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X takes 3 hours more than Y to walk 30 km. But, if X doubles his pace, he is ahead of Y by $1\dfrac{1}{2}$ hours. Find their speed of walking.

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Hint: Assume that the speed of X is x and the speed of Y is y. Use the two conditions given to make two equations in x and y using the formula $time=\dfrac{distance}{speed}$ and with distance as 30 km. Solve the two equations to find the values of x and y.

Complete step-by-step answer:
Let us assume that the speed of X is x and the speed of Y is y. Given that the time taken by X to walk 30 km is 3 hours more than the time taken by Y to walk the same distance. Further using the formula $time=\dfrac{distance}{speed}$, we can make the following equation:
$\begin{align}
 & \dfrac{30}{x}-\dfrac{30}{y}=3 \\
 & \Rightarrow \dfrac{30y-30x}{xy}=3 \\
 & \Rightarrow 30y-30x=3xy \\
 & \Rightarrow 10y-10x=xy\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 1 \right) \\
\end{align}$
If X doubles his speed, his speed becomes 2x. Hence, according to the second condition given, the time taken by Y to walk 30 km at a speed of y is $1\dfrac{1}{2}$ hours more than the time taken by X to walk the same distance at a speed of 2x.
$\begin{align}
 & \dfrac{30}{y}-\dfrac{30}{2x}=1\dfrac{1}{2} \\
 & \Rightarrow \dfrac{30}{y}-\dfrac{30}{2x}=\dfrac{3}{2} \\
 & \Rightarrow \dfrac{60x-30y}{2xy}=\dfrac{3}{2} \\
 & \Rightarrow 60x-30y=\dfrac{3}{2}\left( 2xy \right) \\
 & \Rightarrow 60x-30y=3xy \\
 & \Rightarrow 20x-10y=xy\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 2 \right) \\
\end{align}$
Equating the two values of $xy$ obtained from equations (1) and (2), we get
$\begin{align}
 & 10y-10x=20x-10y \\
 & \Rightarrow 10x+20x=10y+10y \\
 & \Rightarrow 30x=20y \\
 & \Rightarrow 3x=2y \\
 & \Rightarrow x=\dfrac{2y}{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 3 \right) \\
\end{align}$
Using this value of x in equation (1), we get
\[\begin{align}
 & 10y-10x=xy \\
 & \Rightarrow 10y-10\left( \dfrac{2y}{3} \right)=\left( \dfrac{2y}{3} \right)y \\
 & \Rightarrow 10y-\dfrac{20y}{3}=\dfrac{2{{y}^{2}}}{3} \\
 & \Rightarrow \dfrac{30y-20y}{3}=\dfrac{2{{y}^{2}}}{3} \\
 & \Rightarrow 10y=2{{y}^{2}} \\
 & \Rightarrow 2{{y}^{2}}-10y=0 \\
 & \Rightarrow 2y\left( y-5 \right)=0 \\
\end{align}\]
This equation can have 2 solutions, $y=0$ and $y=5$. But the speed cannot be 0. Hence, the speed of Y is 5 km/hr. To find the speed of X, we substitute this value in equation (3) to get
$\begin{align}
& x=\dfrac{2y}{3} \\
& \Rightarrow x=\dfrac{2\times 5}{3} \\
& \Rightarrow x=\dfrac{10}{3} \\
\end{align}$
Thus, the speed of X is 3.33 km/hr.

Note: Observe that in the initial condition that we have formulated, we take x and y in the denominator. This division by x and y is possible only with the assumption that x and y are non-zero. Thus, we can eliminate the solution $y=0$. Such an elimination would not have been possible, had this condition not been there.