A car covered a distance at a uniform speed. If the car had been 5 km/hr…

Concept

In two-speed journey problems, the distance covered is invariant: distance = speed x time. When the speed is altered, the time changes so that speed x time still equals the same original distance. Writing this relation for each altered scenario in terms of the original speed s and original time t gives two linear equations in s and t, which can then be solved simultaneously.

Step-by-step solution

1. Let the original speed be s km/hr and the original time be t hours, so distance d = s x t.

2. Faster case: speed becomes (s+5) and time reduces by 48 minutes = 4/5 hour, so time = (t - 4/5). Since distance is unchanged: (s+5)(t - 4/5) = s x t. Expanding and simplifying gives 5t = 0.8s + 4, i.e. t = 0.16s + 0.8 (Equation A).

3. Slower case: speed becomes (s-10) and time increases by 2 hours, so time = (t+2). Since distance is unchanged: (s-10)(t+2) = s x t. Expanding and simplifying gives 2s - 10t - 20 = 0, i.e. s = 5t + 10 (Equation B).

4. Substitute Equation A into Equation B: s = 5(0.16s + 0.8) + 10 = 0.8s + 14, so 0.2s = 14, giving s = 70 km/hr.

5. From Equation A: t = 0.16(70) + 0.8 = 11.2 + 0.8 = 12 hours.

6. Distance = s x t = 70 x 12 = 840 km.

Cross-check

At speed 75 km/hr (70+5), time = 840 / 75 = 11.2 hours = 11 hours 12 minutes, which is exactly 48 minutes less than 12 hours -- confirmed. At speed 60 km/hr (70-10), time = 840 / 60 = 14 hours, which is exactly 2 hours more than 12 hours -- confirmed.

Answer: Distance = 840 km.