A is faster than B, A and B each walk 24 km. The sum of their speeds is 7 km/hr and the sum of times taken by them is 14 hours. Then A’s speed is equal to.
A. 3 km/hr
B. 4 km/hr
C. 5 km/hr
D. 7 km/hr

Hint: We start solving the problem by assigning variables for the speeds of A and B and the total distance walked. We then find the first equation by using the relation between speeds given in the problem. We then use the fact $\text{Time = }\dfrac{\text{distance}}{\text{speed}}$ to find our second equation using the relation given about times in the problem. We then solve the both equations obtained to get the desired result.

Complete step by step answer:
We have to find the speed of A.
Let us assume speed of A be ${{V}_{A}}\left( km/hr \right)$ and speed of B be ${{V}_{B}}\left( km/hr \right)$.
According to the question, A is faster than B.
$\Rightarrow {{V}_{A}}>{{V}_{B}}...........\left( 1 \right)$
Let the total distance they walked $=x$.
According to the question, $x=24km$.
It is given that the sum of speed of ‘A’ and ‘B’ is 7 km/hr.
$\begin{align}
  & i.e.\ {{V}_{A}}+{{V}_{B}}=7km/hr...........\left( 2 \right) \\
 & \Rightarrow {{V}_{B}}=7-{{V}_{A}} \\
\end{align}$
As, we know that $\text{distance = speed }\!\!\times\!\!\text{ time}$
$\Rightarrow \text{Time = }\dfrac{\text{distance}}{\text{speed}}$
So, Time taken by A $=\dfrac{24}{{{V}_{A}}}$
Similarly, time taken by B $=\dfrac{24}{{{V}_{B}}}$
According to the question, the sum of the times taken by them is 14 hours.
$i.e.\ \dfrac{24}{{{V}_{A}}}+\dfrac{24}{{{V}_{B}}}=14.........\left( 3 \right)$
Putting value of ${{V}_{B}}$ from equation (2) to equation (3);
$\Rightarrow \ \dfrac{24}{{{V}_{A}}}+\dfrac{24}{7-{{V}_{A}}}=14$
Taking LCM and adding the two terms of LHS, we will get;
$\begin{align}
  & \Rightarrow \dfrac{24\left( 7-{{V}_{A}} \right)+24{{V}_{A}}}{\left( {{V}_{A}} \right)\left( 7-{{V}_{A}} \right)}=14 \\
 & \Rightarrow \dfrac{168-24{{V}_{A}}+24{{V}_{A}}}{7{{V}_{A}}-{{V}_{A}}^{2}}=14 \\
\end{align}$
Multiplying both sides by $\left( 7{{V}_{A}}-{{V}_{A}}^{2} \right)$, we will get;
$\Rightarrow \dfrac{168}{14}=14\left( 7{{V}_{A}}-{{V}_{A}}^{2} \right)$
Dividing both sides by ‘14’, we will get;
$\begin{align}
  & \Rightarrow \dfrac{168}{14}=7{{V}_{A}}-{{V}_{A}}^{2} \\
 & \Rightarrow 12=7{{V}_{A}}-{{V}_{A}}^{2} \\
\end{align}$
Taking all the terms to LHS, we will get;
$\Rightarrow {{V}_{A}}^{2}-7{{V}_{A}}+12=0$
Now, we got a quadratic equation. For factoring, let’s split the middle term;
$\Rightarrow {{V}_{A}}^{2}-4{{V}_{A}}-3{{V}_{A}}+12=0$
Taking $''{{V}_{A}}''$ common from first two terms and “-3” common from last two terms,
$\Rightarrow {{V}_{A}}\left( {{V}_{A}}-4 \right)-3\left( {{V}_{A}}-4 \right)=0$
Taking $\left( {{V}_{A}}-4 \right)$common, we will get;
$\begin{align}
  & \Rightarrow \left( {{V}_{A}}-3 \right)\left( {{V}_{A}}-4 \right)=0 \\
 & \Rightarrow {{V}_{A}}=3km/hr\ or\ {{V}_{A}}=4km/hr \\
\end{align}$
If ${{V}_{A}}=3km/hr$, then
${{V}_{B}}=\left( 7-3 \right)=4km/hr\ \ \ \ \ \ \ \ \ \ \ \ \ \left[ From\ equation\ \left( 2 \right) \right]$
Similarly, If ${{V}_{A}}=4km/hr$, then
${{V}_{B}}=\left( 7-4 \right)=3km/hr$
But according to equation (1), ${{V}_{A}}>{{V}_{B}}$;
So, ${{V}_{A}}=4km/hr\ and\ {{V}_{B}}=3km/hr$.
Hence, speed of A is 4 km/hr and option (B) will be the answer.


Note:
In the last step when we got two values of ${{V}_{A}}$ by solving the quadratic equation, a student can do mistake by accepting both of them as answer but when we calculate ${{V}_{B}}$ by using ${{V}_{A}}$, one of them lead to ${{V}_{B}}$ larger than ${{V}_{A}}$, which is not acceptable. As in question, it is clearly given that A is faster than B.