?
A minibus takes 6 hour less to cover 1680 km distance, if its speed is increased by 14 kmph ? What is the usual time of the minibus ?
Answer & Solution
Correct Answer:
Option
D
Let
Usual Speed = x kmph.
Now,
Speed of bus after increasing the speed = (x + 14)kmph. . . . . . . (1)
A ____________1680km ____________B
In first case, Time taken to covered the distance 1680 km = $$\frac{{1680}}{x}$$ . . . . . . . (2) In Second Case, Time Taken to covered the distance 1680 km = $$\frac{{1680}}{{x + 14}}$$ Time difference = 6 Hours. So, $$\eqalign{ & \Rightarrow {\frac{{1680}}{x} - \frac{{1680}}{{ {x + 14} }}} = 6 \cr & \Rightarrow 1680 {\frac{{14}}{{ {{x^2} + 14x} }}} = 6 \cr & \Rightarrow 280 \times 14 = {x^2} + 14x \cr & \Rightarrow {x^2} + 70x - 56x - 3920 = 0 \cr & \Rightarrow x\left( {x + 70} \right) - 56\left( {x + 70} \right) = 0 \cr & \Rightarrow x = - 70,56. \cr & {\text{hence,}}\,{\text{speed}}\,{\text{of}}\,{\text{minibus}}\,{\text{is}}\,56\,km/h \cr & {\text{put}}\,x = 56\,{\text{in}}\,{\text{equation}}\,(2) \cr & T = \frac{{1680}}{{56}} \cr & \,\,\,\,\,\, = 30\,{\text{hours}} \cr} $$
In first case, Time taken to covered the distance 1680 km = $$\frac{{1680}}{x}$$ . . . . . . . (2) In Second Case, Time Taken to covered the distance 1680 km = $$\frac{{1680}}{{x + 14}}$$ Time difference = 6 Hours. So, $$\eqalign{ & \Rightarrow {\frac{{1680}}{x} - \frac{{1680}}{{ {x + 14} }}} = 6 \cr & \Rightarrow 1680 {\frac{{14}}{{ {{x^2} + 14x} }}} = 6 \cr & \Rightarrow 280 \times 14 = {x^2} + 14x \cr & \Rightarrow {x^2} + 70x - 56x - 3920 = 0 \cr & \Rightarrow x\left( {x + 70} \right) - 56\left( {x + 70} \right) = 0 \cr & \Rightarrow x = - 70,56. \cr & {\text{hence,}}\,{\text{speed}}\,{\text{of}}\,{\text{minibus}}\,{\text{is}}\,56\,km/h \cr & {\text{put}}\,x = 56\,{\text{in}}\,{\text{equation}}\,(2) \cr & T = \frac{{1680}}{{56}} \cr & \,\,\,\,\,\, = 30\,{\text{hours}} \cr} $$
Join the Discussion
Login to post a comment or share your explanation.
Logint and T are two time taken in two difference condition.