Question

A man covers a distance of 200km travelling with a uniform speed of ‘x’ km per hour. The distance could have been covered in 2 hours less, had the speed been ( x + 5 ) km per hour. Calculate the value of ‘x’
A. 20
B. 24
C. 28
D. 32

Answer 1

Hint: The distance in both the cases i.e. (before and after the change in speed) is same, but we know that speed is inversely proportional to time taken and therefore as the speed changes from ‘x’ km/hr to ( x + 5 ) km/hr then time taken will also change. And we can solve this problem with the basic formula of \[{\text{speed}} = \dfrac{{{\text{Distance}}}}{{{\text{Time}}}}\].

Complete step-by-step answer:
Let us assume that the time taken to cover the distance of 200km with the speed of ‘x’ kmph is ‘\[{t_1}\]’ hours.
So, the speed of man to travel ‘200’ km in time ‘\[{t_1}\]’ hours is ‘x’ kmph.
\[ \Rightarrow \]So, \[{\text{x 'kmph'}} = \dfrac{{{\text{200km}}}}{{{{\text{t}}_{\text{1}}}{\text{hour}}}}\]
So, cross multiplying above equation to find the value of \[{t_1}\]
\[ \Rightarrow \]So, \[{{\text{t}}_{\text{1}}} = \dfrac{{{\text{200}}}}{{\text{x}}}\] (1)
Now as given in question we know that speed changed to ( x + 5 ) km/hour and then the time will decrease by 2 hours.
\[ \Rightarrow \]Now let the new time be ‘\[{t_2}\]’ hours .
Now putting the values of new speed and new time in formula of \[\left( {{\text{speed }} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}} \right)\]
\[ \Rightarrow \]So \[\left( {{\text{x}} + 5{\text{kmph }} = \dfrac{{{\text{200km}}}}{{{{\text{t}}_{\text{2}}}{\text{hours}}}}} \right)\]
Cross-multiplying above equation
\[ \Rightarrow \]\[\left( {{{\text{t}}_{\text{2}}} = \dfrac{{{\text{200}}}}{{{\text{x + 5}}}}} \right)\] (2)
 From above we come know that the difference between \[{t_1}\] and \[{t_2}\] is 2 hours
\[ \Rightarrow \]So \[\left( {{{\text{t}}_{\text{1}}} - {{\text{t}}_{\text{2}}} = {\text{2hours}}} \right)\] (3)
Now putting the values of equation (1) and equation (2) in equation (3)
\[ \Rightarrow \]\[\left( {\dfrac{{{\text{200}}}}{{\text{x}}} - \dfrac{{{\text{200}}}}{{{\text{x + 5}}}} = {\text{2 hours}}} \right)\]
Now taking L.C.M of L.H.S and solving the above equation.
\[ \Rightarrow \] \[\dfrac{{1000}}{{{{\text{x}}^{\text{2}}}{\text{ + 5x}}}} = 2\]
Cross-multiplying above equation.
\[ \Rightarrow \]\[{\text{1000}} = 2\left( {{{\text{x}}^{\text{2}}} + 5{\text{x}}} \right) \Rightarrow \left( {{\text{500 }} = ({{\text{x}}^{\text{2}}} + 5{\text{x}})} \right)\]
Now further solving the equation by middle term splitting
\[ \Rightarrow {x^2} + 5x - 500 = 0\]
\[ \Rightarrow {x^2} + 25x - 20x - 500 = 0\]
\[ \Rightarrow x(x + 25) - 20(x + 25) = 0\]
Now on comparing ( x + 25 ) and ( x – 20 ) with 0 we will get x = -25 and x = 20.
But as we know that the speed can never be negative
So, the value of ‘x’ must be 20.
So, the original speed of man is 20kmph.
Hence, the correct option will be A.

Note: In such problems all the factors are interconnected with each other so if we want to find any term we have to relate all the three with each other. This problem can also be solved by taking \[{t_2}\]as \[{t_1} - 2\] because it is given that the difference between both the timing is 2hours but this method will elaborate the solution so the above solution is best for such type of problems.