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e-GMAT Representative V
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Every day in the morning Ross cycles for 2 hours  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 63% (02:30) correct 37% (02:45) wrong based on 204 sessions

### HideShow timer Statistics Application of Average Speed in Distance problems - Exercise Question #3

Every day in the morning Ross cycles for 2 hours. He always starts at a constant rate of 1 mile of distance in every 12 minutes. However, he takes different path for onward and return journey – the distance covered in the return journey is double of the distance covered in his onward journey. Also, his speed in the return journey becomes half of his original speed. What is his average speed in the whole journey?

A. 3 mph
B. 4 mph
C. 5 mph
D. 6 mph
E. 7 mph

To solve question 4: Question 4

To read the article: Application of Average Speed in Distance problems

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Originally posted by EgmatQuantExpert on 09 May 2018, 08:36.
Last edited by EgmatQuantExpert on 13 Aug 2018, 06:15, edited 7 times in total.
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 2942
Re: Every day in the morning Ross cycles for 2 hours  [#permalink]

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Solution

Given:
• Every morning Ross cycles for 2 hours
• He always starts at a constant rate of 1 mile of distance in every 12 minutes
• He takes different routes for onward and return journey
o The distance covered in the return journey is double of the distance covered in the onward journey
o The speed in the return journey becomes half of his original speed

To find:
• Average speed of Ross in his whole journey

Approach and Working:
• If Ross covers a constant rate of 1 mile of distance in every 12 minutes, then his speed in the onward journey = $$\frac{1}{12}$$ * 60 mph = 5 mph
• As his speed becomes half in the return journey, his return journey speed = $$\frac{5}{2}$$ mph = 2.5 mph
• Let us assume the distance covered in the onward journey is d miles
o Therefore, the distance covered in the return journey is 2d miles
• Hence, the time taken to complete the onward journey = $$\frac{d}{5}$$ hrs
o And similarly, the time taken to complete the return journey = $$\frac{2d}{2.5}$$ hrs
All this information can be collated in the following table: We already know that the total journey time, combining both onward and return journey, is 2 hrs.
• Hence, we can say, $$\frac{d}{5} + \frac{2d}{2.5}$$ = 2
o Solving, we get d = 2 miles
• Therefore, the average speed of the whole journey = total distance travelled/total time taken
o Or, average speed = (d + 2d)/(d/5 + 2d/2.5)
Replacing the value of d = 2, we get
Average speed = (2 + 4)/(2/5 + 4/2.5) mph = 3 mph

Hence, the correct answer is option A.

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Originally posted by EgmatQuantExpert on 09 May 2018, 08:37.
Last edited by EgmatQuantExpert on 14 May 2018, 00:33, edited 1 time in total.
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Re: Every day in the morning Ross cycles for 2 hours  [#permalink]

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2
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EgmatQuantExpert wrote:
Application of Average Speed in Distance problems - Exercise Question #3

Every day in the morning Ross cycles for 2 hours. He always starts at a constant rate of 1 mile of distance in every 12 minutes. However, he takes different path for onward and return journey – the distance covered in the return journey is double of the distance covered in his onward journey. Also, his speed in the return journey becomes half of his original speed. What is his average speed in the whole journey?

A. 3 mph
B. 4 mph
C. 5 mph
D. 6 mph
E. 7 mph

To solve question 4: Question 4

To read the article: Application of Average Speed in Distance problems

d - distance traveled in onward direction
given speed = 1 miles in 12 minutes = 5 miles in 60 minutes = 5 mph.

Time taken for onward journey = $$\frac{distance traveled}{time taken to travel}$$ = $$\frac{d}{5}$$

Time taken for return journey = $$\frac{twice distance traveled for onward journey}{half the time taken for onward journey}$$ = $$\frac{4d}{5}$$

Total time taken = $$\frac{d}{5}$$ + $$\frac{4d}{5}$$ = d.

Quote:
Every day in the morning Ross cycles for 2 hours
.

it implies d =2.

Average speed = $$\frac{Distance traveled in onward journey + distance traveled in return journey}{Total time taken}$$
= $$\frac{d + 2d}{2}$$
= $$\frac{3d}{2}$$
= $$\frac{6}{2}$$ = 3mph

Ans A
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Re: Every day in the morning Ross cycles for 2 hours  [#permalink]

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1
Another way to look at this:

Return journey compared to onward journey- distance D is double but speed s is half. Since time t = D/s, time in return journey will be 4 times of that in onward journey.

So out of total time of 2 hours or 120 minutes, onward time to return time ratio = 1:4

Divide 120 minutes in this ratio, 24 minutes onward and 96 minutes return.

Onward time is 24 minutes,and covers 1 mile in 12 minutes. So in 24 minutes he covers 2 miles.

This onward distance is 2 miles so return distance must be 4 miles. Total distance= 6 miles.

So 6 miles covered in 2 hours. Average speed = 6/2 = 3mph

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Senior SC Moderator V
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Re: Every day in the morning Ross cycles for 2 hours  [#permalink]

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EgmatQuantExpert wrote:

Every day in the morning Ross cycles for 2 hours. He always starts at a constant rate of 1 mile of distance in every 12 minutes. However, he takes different path for onward and return journey – the distance covered in the return journey is double of the distance covered in his onward journey. Also, his speed in the return journey becomes half of his original speed. What is his average speed in the whole journey?

A. 3 mph
B. 4 mph
C. 5 mph
D. 6 mph
E. 7 mph

I had the same thought as amanvermagmat : Leg 2 is twice the distance of Leg 1, and Leg 2's speed is half that of Leg 1. Slightly different approach.

Base D, S, and T on Leg 1.
Direct proportion between time and distance. Twice the distance = twice the time
Inverse proportion between speed and time. Half the speed = twice the time. Ultimately:

Leg 1: D = 10, S = 5, T = 2
Leg 2: D = 20, S = 2.5, T = 8

Leg 1
Leg 1, speed: $$(\frac{1mi}{12min}*60)=\frac{5mi}{1hour}$$

Leg 1, Time: Leg 1 gets overall time: 2 hours

Leg 1, D: If the whole trip were done at Leg 1's speed, he would cover (5 mph * 2 hrs) = 10 miles

Leg 2
Leg 2, D = (2 * Leg 1): 20 miles
Leg 2, Speed: half of Leg 1: $$\frac{5}{2}$$ mph

Leg 2, Time? Will be FOUR times that of Leg 1's time.
Let $$S$$ = Leg 1 speed

Leg 1, D: 10 miles
Leg 2, D: 20 miles

Leg 1, speed: S
Leg 2, speed: 0.5S

Distance = Speed*Time
10 miles = (S)(2 hours)
20 miles = (0.5S)(T2 = ?)

$$\frac{(T2*0.5S)}{(2S)}$$ = $$\frac{20}{10}$$
$$(10)*(T2)*(0.5S) = 40S$$
$$T2 = \frac{40S}{5S}$$
$$T2 = 8$$
hours

Average speed
Leg 1: D = 10, T = 2
Leg 2: D = 20, T = 8

Average speed: $$\frac{TotalDistance}{TotalTime}$$
Total D: 10 + 20 = 30 mi
Total T: 2 + 8 = 10 hrs
Average speed:$$\frac{30}{10}=3$$ mph

*OR Leg 2, T = FOUR times that of Leg 1 (= 2 hrs)
At Leg 1's speed, if D = 20, time would = 4 hrs
At half that speed, Leg 2 = twice that time: 8 hours
$$\frac{20}{(\frac{5}{2})}=(20*\frac{2}{5})=8$$ hrs
Leg 2, T: 8 hours

With thanks to hazelnut , who cleared a doubt and improved the answer yesterday.

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Re: Every day in the morning Ross cycles for 2 hours  [#permalink]

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EgmatQuantExpert wrote:
Application of Average Speed in Distance problems - Exercise Question #3

Every day in the morning Ross cycles for 2 hours. He always starts at a constant rate of 1 mile of distance in every 12 minutes. However, he takes different path for onward and return journey – the distance covered in the return journey is double of the distance covered in his onward journey. Also, his speed in the return journey becomes half of his original speed. What is his average speed in the whole journey?

[list]A. 3 mph
B. 4 mph
C. 5 mph
D. 6 mph
E. 7 mph

let total distance=3d
d/5+2d/2.5=2 hrs
d=2 miles
3d=6 miles
6 miles/2 hrs=3 mph average speed
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Re: Every day in the morning Ross cycles for 2 hours  [#permalink]

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Re: Every day in the morning Ross cycles for 2 hours  [#permalink]

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LeonidK wrote:

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Done. Thank you.
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Re: Every day in the morning Ross cycles for 2 hours  [#permalink]

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EgmatQuantExpert wrote:
Application of Average Speed in Distance problems - Exercise Question #3

Every day in the morning Ross cycles for 2 hours. He always starts at a constant rate of 1 mile of distance in every 12 minutes. However, he takes different path for onward and return journey – the distance covered in the return journey is double of the distance covered in his onward journey. Also, his speed in the return journey becomes half of his original speed. What is his average speed in the whole journey?

A. 3 mph
B. 4 mph
C. 5 mph
D. 6 mph
E. 7 mph

The information in red is irrelevant.
Only the following facts matter:
Ross travels onward at a rate of 1 mile every 12 minutes.
He travels home for twice the onward distance at half the onward speed.

Onward speed = 1 mile per 12 minutes = 5 miles per 60 minutes = 5 mph.
Let the onward distance = 5 miles.
Time to travel 5 miles onward at a rate of 5 mph = d/r = 5/5 = 1 hour.
Time to travel 10 miles home at a rate of 2.5 mph = d/r = 10/2.5 = 100/25 = 4 hours.
Since the total time = 1+4 = 5 hours, the average speed for the entire 15-mile trip = (total distance)/(total time) = 15/5 = 3 mph.

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Re: Every day in the morning Ross cycles for 2 hours  [#permalink]

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EgmatQuantExpert wrote:
Application of Average Speed in Distance problems - Exercise Question #3

Every day in the morning Ross cycles for 2 hours. He always starts at a constant rate of 1 mile of distance in every 12 minutes. However, he takes different path for onward and return journey – the distance covered in the return journey is double of the distance covered in his onward journey. Also, his speed in the return journey becomes half of his original speed. What is his average speed in the whole journey?

A. 3 mph
B. 4 mph
C. 5 mph
D. 6 mph
E. 7 mph

My thought process if it helps anyone:

Total time = 2 hours

Rate starting is 1 mile/12 min OR 5 miles/hr
Rate going back is half of the rate starting so 5/2miles/hr

Time = distance/rate AND

time going there + time coming back = 2 hours

\frac{d}{5} + \frac{4d}{5} = 2hours

d = 2

average speed = total distance/total time

total distance = 3(d) = 6 miles

Average speed = 6 miles/2hours = 3miles/hour Re: Every day in the morning Ross cycles for 2 hours   [#permalink] 20 Jan 2019, 14:27
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