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**vgopalak**- Replies: 4

A car travels from city A to city B. After two hours they had a problem and hence reduced the speed to 30 mph. As a result they were late by an hour. If they faced the problem 15 miles further down, they would have been 40 minutes late. Determine the distance between the two cities.

ganesh wrote:

SDT # 6

A man riding a cycle at 12 km/hour can reach his village in 4½ hours. If he is delayed by 1½hour at the start, in order to reach his destination in time, at what speed should he ride?

[Solution]

Total distance = 12 X 4.5 = 54 Kms

If he starts 1.5 hrs late, then he has only (4.5 - 1.5) = 3 Hrs to commute 54 Kms.

So he has to cycle at speed of :

**V = 54/3 = 18 Km/hour**

ganesh wrote:

SDT # 4

X is picked up by his father by car from school everyday and they reach home at 5.00 p.m. One day, since the school got over an hour earlier than usual, he started walking towards home at 3 km/hr. He met his father on the way and they reached home 15 minutes earlier than their usual time. What is the speed of the car?

[Solution]

H=Home, S=School, M=Father and Son meet

H========================================M=============S

(Time = ?) (Car arrives at 5.00 pm)

The car arrives at 5.00 pm at point 'S'. But since Mr X started walking at

4.00 pm, they will meet at a point 'M'.

The car saved travel distance M -> S and S -> M. This resulted in 15 minutes

savings. Assuming uniform speed, it takes 7 1/2 minutes from 'M' to 'S' and

vice-versa. So both meet at 4:52 1/2 pm. The Son had been walking from 4.00 pm

till this time, which is 52 1/2 min.

The same distance is covered by the car in just 7 1/2 min. Therefore the car

travels at 52.5/7.5 times the walking speed of Mr X (3 km/hr).

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Speed of the car = 3 X 52.5/7.5 = 21 km/hr

Distance from S = 21 X 7.5/60 = 2 5/8 km (2.625 km)

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