Car A takes 1 hour more than car B, which travels at a speed of 60 km per hour, to cover some fixed distance. If car A had doubled its speed, it could cover the distance in 1 hour less time than car B travelling at 60 km per hour. What is the original speed of car A in km per hour ?
Time taken by car B = T_B = D/60 hours.
Time taken by car A originally = T_A = D/S_A hours.
According to the first condition: T_A = T_B + 1 => D/S_A = D/60 + 1 (Equation 1)
If car A doubles its speed (2*S_A), the new time taken is T_A’ = D/(2*S_A).
According to the second condition: T_A’ = T_B – 1 => D/(2*S_A) = D/60 – 1 (Equation 2)
Multiply Equation 2 by 2:
2 * [D/(2*S_A)] = 2 * [D/60 – 1]
D/S_A = D/30 – 2
Now we have two expressions for D/S_A:
From Eq 1: D/S_A = D/60 + 1
From the modified Eq 2: D/S_A = D/30 – 2
Equating the two expressions:
D/60 + 1 = D/30 – 2
Rearrange the terms to solve for D:
1 + 2 = D/30 – D/60
3 = (2D – D) / 60
3 = D/60
D = 180 km.
Now substitute the value of D back into Equation 1 to find S_A:
180/S_A = 180/60 + 1
180/S_A = 3 + 1
180/S_A = 4
S_A = 180 / 4
S_A = 45 km/hr.
Car B time: 180/60 = 3 hours.
Original Car A time: 180/45 = 4 hours. 4 hours is 1 hour more than 3 hours. Condition 1 holds.
Car A doubled speed: 2 * 45 = 90 km/hr.
Car A new time: 180/90 = 2 hours. 2 hours is 1 hour less than 3 hours. Condition 2 holds.
The original speed of car A is 45 km/hr.