2002 AMC 8 Problems/Problem 18: Difference between revisions

Revision as of 23:45, 4 July 2013

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

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

@@ Line 10: / Line 10: @@
 ==See Also==
 {{AMC8 box|year=2002|num-b=17|num-a=19}}
+{{MAA Notice}}

Page

Toolbox

Search

2002 AMC 8 Problems/Problem 18: Difference between revisions

Revision as of 23:45, 4 July 2013

Problem

Solution

See Also