Problem on ratio wrt time distance

Question

This is a rough translation from a local language so please bear with it,

Say we have two people, police $p1$ and thief $p2$.

$p1$ takes $4$ $steps$ and $p2$ takes $5$ $steps$ in the same amount of time. Also the distance $p1$ covers in $6$ $steps$ is equal to the distance $p2$ covers in $8$ $steps$.

What is the ratio of their velocities?

What I tried:

Let $p1$ and $p2$ cover $d1$ and $d2$ distance in 4 and 5 steps, also let say it takes $t1$ time for $p1$ to take 6 steps and $t2$ time for $p2$ to take 8 steps,
so $d1/d2 = 4/5$ and $t1/t2 = 6/8$, so I get $(d1/t1)/(d2/t2) = 16/15$

In my friend circle we are getting another result $15/16$ (if required I will add the procedure).

I just can't wrap my head around the relation of steps with the other units.

Answer 1

Let the distance covered by $p_1$ in $6$ steps or $p_2$ in $8$ steps be $L$.

Then after one time unit, $p_1$ has travelled $\frac{4L}{6}$, and $p_2$ has travelled $\frac{5L}{8}$.

Velocity = Distance / Time, so the ratio of velocities is:

$$ \frac{p_1}{p_2} =\frac{\frac{4L}{6}}{\frac{5L}{8}} =\frac{\frac{2}{3}}{\frac{5}{8}} =\frac{2}{3}\cdot\frac{8}{5} =\frac{16}{15} $$

Answer 2

Although your answer is correct, your method looks wrong, I believe, because $\frac{d_1}{d_2}\neq \frac45$. $p_1$ takes $\frac86\times$ longer steps than $p_2$, so the distance covered by $p_2$ in time $t$ is $5D$, where $D$ is the length of a step $p_2$ takes, and that covered by $p_1$ in same time is $\frac86\times4D$. So, $\frac{v_1}{v_2}=\frac{d_1/t}{d_2/t}=\frac{4\times\frac86}{5}=\frac{d_1}{d_2}=\frac{16}{15}$.

Answer 3

You're correct. I think you and your friends will find this way of looking at it convincing:

In $24$ times units, $p_1$ takes $24\cdot4=96$ steps, and so covers $96/6=16$ units of distance. In the same time, $p_2$ takes $24\cdot5=120$ steps, and covers $120/8=15$ units of distance.

Answer 4

Let the police have run step length l1 and the thief l2;

Let the police have run speed v1 thief v2;

we have $$\dfrac{4\; l1}{v1}= \dfrac{5\; l2}{v2} \tag1$$ and $$ 6 \;l1 = 8\; l2 \tag2$$ Divide and simplify $$\dfrac{v1}{v2}= \dfrac{16}{15} >1 \tag3$$ So the police can catch the thief.