 85 minutes. And for the nest three days, you are 5 minutes above 85 minutes. So in total you are missing 3*5-5*10 which equals to negative 35. So on the ninth day, to have an average of 85 minutes, you need to skate for 85+35 minutes, which is 120 minutes, or $\boxed{\text{(E)}\ 2\ \text{hr}}$.

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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