Question

Problem ⤵️

A car left town A towards town B driving at a speed of V = 32 km/hr. After 3 hours on the road the driver stopped for 15 min in town C. Because of a closed road he had to change his route, making the trip 28 km longer. He increased his speed to V = 40 km/hr but still he was 30 min late. Find:

a) The distance the car has covered.

b) The time that took it to get from C to B.​

Answer 1

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Answer 2

Solution:

From the statement of the problem we don't know if the 15 min stop in town C was planned or it was unexpected. So we have to consider both cases.

1st case. The stop was planned. Let us consider only the trip from C to B, and let xx be the number of hours the driver spent on this trip.

Then the distance from C to B is S=40⋅x km. If the driver could use the initial route, it would take him x− 30/60 = x−12x−30/60 = x−12 hours to drive from C to B.

The distance from C to B according to the initially itinerary was (x−12)⋅32 km, and this distance is 2828 km shorter than 40⋅x km. Then we have the equation

(x−1/2)⋅32+28 = 40x

32x−16+28 = 40x

8x = 12

x = 12/8

x =1 (4/12) =1 (20/60) = 1 hr 20 min.

So, the car covered the distance between C and B in 1 hour and 20 min.

The distance from A to B is 3⋅32+12/8⋅40

= 96+60

= 156 km.

2nd case.

The driver did not plan the stop at C. Suppose it took xx hours for him to get from C to B. Then the distance is S = 40⋅x km.

It took x−30/60−15/60 = x−45/60 = x−3/4 h to drive from C to B.

The distance from C to B is 32(x−3/4) km, which is 28 km shorter than 40⋅x, i.e.

32(x−3/4)+28 = 40x

32x−24+28 = 40x

4 = 8x

x = 1/2hr⋅x=30min

Then the time of the trip from C to B was 30 min.

The distance covered equals

3⋅32+1/2⋅40 = 96+20

= 116km