Every day in the morning Ross cycles for 2 hours

Question

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.

B.

C.

D.

E.

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

EgmatQuantExpert · Answer

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:
 https://e-gmat.com/blogs/wp-content/uploads/2018/05/FollowupQ3_1.png

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.

Answer: A

SSM700plus · Answer

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.

.

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

amanvermagmat · Answer

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

Posted from my mobile device

generis · Answer

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

Answer A

*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.

gracie · Answer

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

LeonidK · Answer

Please add official answer to timer

Bunuel · Answer

_______________________
Done. Thank you.

GMATGuruNY · Answer

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.

A

NinetyFour · Answer

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

fireagablast · Answer

Some length approaches here..

Given 1 mile every 12 minutes = 5 miles per hour
He travels for two hours
5*2=10 miles

Return trip is double the distance at half the speed
10*2 = 2.5t
t=8

Average speed for the trip is the total distance over the total time
(20+10)/(2+8) = 30/10 = 3

bumpbot · Answer

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Every day in the morning Ross cycles for 2 hours

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