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 ?<\/p>\n
[amp_mcq option1=”30″ option2=”40″ option3=”45″ option4=”50″ correct=”option3″]<\/p>\n
If car A doubles its speed (2*S_A), the new time taken is T_A’ = D\/(2*S_A).
\nAccording to the second condition: T_A’ = T_B – 1 => D\/(2*S_A) = D\/60 – 1 (Equation 2)<\/p>\n
Multiply Equation 2 by 2:
\n2 * [D\/(2*S_A)] = 2 * [D\/60 – 1]
\nD\/S_A = D\/30 – 2<\/p>\n
Now we have two expressions for D\/S_A:
\nFrom Eq 1: D\/S_A = D\/60 + 1
\nFrom the modified Eq 2: D\/S_A = D\/30 – 2<\/p>\n
Equating the two expressions:
\nD\/60 + 1 = D\/30 – 2
\nRearrange the terms to solve for D:
\n1 + 2 = D\/30 – D\/60
\n3 = (2D – D) \/ 60
\n3 = D\/60
\nD = 180 km.<\/p>\n