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

Latest revision as of 13:32, 29 October 2024

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

Video Solution by WhyMath

https://youtu.be/_w_u7qoayIY

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 12: / Line 12: @@
 ==Video Solution==
 https://www.youtube.com/watch?v=Ea1C64_qBH4  ~David
+==Video Solution by WhyMath==
+https://youtu.be/_w_u7qoayIY
 ==See Also==
 {{AMC8 box|year=2002|num-b=17|num-a=19}}
 {{MAA Notice}}

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