vikasp99 wrote:
Clarissa spent all day on a sightseeing trip in Britain. Starting from her hotel, Clarissa boarded a bus, which traveled at an average speed of 15 miles per hour through a 30 mile section of the countryside. The bus then stopped for lunch in London before continuing on a 3 hour tour of the city's sights at a speed of 10mph. Finally, the bus left the city and drove 40 miles straight back to the hotel. Clarissa arrived at her hotel exactly 2 hours after leaving London. What was the bus's average rate, approximately, for the entire journey?
A. 8
B. 14
C. 21
D. 25
E. 30
Dear
vikasp99,
I'm happy to respond.
I like the fact that the woman in the problem has such an
archetypally British name.
Because we are asking for the
average speed of the bus, we don't count the lunch break of unspecified time because presumably Clarissa wasn't on the bus for that.
For multiple legs of a trip, we have to figure out distance and time for each one.
1)
Leg #1 (the countryside)
R = 15 mph, D = 30 mi, so T = 2 hr
2)
Leg #2 within London
T = 3 hr, R = 10 mph, so D = 30 mi
3)
Leg #3 return from London
D = 40 mi, T = 2 hr
Now that we have the distance and time for each leg, we add to find the total distance and the total time.
total distance = 30 + 30 + 40 = 100 mi
total time = 2 + 3 + 2 = 7 hr
average velocity = (total distance)/(total time)
average velocity = \(\frac{(100 mi)}{(7 hr)} \approx 14 mph\)
OA =
(B) Does all this make sense?
Mike
_________________
Mike McGarry
Magoosh Test PrepEducation is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)