How many times faster is Y than X?

Question

X starts running from point A to point B, and at the same moment, Y starts running from point B to point A.

Once the distance between X and Y gets four times smaller, Z started running from point A to point B.

Later, X, Y and Z met at the same moment at the same point.

Later, Y reached to A at the same moment Z reached to B.

How much faster is Y than X?

Answer 1

Let the distance between points $A$ and $B$ is $d$.

Therefore, $Z$ starts running when $X$ and $Y$ are $\frac{d}{4}$ apart, say time $t$.

$$v_xt+v_yt=\frac{3d}{4}$$

Now, $X,Y,Z$ meet at time $t'$

$$v_xt'+v_yt'=d$$

Comparing, $t=\frac{3t'}{4}$

Distance covered by $X$ in this time is also equal to distance covered by $Z$

$$v_xt'=v_z(t'-t)$$ $$v_z=4v_x$$

Now, $Y$ reaches $A$ and $Z$ reaches $B$ say in time $t_c$, ($Z$ covers the distance $Y$ covered in $t'$ and $Y$ covers the distance $Z$ covered in $(t'-t)$)

$$v_yt_c=v_z(t'-t)$$ $$v_zt_c=v_yt'$$

Solving, $v_z=2v_y$

Therefore,

$$v_y=2v_x$$