Question

A car travels a distance of 840 kms at a uniform speed. If the speed of a car is 10 kmph more it takes two hours less to cover the same distance. The original speed of the car is
A. 8Okmph
B. 50kmph
C. 75kmph
D. 60kmph

Answer 1

Hint: The distance in both the cases i.e. (before and after the change in speed), but we know that speed is inversely proportional to time taken and therefore as the speed is increasing the time taken must be decreasing . 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 original speed of the car is ‘x’ kmph.
And the time taken to cover the distance of 840km with the speed of ‘x’ kmph is ‘\[{t_1}\]’ hours.
\[ \Rightarrow \]As we know from the formula that \[{\text{speed}} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}\]
So, the speed of a car to travel ‘840’ km in time ‘\[{t_1}\]’ hours is ‘x’ kmph.
\[ \Rightarrow \]So, \[{\text{x 'kmph'}} = \dfrac{{{\text{840km}}}}{{{{\text{t}}_{\text{1}}}{\text{hour}}}}\]
So, cross multiplying above equation to find the value of \[{t_1}\] in terms of distance and speed
\[ \Rightarrow \]So, \[{{\text{t}}_{\text{1}}} = \dfrac{{{\text{840}}}}{{\text{x}}}\] (1)
Now as given in question we know that speed is increasing by 10kmph and then the speed will decrease by 2 hours.
\[ \Rightarrow \]So the new speed will become ( ‘ x’ + 10 ) kmph.
\[ \Rightarrow \]Now let the new time be ‘\[{t_2}\]’ hours .
Now putting the values of new speed and new time in formula of \[\left( {{\text{speed }} = \dfrac{{{\text{distance}}}}{{{\text{time}}}}} \right)\]
We will get that to cover the distance of 840 km in the time \[{t_2}\] the speed must be x + 10 kmph.
\[ \Rightarrow \]So \[\left( {{\text{x}} + {\text{10kmph }} = \dfrac{{{\text{840km}}}}{{{{\text{t}}_{\text{2}}}{\text{hours}}}}} \right)\]
Cross-multiplying above equation
\[ \Rightarrow \]\[\left( {{{\text{t}}_{\text{2}}} = \dfrac{{{\text{840}}}}{{{\text{x + 10}}}}} \right)\] (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)
\[ \Rightarrow \]\[\left( {\dfrac{{{\text{840}}}}{{\text{x}}} - \dfrac{{{\text{840}}}}{{{\text{x + 10}}}} = {\text{2hours}}} \right)\]
Now taking L.C.M of L.H.S and solving the above equation.
\[ \Rightarrow \]\[\dfrac{{{\text{8400}}}}{{{{\text{x}}^{\text{2}}}{\text{ + 10x}}}} = 2\]
Cross-multiplying above equation.
\[ \Rightarrow \]\[{\text{8400}} = 2\left( {{{\text{x}}^{\text{2}}} + {\text{10x}}} \right) \]
\[ \Rightarrow \]\[\left( {{\text{4200 }} = ({{\text{x}}^{\text{2}}} + {\text{10x}})} \right)\]
On further solving above equations to find the value of x ( i.e. original speed of the car )
\[ \Rightarrow \]\[({\text{x}} + 70)({\text{x}} - 60) = {\text{0}}\]
\[ \Rightarrow \]So, x = -70 or 60
And we all know that the speed can never be negative
So the value of ‘x’ must be 60.
Hence, the original speed of the car is 60kmph. So, the correct option will be D.

NOTE :- In such problems all the factors are interconnected with each other so to find anyone’s value we had 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.