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Walking at 4 ½ km/hr a worker reaches the factory 15 mins late. If he walks at 6 km/hr he will be 7 ½ minutes early. The distance of the factory from his house is:

Question

Walking at 4 ½ km/hr a worker reaches the factory 15 mins late. If he walks at 6 km/hr he will be 7 ½ minutes early. The distance of the factory from his house is:

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Solution

Let's denote the distance between the worker's house and the factory as D (in km) and the time he should take to reach the factory on time as T (in hours).

From the problem, we have two equations:

  1. D = (4.5 km/hr) * (T + 15/60 hr) (since he is 15 minutes late when walking at 4.5 km/hr)

  2. D = (6 km/hr) * (T - 7.5/60 hr) (since he is 7.5 minutes early when walking at 6 km/hr)

We can solve these two equations to find the values of D and T.

First, let's simplify the equations:

  1. D = 4.5T + 1.125

  2. D = 6T - 0.75

Setting the two equations equal to each other gives:

4.5T + 1.125 = 6T - 0.75

Solving for T gives:

T = 0.375 hours = 22.5 minutes

Substituting T = 0.375 hours into the first equation gives:

D = 4.5 * 0.375 + 1.125 = 2.8125 km

So, the distance of the factory from his house is approximately 2.8125 km.

This problem has been solved

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