Difference between revisions of "2002 AMC 8 Problems/Problem 18"
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==Problem== | ==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? | + | 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 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>. | ||
− | + | ==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 | ||
==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}} |
Latest revision as of 20:01, 15 April 2023
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
,
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
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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