# distance

• Feb 28th 2010, 03:17 PM
sri340
distance
If a car starts from A towards B with some velocity due to some problem in the engine after traveling 30km.If the car goes with 4/5th of its actual velocity the car reaches B 45min later to the actual time. If the car engine fails after traveling 45km, the car reaches the destination B 36min late to the actual time, what is the initial velocity of car and what is the distance between A and B in km ?

(a) 40 & 100
(b) 10 & 150
(c) 20 & 130
(d) 50
• Mar 1st 2010, 12:04 AM
earboth
Quote:

Originally Posted by sri340
If a car starts from A towards B with some velocity due to some problem in the engine after traveling 30km.If the car goes with 4/5th of its actual velocity the car reaches B 45min later to the actual time. If the car engine fails after traveling 45km, the car reaches the destination B 36min late to the actual time, what is the initial velocity of car and what is the distance between A and B in km ?

1. Let d denote the distance between A and B, v the original speed of the car and t the original traveling time.

2. According to the text of the question I've got a system of simultaneous equations:

$\left|\begin{array}{rcl}\dfrac dv&=&t \\ \\ \dfrac{30}v + \dfrac{d-30}{0.8 \cdot v} &=& t + \dfrac34 \\ \\ \dfrac{45}v + \dfrac{d-45}{0.8 \cdot v} &=& t+\dfrac35 \end{array}\right.$

3. Unfortunately the solution of this system doesn't match any of the given results (Crying) so either I didn't understand the question correctly or the given results are wrong ... ?

My result: $(d, t, v)= \left(105, \frac{21}5, 25 \right)$