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A hare and a Jackal are running a race. Three leaps of the hare are equal to four leaps of Jackal. For every six leaps of the hare, the jackal takes 5 leaps. Find the ratio of their speeds?

I have tried to solve the question like:

speed of Hare = X/3 Speed of Jackal = X/4

so ratio should be 6x/3:5x/4 or 2x:1.25x or 2:1.25

Now I am not very sure that I am following the right process?

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up vote 2 down vote accepted

Let $l_h,l_j$ be the leap lengths of the hare & jackal respectively. Let $r_h, r_j$ be the number of leaps per second (rate).

We have $3 l_h = 4 l_j$, or $\frac{l_h}{l_j} = \frac{4}{3}$.

We also have $r_h = \frac{6}{5} r_j$, or $\frac{r_h}{r_j} = \frac{6}{5}$.

Their speeds are given by the leap rate times the leap length, so $\frac{s_h}{s_j} = \frac{l_h r_h}{l_j r_j}= \frac{l_h}{l_j} \frac{r_h}{r_j} = \frac{4}{3} \frac{6}{5} = \frac{8}{5}$.

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Essentially, yes, your process was correct.

Let's simplify the problem and say that the length of every leap the hare takes is $\frac{4 \textrm{distance units}}{1 \textrm{leap}}$, and the length of a jackal's leap is $\frac{3 \textrm{distance units}}{1 \textrm{leap}}$. The pace of the hare is $\frac{1 \textrm{leap}}{5 \textrm{time units}}$, while the pace of the jackal is $\frac{1 \textrm{leap}}{6 \textrm{time units}}$. Now we can multiply the rates to get the hare's speed as $\frac{4 \textrm{distance units}}{1 \textrm{leap}} \cdot \frac{1 \textrm{leap}}{5 \textrm{time units}} = \frac{4 \textrm{distance units}}{5 \textrm{time units}}$ and the jackal's speed as $\frac{3 \textrm{distance units}}{1 \textrm{leap}} \cdot \frac{1 \textrm{leap}}{6 \textrm{time units}} = \frac{3 \textrm{distance units}}{6 \textrm{time units}}$. Now it is a simple task to find the ratio of their speeds, that is, the ratio of the hare's speed to the jackal's is $\frac{\frac{4}{5}}{\frac{3}{6}} = \frac{8}{5} = 1.6$

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