Question

A car travels a total distance of 300 km. After
travelling a part of the distance without any trouble,
the car develops an engine problem and proceeds
at 3/4th of its former speed and arrives at the
destination 80 minutes late. Had the problem
developed 50 km further on, the car would have
arrived 20 minutes sooner. Find the original distance
it travelled without any problem, and the speed over
that part of the journey.​

Answer 1

Step by step explanation:

• So according to the question we have let the car travel from x to y that is the distance 300 km. This it travels till z without any trouble. It travels with a speed v and covers the remaining distance from z to y with a speed of 3v/4.
• Now the engine develops a problem and reaches y 80 minutes late.
• Also the car travels an extra 50 km and reaches m.
• Here (m = z + 50) with speed v and then travels from m to y with speed 3v/4 and reaches 20 minutes early
• So it is 60 minutes late.
• Therefore we have
•       50/(3v/4) – 50/v = 20/60
•        200/3v – 50/v = 1/3
•           200 – 150 / 3v = 1/3
•                  50/v = 1
•               Or v = 50 kmph
• Now for 50 km,20 minutes is made up of the entire 80 minutes.
• Therefore we get 50 x 4 = 200 km
• Now total distance is 300 km
• So 300 – 200 = 100 km

Reference link will be

https://brainly.in/question/15883513

Answer 2

The original distance it travelled without any problem, and the speed over that part of the journey are v = 50 kmph and 100 km.

What is distance Travelled?

• The distance of an object can be defined as the complete path travelled by an object.
• For example. if a car travels east for 5 km and takes a turn to travel north for another 8 km, the total distance travelled by car shall be 13 km.

So, according to the question we have let the car travel from x to y that is the distance 300 km. This it travels till z without any trouble. It travels with a speed v and covers the remaining distance from z to y with a speed of 3v/4.

Now the engine develops a problem and reaches y 80 minutes late.

Also the car travels an extra 50 km and reaches m.

Here (m = z + 50) with speed v and then travels from m to y with speed 3v/4 and reaches 20 minutes early

So, it is 60 minutes late.

Therefore we have:

     50/(3v/4) – 50/v = 20/60

      200/3v – 50/v = 1/3

         200 – 150 / 3v = 1/3

                50/v = 1

            Or v = 50 kmph.

Now for 50 km,20 minutes is made up of the entire 80 minutes.

Therefore we get 50 x 4 = 200 km

Now total distance is 300 km

So, 300 – 200 = 100 km.

Hence, It travelled the original distance without difficulty, and the speed over that portion of the journey was v = 50 kmph and 100 km.

Learn more about distance here,

https://brainly.in/question/7737285?msp_poc_exp=5

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