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

Revision as of 19:01, 15 April 2023

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 next 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, Gage need to skate for $85+35$ minutes, which is $120$ minutes, or $\boxed{\text{(E)}\ 2\ \text{hr}}$ .

Video Solution

https://www.youtube.com/watch?v=Ea1C64_qBH4 ~David

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.

@@ Line 1: / Line 1: @@
 ==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>
-==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 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>.
+==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==
 {{AMC8 box|year=2002|num-b=17|num-a=19}}
+{{MAA Notice}}

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