Question

A man walks half of the journey at 4km/h by cycle, does one-third of journey at 12 km/h and rides the remainder journey in a horse cart at 9 km/h, thus completing the whole journey in 6 hrs and 12 min. The length of the journey is:

Answer 1

Answer:

A man walks half of the journey at 4km/h by cycle, does one-third of journey at 12 km/h and rides the remainder journey in a horse cart at 9 km/h, thus completing the whole journey in 6 hrs and 12 min. The length of the journey is:

Step-by-step explanation:

A man goes to a place at the rate of 4 kmph. He comes back on a bicycle at 16kmph. His average speed for the entire journey is how much ?

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It's similar to parallel resistors or pipes filling tanks, but times 2

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Avg = 2*product/sum

= 2*4*16/(4 + 16) = 128/20

= 6.4 km/hr

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It's always less than the average of the 2 speeds, 4 & 16 --> (10), because more time is spent at the lower speed.

Also notice that the distance is irrelevant.

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I'll work it out:

r1 = rate going, r2 = rate returning, d = distance one-way

t = d/r1 + d/r2 = d(r1 + r2)/r1*r2

t/d = (r1 + r2)/r1*r2

d/t = r1*r2/(r1 + r2)

Avg speed = 2d/t for the round trip

Avg speed = 2*r1*r2/(r1 + r2)

Answer 2

Answer:

The length of the journey is 46.48km.

Step-by-step explanation:

Given that,

A man walks half of the journey at 4km/h by cycle, does one-third of journey at 12 km/h and rides the remainder journey in a horse cart at 9 km/h, thus completing the whole journey in 6 hrs and 12 min.

We are required to find the total distance covered during the journey.

The journey is divided into three parts in which the time taken for each part and speed is written as shown below-

$v_{1}=4km/h,t_{1}=\frac{6hr,12min}{2} =3h,6min=3.1hrs\\v_{2}=12km/h,t_{2}=\frac{6hr,12min}{3} =2hr,4min=2.06hr\\v_{3}=9km/h,t_{3}=6.2-(3.1+2.06)=1.04hrs$

We know that,

Speed is the ratio of distance to the time.Thus distance is the product of speed and time.Hence,

The total distance covered is $d=v_{1}t_{1}+v_{2}t_{2}+v_{3}t_{3}$

Substituting the known values in the above relation,we get

$d=(4)(3.1)+(12)(2.06)+9(1.04)=46.48km$

Therefore,the length of the journey is 46.48km.

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