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. (A) 100 km, 50 kmph (B) 50 km, 75 kmph (C) 150 km, 50 kmph (D) 50 km, 100 kmph​

Answer 1

Answer:

Step-by-step explanation:

Train proceeds at 3/4 S where S is his usual speed means 1/4 decrease in the speed which will lead to 1/3 increase in the time.

[ As S *T = D, Speed is inversely proportional to time when distance is constant].

Now, the main difference come in those 24 km and the change in difference of time = 35 - 25 = 10 m

= 1 / 3 *T = 10 where T is the time required to cover the distance of (50 - 74) = 24 km

T = 30 min = 0.5 hours

Speed of the train = 24 / (0.5) = 48 kmph.

Answer 2

Answer:

Original distance is 60 km and speed is 15 km/hr.

(here all options are wrong)

Step-by-step explanation:

Let the speed of car be x kmph.

Let the distance between the point of engine failure and destination be y km

If travelled through original speed, time =$\frac{y}{x}$hours.

But since speed is 3/4th of original. So, time= $\frac{4y}{3x}$hours

According to given conditions:

$\frac{4y}{3x} - \frac{y}{x} = \frac{80}{60} \: hours$

$\frac{4y - 3y}{3x} = \frac{4}{3} \\ \frac{y}{3x} = \frac{4}{3} \\ \frac{y}{x} = 4...(1)$

Also,

$\frac{4(y - 50)}{3x} - \frac{(y - 50)}{x} = \frac{60 - 20}{60} \: hours \\ \frac{4y - 200}{3x} - \frac{y - 50}{x} = \frac{40}{60} \\ \frac{4y - 200 - y + 50}{3x} = \frac{2}{3} \\ \frac{3y - 150}{3x} = \frac{2}{3} \\ \frac{3y - 150}{x} = 2 \\ 3 \times \frac{y}{x} - \frac{150}{x} = 2.....(2)$

We are putting $\frac{y}{x} = 4$in equation (2),

$3 \times 4 - \frac{150}{x} = 2 \\ 12 - \frac{150}{x} = 2 \\ \frac{150}{x} = 10 \\ x = \frac{150}{10} \\ x = 15$

and from equation (1),

$\frac{y}{15} = 4 \\ y = 4 \times 15 \\ y = 60$

This is a problem of Speed and Distance chapter,

Know more about speed and distance -

1) https://brainly.in/question/47777661

2) https://brainly.in/question/40963838