From points A and B, the distance between which is 50 km, two motorcyclists left at the same time towards

From points A and B, the distance between which is 50 km, two motorcyclists left at the same time towards each other and met in 30 minutes to find the speed of each if it is known that one of them arrived at A 25 minutes earlier than the other at B.

It is known from the problem statement that the distance between points A and B is 50 kilometers. Two motorcyclists who left each point at the same time met 30 minutes after leaving. One of them arrived 25 minutes earlier than the other. we will solve this problem with the help of an equation and compose a system.

Let x be the speed of 1 motorcyclist.
Let y be the speed of 2 motorcyclists.
30 minutes = 0.5 hours.
0.5x + 0.5y = 50, but it can be simplified: x + y = 100.
Let 50 / x be the time of the first.
let 50 / y be the time of the second.

Let’s compose a system.

{x + y = 100;
{50 / x + 25/60 = 50 / y;

{y = 100 – x;
{2 / x + 1/60 – 2/100 – x = 0;

{y = 100 – x;
{12000 – 120x + 100x – x ^ 2 – 120x / 60x * (100 – x) = 0

x ^ 2 + 140x – 12000 = 0

Next, we solve the problem through the discriminant.

D = 4900 + 12000 = 16900 = 130 ^ 2;

x1 = – 70 + 130 = 60 (km / h) – the speed of the first.
x2 = – 70 + 340: 2 – 60 = 270: 2 – 60 = 100 – 60 = 40 (km / h) – speed of the second.

Answer: the speed of the first rider is 60 km / h, the speed of the second rider is 40 km / h.