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Clarissa spent all day on a sightseeing trip in Britain

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Clarissa spent all day on a sightseeing trip in Britain [#permalink]

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New post 28 Feb 2017, 02:57
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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
[Reveal] Spoiler: OA
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Re: Clarissa spent all day on a sightseeing trip in Britain [#permalink]

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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 :-)
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Re: Clarissa spent all day on a sightseeing trip in Britain [#permalink]

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New post 02 Mar 2017, 16:58
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


We can use the following formula:

average rate = total distance/total time

We need to find the distance for each leg of the trip and the time for each leg of the trip.

First leg (countryside): Since the bus traveled 15 mph for 30 miles, the time for that stretch was 30/15 = 2 hours.

Second leg (London tour): Since the bus traveled for 3 hours at a rate of 10 mph, the distance traveled for that stretch was 3 x 10 = 30 miles.

Third leg (return to hotel): Finally, the bus took 2 hours and drove for 40 miles to get back to the hotel. Thus:

average rate = (30 + 30 + 40)/(2 + 3 + 2)

average rate = 100/7, which is closest to 14.

Answer: B
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Re: Clarissa spent all day on a sightseeing trip in Britain   [#permalink] 02 Mar 2017, 16:58
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