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A girl rode her bicycle from home to school, a distance of 1 5 miles,

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Bunuel
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A girl rode her bicycle from home to school, a distance of 1 5 miles, [#permalink] New post   12 Feb 2018, 00:08
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A girl rode her bicycle from home to school, a distance of 15 miles, at an average speed of 15 miles per hour. She returned home from school by walking at an average speed of 5 miles per hour. What was her average speed for the round trip if she took the same route in both directions?

(A) 7.5 miles per hour

(B) 10 miles per hour

(C) 12.5 miles per hour

(D) 13 miles per hour

(E) 25 miles per hour
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Re: A girl rode her bicycle from home to school, a distance of 1 5 miles, [#permalink] New post   12 Feb 2018, 01:13
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Time taken to ride from home to school= \(\frac{15}{15}\)= 1 hour

Time taken to walk from school to home= \(\frac{15}{5}\)= 3 hour

Total distance= 15+15= 30 miles

Total time= 3+1= 4 hours

Average speed= \(\frac{Total Distance}{Total Time}\)= \(\frac{30}{4}\)= 7.5 miles per hour

Answer:A.
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Re: A girl rode her bicycle from home to school, a distance of 1 5 miles, [#permalink] New post   12 Feb 2018, 02:32
Bunuel wrote:
A girl rode her bicycle from home to school, a distance of 15 miles, at an average speed of 15 miles per hour. She returned home from school by walking at an average speed of 5 miles per hour. What was her average speed for the round trip if she took the same route in both directions?

(A) 7.5 miles per hour

(B) 10 miles per hour

(C) 12.5 miles per hour

(D) 13 miles per hour

(E) 25 miles per hour


In addition to the approach shown above, we can also use the properties of ratios.
This is a Logical methodology.

Since she walked home at 1/3 of the speed, the walk took 3 times the time.
Then the ratio of times is 1:3 giving a total of 4 'time units'.
Our two speeds are 15 and 5 which have a difference of 10.
Dividing 10 into the 4 'units' gives 2.5.
Then we need to add one 'unit' of 2.5 to the more dominant speed (which is to say, the one she spent more time at).
This is 5 + 2.5 = 7.5
(A) is our answer.
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Re: A girl rode her bicycle from home to school, a distance of 1 5 miles, [#permalink] New post   13 Feb 2018, 18:00
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Bunuel wrote:
A girl rode her bicycle from home to school, a distance of 15 miles, at an average speed of 15 miles per hour. She returned home from school by walking at an average speed of 5 miles per hour. What was her average speed for the round trip if she took the same route in both directions?

(A) 7.5 miles per hour

(B) 10 miles per hour

(C) 12.5 miles per hour

(D) 13 miles per hour

(E) 25 miles per hour


We can calculate average speed by using the formula average speed = total distance/total time.

The total distance she traveled was 15 x 2 = 30 miles. The time going to school was 15/15 = 1 hour. The time coming home was 15/5 = 3 hours.

Thus, the average speed for the round trip was:

30/4 = 7.5 mph

Answer: A
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Re: A girl rode her bicycle from home to school, a distance of 1 5 miles, [#permalink] New post   30 Mar 2020, 10:28
Bunuel wrote:
A girl rode her bicycle from home to school, a distance of 15 miles, at an average speed of 15 miles per hour. She returned home from school by walking at an average speed of 5 miles per hour. What was her average speed for the round trip if she took the same route in both directions?

(A) 7.5 miles per hour

(B) 10 miles per hour

(C) 12.5 miles per hour

(D) 13 miles per hour

(E) 25 miles per hour


Given:
1. A girl rode her bicycle from home to school, a distance of 15 miles, at an average speed of 15 miles per hour.
2. She returned home from school by walking at an average speed of 5 miles per hour.

Asked: What was her average speed for the round trip if she took the same route in both directions?

Total distance = 30 miles
Total time = 1 hr + 3 hr = 4hr
Average speed = 30 miles / 4hr = 75 miles/hr

IMO A
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Re: A girl rode her bicycle from home to school, a distance of 1 5 miles, [#permalink] New post   26 Apr 2022, 04:59
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Re: A girl rode her bicycle from home to school, a distance of 1 5 miles, [#permalink]
26 Apr 2022, 04:59
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