# On a stretch of 600 km, after passing 1/4 of the way, the train was delayed for 1 hour 30 minutes.

**On a stretch of 600 km, after passing 1/4 of the way, the train was delayed for 1 hour 30 minutes. To arrive at the terminal station on time, the driver increased the train speed by 15 km / h. How long was the train on the way?**

Solution:

Let’s denote the unknown speed by x km / h. The train had to stay on the way for 600 / x hours. But at this speed he was driving only 150 kilometers, the rest of the distance he was driving at a speed of x + 15 km / h. According to the condition of the problem, an equation was drawn up:

150 / x + 450 / (x + 15) + 1.5 = 600 / x;

450 / (x + 15) – 450 / x + 1.5 = 0;

Now you need to equate the fractions to a common denominator:

(450x – 450 * (x + 15)) / (x ^ 2 + 15x) + 1.5 = 0;

(450x – 450 * (x + 15) + 1.5 * (x ^ 2 + 15x)) / (x ^ 2 + 15x) = 0;

(450x – 450x – 6750 + 1.5x ^ 2 + 22.5x) / (x ^ 2 + 15x) = 0;

(1.5x ^ 2 + 22.5x – 6750) / (x ^ 2 + 15x) = 0;

The fraction is 0 when the numerator is 0 and the denominator is not 0.

1.5x ^ 2 + 22.5x – 6750 = 0;

Discriminant = 22.5 * 22.5 + 4 * 1.5 * 6750 = 41006.25; The root of this number is 202.5.

x = (-22.5 + 202.5) / 3 or x = (-22.5 – 202.5) / 3

x = 60 or x = -75

The speed cannot be negative, so it is 60 km / h.

This means that the train was on the way 600/60 – 1.5 = 10 – 1.5 = 8.5 hours.

Answer: the train was on the way for 8.5 hours.