Question

Bablu has to travel from point A to D . He decided to divide journey in 3 equel parts. For first part of travel he walks with a speed of a km/hr and take a rest for a hrs , for second part he doubles his speed and rest time also for third part he again doubles his previous speed and thus reaches D in 16 hours. Find the maximum speed with which he started to travel if the distance between A and D is 36 km.
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Answer 1

Answer:

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Answer 2

The maximum speed with which Bablu started to travel is equal to 3 km/hr.

Given:

The speed with which the man started is equal to 'a'

The total distance traveled by the man be 'd' is equal to 36 km.

Total time taken (T) = 16 hours

To Find:

The maximum speed (in km/hr) with which Bablu started to travel.

Solution:

The total distance traveled by the man be 'd' is equal to 36 km.

Distance traveled by the man during each interval = d/3 = 12 km

Speed during the first interval = a

Speed during the second interval = 2a

Speed during the third interval = 2 × (2a) = 4a

The time taken for rest after the first interval = a

The time taken for rest after the second interval = 2a

→ Calculating the time taken for each interval and adding them to the time taken for rest to get the total time taken:

   $\because t_{1}+ a+t_{2}+2a+ t_{3}=T\\\\\therefore \frac{12}{a} +a+\frac{12}{2a}+ 2a+\frac{12}{4a} =16\\\\\therefore \frac{12}{a}+ \frac{6}{a}+ \frac{3}{a} +3a=16\\\\\therefore \frac{21}{a}+3a =16\\\\\therefore3a^{2} -16a+21=0\\\\\therefore 3a^{2} -9a-7a+21=0\\\\\therefore 3a(a-3)-7(a-3)=0\\\\\therefore a=7/3\:km/hr\:or\:a=3\:km/hr\\\\But\:the\:maximum\:starting\:speed\:would\:be\:equal\:to\:3\:km/hr.$

Therefore the maximum speed with which Bablu started to travel is equal to 3 km/hr.

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