A man leaves his office to attend the Teacher-Parents' meet in his son's school. The school is distance of 8 km away from his office and he travels at a constant speed. He travelled half the way with the specified speed and stopped at a stationery shop for 10 minutes. To reach the school on time, he had to increase its speed by 2 km/h for the rest of the way. Next time he stopped at stationery shop for 15 minutes. By what value must he increase his speed for the remaining half of the distance to reach the school as per the schedule?
Solution
Correct Answer: Option C
Let the regular speed of the man = x km/hr.
Now, time = distance/speed
∴ regular time taken by the man to travel the full distance = (8/x) hrs
Case-1:
Distance travelled at regular speed = 8/2= 4 km
∴ time taken to travel this distance = (4/x) hrs ---(1)
Waiting time = 10 mins = 1/6 hrs
⇒ for the remaining 4 kms, the man increases his speed by 2 km/h (i.e. x + 2)
∴ time taken to travel last 4 km = 4/(x + 2) ---(2)
From the given data, we can clearly deduce that:
Regular time = (time taken to travel first 4 km) + (waiting time) + (time taken to travel last 4 km)
\(\begin{array}{l} \Rightarrow \frac{8}{x} = \frac{4}{x} + \frac{1}{6} + \frac{4}{{x + 2}}\\ \Rightarrow \frac{4}{x} - \frac{4}{{x + 2}} = \frac{1}{6}\\ \Rightarrow \frac{1}{{{x^2} + 2x}} = \frac{1}{{48}} \end{array}\)
⇒ x2 + 2x – 48 = 0
⇒ (x + 8) (x - 6) = 0
Omitting the negative value of speed,
x = 6 km/h
Now, in the second case:
The man travels first 4 kms at its regular speed and waits for 15 mins.
In this case: waiting time = ¼ hrs
Let y = the speed with which he travels the remaining distance.
Again, he reaches the destination in regular time.
\(\begin{array}{l} \Rightarrow \frac{8}{x} = \frac{4}{x} + \frac{1}{4} + \frac{4}{y}\\ \Rightarrow \frac{4}{x} = \frac{1}{4} + \frac{4}{y} \end{array}\)
Substituting x = 6,
⇒ y = 9.6 km/h
Increase in usual speed = y – x = 9.6 – 6 = 3.6 km/hr
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