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 1Leg 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 2Leg 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 speedLeg 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.

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