A lady reaches her office 20 mins late by travelling at a speed of 20 km/h and reaches 15 mins early by travelling at a speed of 30 km/h. By how much time(early or late) will she reach her office if she travels at 25 km/h?

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A lady reaches her office 20 mins late by travelling at a speed of 20 km/h and reaches 15 mins early by travelling at a speed of 30 km/h. By how much time(early or late) will she reach her office if she travels at 25 km/h?

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Answer 1 · 2016-11-27T11:55:20

Travelling at $25$ $25$ km/hr she will reach $1$ $1$ mt early.

Explanation:

Let the distance of the office be $x$ $x$ Km
Time taken at $20$ $20$ km/hr is $\frac{x}{20}$ $x/20$ hr; reaches $\frac{20}{60} = \frac{1}{3}$ $20/60=1/3$ hr=20mt late
Time taken at $30$ $30$ km/hr is $\frac{x}{30}$ $x/30$ hr; reaches $\frac{15}{60} = \frac{1}{4}$ $15/60=1/4$ hr=15mt early
Therefore $\frac{x}{20} - \frac{x}{30} = \frac{1}{3} + \frac{1}{4} or \frac{x}{60} = \frac{7}{12} or x = \frac{7}{12} \cdot 60 = 35$ $x/20-x/30=1/3+1/4 or x/60=7/12 or x=7/12*60=35$ Km
Time taken at $30$ $30$ km/hr is $\frac{35}{30} = \frac{35}{30} \cdot 60 = 70$ $35/30=35/30*60 =70$ mt
Time taken at $25$ $25$ km/hr is $\frac{35}{25} = \frac{7}{5} h r = \frac{7}{5} \cdot 60 = 84 m t$ $35/25 =7/5 hr=7/5*60=84 mt$ ;
So travelling at $25$ $25$ km/hr she will reach $70 + 15 - 84 = 1$ $70+15-84 =1$ mt early.[Ans]

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A lady reaches her office 20 mins late by travelling at a speed of 20 km/h and reaches 15 mins early by travelling at a speed of 30 km/h. By how much time(early or late) will she reach her office if she travels at 25 km/h?

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Explanation:

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