The distance between towns M and N is 280 km. A car and a lorry travel from M to N.

The average speed of the lorry is 20 km/h less than that of the car. The lorry takes 1 h 10 min more than the car to travel from M and N.
(a) If the speed of the lorry is x km/h, find x
(b) The lorry left town M at 8: 15 a.m. The car left town M later and overtook the lorry at 12.15 p.m. Calculate the time the car left town M.

lneed answer

To solve this problem, we can use the formula: Speed = Distance/Time.

(a) Let's first find the speed of the car. We know that the lorry's speed is 20 km/h less than the car's speed. So, if the speed of the lorry is x km/h, the speed of the car will be (x + 20) km/h.

The lorry takes 1 hour and 10 minutes (or 1.17 hours) longer than the car to travel from M to N. We can set up the following equation:

280/(x + 20) = 280/x + 1.17

To solve this equation, we need to cross-multiply and solve for x.

280x = 280(x + 20) + 1.17x(x + 20)
280x = 280x + 5600 + 1.17x^2 + 23.4x
0 = 1.17x^2 + 303.40x + 5600

Now, we can solve this quadratic equation using factoring or the quadratic formula. Factoring might be difficult in this case, so let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our quadratic equation, a = 1.17, b = 303.40, and c = 5600. Plugging in these values, we get:
x = (-303.40 ± √(303.40^2 - 4 * 1.17 * 5600)) / (2 * 1.17)

After evaluating this equation, we find two possible solutions for x: x ≈ -11.96 and x ≈ -244.67. However, since speed cannot be negative, we disregard the negative solution. Therefore, the speed of the lorry is approximately x ≈ 244.67 km/h.

(b) The lorry left town M at 8:15 a.m., and the car overtook the lorry at 12:15 p.m., which is 4 hours later. Therefore, the car started its journey 4 hours after the lorry.

If we subtract 4 hours from 8:15 a.m., we get the time the car left town M:
8:15 a.m. - 4 hours = 4:15 a.m.

Hence, the car left town M at approximately 4:15 a.m.