Question

Walking at three-fourth of his normal speed a man is late by \[2.5\]hr. Then his usual time is:
A) \[7.5{\rm{hr}}\]
B) \[3.5{\rm{hr}}\]
C) \[3.25{\rm{hr}}\]
D) \[8{\rm{hr}}\]

Answer 1

Hint:
Here, we have to find the usual time of the man. We will assume the normal speed of the man to be some variable. Then we will find the new speed using the given information. We will then find the distance for both normal speed and the new speed by using the given speed and distance. By assuming that the distances are equal, we will equate both the distances. We will solve it to find the usual time. Distance is defined as the length between the starting point and the initial point.

Formula Used:
We will use the formula \[{\rm{Distance}} = {\rm{Speed}} \times {\rm{Time}}\].

Complete Step by Step Solution:
Let \[x\] be the normal speed of the man at the time \[t\].
We are given that the man is walking at three-fourth of his normal speed and he is late by \[2.5\]hr.
Since the man is walking at three- fourth of his normal speed.
So, now the speed of the man \[ = \dfrac{3}{4}x\]
Since the man is late by \[2.5\] hours than his normal speed. So .
The time taken by the man when he is late \[ = t + 2.5\]
Now we will find the distance covered by a man using the formula \[{\rm{Distance}} = {\rm{Speed}} \times {\rm{Time}}\].
When the man travels at a speed of \[x\] at time \[t\], then the distance covered by the man is given by the distance formula
Distance covered by the man at his normal speed \[ = xt\]…………………………………………………………\[\left( 1 \right)\]
When the man travels at a speed of \[\dfrac{3}{4}x\] at time \[t + 2.5\], then the distance covered by the man is given by the distance formula
\[ \Rightarrow \] Distance covered by the man at the given speed \[ = \dfrac{3}{4}x \times \left( {t + 2.5} \right)\]
 \[ \Rightarrow \] Distance covered by the man at the given speed \[ = \dfrac{3}{4}xt + \dfrac{3}{4}x\left( {2.5} \right)\]…………………………………………\[\left( 2 \right)\]
We assume that the distance covered by the man is the same in both instances.

So, now we will equate the equations \[\left( 1 \right)\]and \[\left( 2 \right)\]. Therefore, we get
\[xt = \dfrac{3}{4}xt + \dfrac{3}{4}x\left( {2.5} \right)\]
By rewriting the equation, we get
\[ \Rightarrow xt - \dfrac{3}{4}xt = \dfrac{3}{4}x\left( {2.5} \right)\]
By taking the L.C.M, we get
\[ \Rightarrow \dfrac{{4xt}}{4} - \dfrac{3}{4}xt = \dfrac{3}{4}x\left( {2.5} \right)\]
\[ \Rightarrow \dfrac{{xt}}{4} = \dfrac{3}{4}x\left( {2.5} \right)\]
By cancelling the similar terms, we get
\[ \Rightarrow t = 3\left( {2.5} \right)\]
\[ \Rightarrow t = 7.5{\rm{hr}}\]

Therefore, the usual time of the man is \[7.5{\rm{hr}}\]. Thus Option(A) is the correct answer.

Note:
We can also solve the problem by another method.
From the given, we get
Speed at present \[ = \dfrac{3}{4}x\]
We know that the speed is inversely proportional to the time. So,
Time at present \[ = \dfrac{4}{3}x\]
Since he is late by \[2.5\]hr, we get
Time at present \[ - \] Usual time \[ = 2.5\]
Substituting the values of the present time and usual time in the above equation, we get
\[ \Rightarrow \dfrac{4}{3}x - x = 2.5\]
Taking LCM on LHS, we get
\[ \Rightarrow \dfrac{4}{3}x - \dfrac{3}{3}x = 2.5\]
Subtracting the terms, we get
\[ \Rightarrow x = \left( {2.5} \right)3\]
\[ \Rightarrow x = 7.5{\rm{hr}}\]
Therefore, the usual time of the man is \[7.5{\rm{hr}}\].