Question

A man covered a certain distance at some speed. Had he moved 3 km/hr faster he would have taken 40 minutes less. If he had moved 2 km/hr slower he would have taken 40 minutes more. The distance (in km) is:
(a) 35
(b) \[36\dfrac{2}{3}\]
(c) \[37\dfrac{1}{2}\]
(d) 40

Answer 1

Hint: We will use the formula for \[\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}\] or \[\text{Time}=\dfrac{\text{Distance}}{\text{Speed}}\]. Also, for forming the equation with three different times, we can write \[{{t}_{1}}+{{t}_{2}}={{t}_{3}}\]. Here, \[{{t}_{3}}\] is the time that an object took as a whole.

Complete step-by-step answer:
We will consider the distance covered by a man as x km and the time taken for this distance is time t. Also, we can say that he covered a distance x km with speed y km. According to the question, we have a condition that he moved 3 km/hr faster then he must have taken 40 minutes less. Therefore, we can say that he moved 3 km/hr faster than the default speed which is y km/hr. Thus, the new speed here is given by (y + 3) km/hr.
Now, we will substitute in the formula,
\[\text{Time}=\dfrac{\text{Distance}}{\text{Speed}}\]
\[Time=\dfrac{x\text{ km}}{\left( y+3 \right)\text{ km/hr}}\]
\[Time=\dfrac{x}{y+3}hr....\left( i \right)\]
Now, the actual time is given by \[\text{Time}=\dfrac{\text{Distance}}{\text{Speed}}\]
Here, the distance covered is x km and we will write the speed as y km/hr. Therefore, we have the value of time \[=\dfrac{x\text{ km}}{y\text{ km/hr}}\]
\[Time=\dfrac{x}{y}\text{hr}\]
Now, we will consider this time as \[{{t}_{1}}\]. So, we get,
\[{{t}_{1}}=\dfrac{x}{y}hr\]
And in equation (i), we will denote the time as \[{{t}_{2}}\]. So, we get,
\[{{t}_{2}}=\dfrac{x}{y+3}\]
Now, we will consider the statement of the question which says that he moved 3 km/hr faster if he took 40 minutes less out of 60 minutes. Thus, we have
\[{{t}_{1}}-{{t}_{2}}={{t}_{3}}\]
where \[{{t}_{3}}\] is given by \[\dfrac{40\min }{60\min }\]
\[\dfrac{x}{y}-\dfrac{x}{y+3}=\dfrac{40}{60}....\left( ii \right)\]
Similarly, we have a condition that if he had moved 2 km/hr slower, then he would have taken 40 minutes more. In this case, we will consider the time denoted as
\[{{t}_{4}}=\dfrac{\text{Distance}}{\text{Speed}}\]
\[{{t}_{4}}=\dfrac{x}{y-2}\]
where – 2 denotes that he moved 2 km/hr slower.
So, we have equations with time \[{{t}_{4}},{{t}_{1}}\] and \[{{t}_{3}}\]. According to the question, the person had moved 2 km/hr slower if he had taken 40 minutes more. Therefore by the statement, it means that \[{{t}_{4}}-{{t}_{1}}={{t}_{3}}\] where \[{{t}_{3}}\] is given by \[\dfrac{40\min }{60\min }\]
\[\dfrac{x}{y-2}-\dfrac{x}{y}=\dfrac{40}{60}....\left( iii \right)\]
Now we will simplify equations (ii) and (iii), for that we will first consider equation (ii) given by
\[\dfrac{x}{y}-\dfrac{x}{y+3}=\dfrac{40}{60}\]
\[\dfrac{x\left( y+3 \right)-xy}{y\left( y+3 \right)}=\dfrac{2}{3}\]
\[\Rightarrow \dfrac{xy+3x-xy}{y\left( y+3 \right)}=\dfrac{2}{3}\]
By canceling the like terms in the numerator, we get,
\[\Rightarrow \dfrac{3x}{y\left( y+3 \right)}=\dfrac{2}{3}\]
\[\Rightarrow x=\dfrac{2y\left( y+3 \right)}{9}....\left( iv \right)\]
Now we will consider equation (iii) and simplify it.
\[\dfrac{x}{y-2}-\dfrac{x}{y}=\dfrac{40}{60}\]
\[\Rightarrow \dfrac{xy-x\left( y-2 \right)}{y\left( y-2 \right)}=\dfrac{2}{3}\]
\[\Rightarrow \dfrac{xy-xy+2x}{y\left( y-2 \right)}=\dfrac{2}{3}\]
By canceling the like terms we get,
\[\Rightarrow \dfrac{2x}{y\left( y-2 \right)}=\dfrac{2}{3}\]
\[\Rightarrow x=\dfrac{2y\left( y-2 \right)}{6}....\left( v \right)\]
Since the left-hand side of the equation (iv) and (v) are equal. Thus, we will also have the right side of the equations equal. Therefore, we have,
\[\dfrac{2y\left( y+3 \right)}{9}=\dfrac{2y\left( y-2 \right)}{6}\]
Now, we will cancel the common terms on both the sides of the equation.
\[\dfrac{\left( y+3 \right)}{9}=\dfrac{\left( y-2 \right)}{6}\]
\[6\left( y+3 \right)=9\left( y-2 \right)\]
\[\Rightarrow 2\left( y+3 \right)=3\left( y-2 \right)\]
\[\Rightarrow 2y+6=3y-6\]
\[\Rightarrow 3y-2y=6+6\]
\[\Rightarrow y=12\]
Therefore, the speed is given by 12 km/hr. Now, we will substitute the value of y in equation (v). Thus, we have
\[x=\dfrac{2y\left( y-2 \right)}{6}\]
\[x=\dfrac{12\left( 12-2 \right)}{3}\]
\[x=\dfrac{12\times 10}{3}\]
\[x=4\times 10\]
x = 40
Therefore, the distance covered is 40 km.
Hence, the correct option is (d).

Note: Alternatively, we could have used one shortcut method of average velocity here. It is given by
Average velocity \[=\dfrac{2{{v}_{1}}{{v}_{2}}}{{{v}_{1}}+{{v}_{2}}}\]
In this formula, we get the time to take to cover the first half of the distance given by \[\dfrac{x}{2{{v}_{1}}}=\left( \dfrac{x}{2} \right)\dfrac{1}{{{v}_{1}}}\] where x is the distance covered. Also, average velocity is the total distance divided by total time. Remember to solve the equations into simpler forms. After simplification, we can equate the variable and find the relevant answer. We are not supposed to apply any elimination method here as it will make the solution more complex. Understand the language of the questions to make simpler equations and be careful with the signs, as a sign change can turn the entire solution wrong.