I am currently helping a friend's child with his schoolwork. He is currently in primary school and being taught the topic of speed. I would like to give him the following problem as practice but I am not sure how to explain the solution to him at his level:

A and B competed in a race. Every time A ran $7$ m, B ran $3$ m. If B took $14$ minutes to complete the race, how long did A take?

My proposed solution: the ratio of A's speed to B's speed is $7:3$. So A will take a shorter time than B. Since distance $=$ speed $\times$ time, we have distance covered by A $=$ A's speed $\times$ A's time $=$ distance covered by B $=$ B's speed $\times$ B's time, i.e. $7$ $\times$ A's time is equal to $3$ $\times$ B's time. Hence the ratio of A's time to B's time is $3:7$. So, $7u=14,\ 1u=2$ and hence $3u=6$.

How do I rephrase my solution to make it more student-friendly? I am trying not to use any formula.

I think many students have difficulty grappling with the fact that speed is inversely proportional to time. If A's speed is $x$ times B's speed, where $x$ is whole number, then most students can easily see that A's time is $1/x$ times of B's time. But if $x$ is not a whole number, most will get stuck.

I appreciate all advice. Thank you.

• Are you opposed to giving him the problem to see how he broaches it - without having a pre-planned digestable solution? Alternatively: Can you give an example of his work to indicate the current stage of his proportional reasoning? I am not sure what would be "student-friendly" without more info on what the student knows/can do. Nov 14, 2015 at 22:44
• You say you gave this problem. Are you sure the problem is at his level? There are a wide range of problems related to speed and I think this may be harder than what he is learning. When my 11 year old students learned speed, a typical problem was given two of the three paremeters (speed, time, distance), find the third parameter. This is much harder than that. Nov 15, 2015 at 21:28

We are not given the speeds of A and of B. But note that the speed of A is $7$ meters for some unknown but fixed unit of time, and the speed of B is $3$ meters for the same unit of time. Thus, the distance traveled by A is $\frac{7\mathsf{~meters}}{\mathsf{unit~of~time}}\times\mathsf{?~minutes}$; and the distance traveled by B is $\frac{3\mathsf{~meters}}{\mathsf{unit~of~time}}\times 14\mathsf{~minutes}$.
Noting that $3\times 14=3\times 2\times 7=6\times 7=7\times 6$, it can be seen that A would take $6$ minutes. (For brevity, I skipped a lot of steps here. You should supply the missing steps.)