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Laps

Age 14 to 16
ShortChallenge Level Yellow starYellow starYellow star
Secondary curriculum
  • Problem
  • Solutions

Answer: On the hare's 13th lap


How far the tortoise goes for each hare lap using speed distance time
Hare  15 km  in   1 hour
         1 km    in 4 minutes
         250 m  in 1 minute
         300 m  in 1$\frac15$ minutes

Tortoise  13.8 km     in   1 hour
             13800 m    in   1 hour
         13800$\div$60 m  in 1 minute
                230 m     in 1 minute
230 + $\frac{230}5$ = 276 m  in 1$\frac15$ minutes

The tortoise loses 24 m every time the hare runs a lap

  
24 $\times$ 12 = 288 $\lt$ 300

24 $\times$ 13 $\gt$ 300

  






So the tortoise does not reach the starting point on the hare's 13th lap

So the hare passes the tortoise on the hare's 13th lap.


How far the tortoise goes for each hare lap using inverse proportion
hare speed : tortoise speed is $15 : 13.8$ 
hare distance : tortoise distance is $13.8:15$, which is equivalent to $23:25$
So every time the hare runs a lap, the tortoise runs $\frac{23}{35}$ of a lap
 
$\frac2 {25}\times12 \lt 1$
$\frac2{25}\times13\gt1$  










So the tortoise does not reach the starting point on the hare's 13th lap

So the hare passes the tortoise on the hare's 13th lap.


After how long has the hare travelled 300 m further than the tortoise?
$t$ time (in hours)
hare distance $=15t$
tortoise distance $=13.8t$
hare first passes tortoise when hare is $300$ metres = $0.3$ km ahead $$\begin{align} &15t-13.8t=0.3\\
\Rightarrow &1.2t = 0.3\\
\Rightarrow & t= \tfrac{0.3}{1.2}=\tfrac14\end{align}$$ When $t=\frac14$ the hare has run $15\times\frac14 = 3\frac34$ km or $3750$ metres
$3600$ m $= 12$ laps
$3900$ m $=13$ laps
So the hare was on her thirteenth lap.


Working out how long it takes each runner to run $k$ laps
Hare runs 15km   in 1 hour
            15000 m in 1 hour
              300 m   in $\frac{1}{50}$ hours.

Tortoise runs 13.8 km   in 1 hour
                    13800 m in 1 hour
                      300 m   in $\frac{1}{46}$ hours.

Hare runs $k$ laps in $\frac{1}{50}\times k$ hours.
When the hare passes the tortoise, the tortoise has only run $k-1$ laps.
Tortoise runs $k-1$ laps in $\frac{1}{46}\times(k-1)$ hours. $$\begin{align}&\tfrac{1}{50}k=\tfrac{1}{46}(k-1)\\
\Rightarrow&46k=50(k-1)\\
\Rightarrow&46k=50k-50\\
\Rightarrow &40=4k\\
\Rightarrow &12.5=k\end{align}$$So the hare has run $12.5$ laps and is on her thirteenth when she passes the tortoise.



You can find more short problems, arranged by curriculum topic, in our short problems collection.

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