Question

A man has to travel 50km in two hours. He could cover 20km in one hour, and then had to stop for 10
minutes for refueling. By what factor should he increase his speed with reference to that during the first hour
so as to be able to complete the journey as per schedule?​

Answer 1

Answer:

1.8

Step-by-step explanation:

total time he has = 2 hrs

total distance he has to covered = 50 km

in first 1 hr he completed 20 km

so S1= 20 km/hr

so remaining distance = 50-20 = 30 Km

and remaining time = 1 hr

now he spent 10 minutes for refueling

so total remaining time he has = 60-10 = 50 minutes

he has to complete 30 km in this time

speed = 30\(10/60) km/hr

speed = 36 km/hr

for making a factor of initial speed as 20 km/hr

multiply 36 with 20/20

speed = 36x(20/20)

=(1.8x S1 )km\hr

= 1.8 km/hr ---- Ans (As S1 is 20km/hr)

Answer 2

Given :- A man has to travel 50km in two hours. He could cover 20km in one hour, and then had to stop for 10 minutes for refueling. By what factor should he increase his speed with reference to that during the first hour so as to be able to complete the journey as per schedule?

Solution :-

given that,

→ Total distance covered by man in 2 hours = 50km.

→ in first hour he cover = 20km.

So,

→ Distance left to be cover = 50 - 20 = 30km.

→ Time left = 2 - 1 = 1 hour.

now, given that, he had to stop for 10 minutes for refueling.

So,

→ Time left for travel = 1 hour - 10 min. = 60 - 10 = 50 minutes.

Therefore,

→ Distance to be covered now = 30km.

→ Time left = 50 minutes = (50/60) = (5/6) hours.

→ New speed = Distance/Time = 30/(5/6) = 30 * (6/5) = 36 km/h.

Hence,

→ Speed in first hour = 20km/h .

→ speed in second time = 36km/h.

→ Increased speed in factor = (36 /20) = 1.8 times. (Ans.)

∴ He has to increased his speed 1.8 times in order to complete the journey as per schedule time.

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