Question

A man walks at 6 kmph and runs at 8kmph. he covers 6 km in 50 min partly by running and partly by walking.
(i) For how long did he walk?
(ii) What distance did he cover walking?

Answer 1

Hint: Here in this question, we have to find the distance covered by the man through walk and find how much time he walked. To solve this, let us take a distance covered by a man through walk x and further simplify by using the relation between distance-speed and time and by basic arithmetic operation we get the required solution.

Complete step-by-step answer:
Now, consider the question:
A man walks at the speed of \[6\,kmph\] and runs at the speed of \[8\,kmph\].
He covers 6 km distance in time of 50 minutes partly by running and partly by waking
Let us take the distance covered by man through walk be \[x\,km\].
The total distance covered by man is \[6\,km\].
The distance covered by man through running is \[\left( {6 - x} \right)\,\,km\]
As we know, the relation between distance-speed and time is: \[Speed = \dfrac{{distance}}{{time}}\] or\[time = \dfrac{{distance}}{{speed}}\].
Given the time to cover the total distance is \[50\,\min \] which speed is \[\dfrac{{50}}{{60}}\], then
\[ \Rightarrow \,\,\dfrac{x}{6} + \dfrac{{6 - x}}{8} = \dfrac{{50}}{{60}}\]
Take 24 as LCM in LHS, then
\[ \Rightarrow \,\,\dfrac{{4x + 3\left( {6 - x} \right)}}{{24}} = \dfrac{{50}}{{60}}\]
Divide both numerator and denominator of RHS by 10, then
\[ \Rightarrow \,\,\dfrac{{4x + 18 - 3x}}{{24}} = \dfrac{5}{6}\]
On simplification, we have
\[ \Rightarrow \,\,\dfrac{{x + 18}}{{24}} = \dfrac{5}{6}\]
On cross multiplication. We have
\[ \Rightarrow \,\,6\left( {x + 18} \right) = 5 \times 24\]
\[ \Rightarrow \,\,6x + 108 = 120\]
Subtract 108 by both side, then
\[ \Rightarrow \,\,6x = 120 - 108\]
\[ \Rightarrow \,\,6x = 12\]
Divide both side by 6, then
\[ \Rightarrow \,\,x = 2\,km\]
The distance by man through walk is 2 km.
Time taken for walk is \[ = \dfrac{x}{6} = \dfrac{2}{6} = \dfrac{1}{3}h\]
1 hour has 60 mins then
\[ \Rightarrow \dfrac{1}{3} \times \,60\,\min \]
\[ \Rightarrow 20\,\min \]

Note: Distance, speed and time are one of the most common topics based on which questions are asked in the various Government exams in the quantitative aptitude section. To solve this, we have to know the relations between the physical quantities. And don’t forget to mention the units for physical quantities.