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> | ||
| − | ==Solution== | + | ==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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | For the first five days, each day you are <math>10</math> minutes short of <math>85</math> minutes. And for the next three days, you are <math>5</math> minutes above <math>85</math> minutes. So in total you are missing <math>3*5-5*10</math>, which equals to negative <math>35</math>. So on the ninth day, to have an average of <math>85</math> minutes, Gage need to skate for <math>85+35</math> minutes, which is <math>120</math> minutes, or <math>\boxed{\text{(E)}\ 2\ \text{hr}}</math>. | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=Ea1C64_qBH4 ~David | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/_w_u7qoayIY | ||
==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}} | ||
Latest revision as of 13:32, 29 October 2024
Contents
Problem
Gage skated 
hr 
min each day for 
days and hr 
min each day for 
days. How long would he have to skate the ninth day in order to average 
minutes of skating each day for the entire time?
Solution 1
Converting into minutes and adding, we get that she skated 
minutes total, where 
is the amount she skated on day 
. Dividing by to get the average, we get 
. Solving for , ![\[645+x=765\]](http://latex.artofproblemsolving.com/1/4/2/1427cf5f8cb037657e41bdb1630cc14a495e7c2b.png)
![\[x=120\]](http://latex.artofproblemsolving.com/b/d/9/bd9f7a95736540cfe5ac1d90c69c4c2801cbf11f.png)
Now we convert back into hours and minutes to get 
.
Solution 2
For the first five days, each day you are 
minutes short of minutes. And for the next three days, you are minutes above minutes. So in total you are missing 
, which equals to negative 
. So on the ninth day, to have an average of minutes, Gage need to skate for 
minutes, which is 
minutes, or .
Video Solution
https://www.youtube.com/watch?v=Ea1C64_qBH4 ~David
Video Solution by WhyMath
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
