A fish leaves a spot swimming 18kph. A shark leaves the same

[OMGSHELA]

A fish leaves the reef swimming 18km/h. A shark leaves the same spot on the reef 15 minutes later swimming 23 km/h. How long will it take the shark to catch and eat the fish?

 

[Mrspi]

Re: A fish leaves a spot swimming 18kph. A shark leaves the

A fish leaves the reef swimming 18km/h. A shark leaves the same spot on the reef 15 minutes later swimming 23 km/h. How long will it take the shark to catch and eat the fish?

Let t = number of hours the fish swims

At 18 km/hr, the fish will cover 18t km.

The shark leaves 15 minutes, or 1/4 hour later. So, the shark swims for [t - (1/4)] hours and at 23 km/hr, it will cover 23[t - (1/4)] km.

When the shark catches the fish, each will have covered the same distance. So,

distance for fish = distance for shark

18(t) = 23 [t - (1/4) ]

 

[soroban]

Re: A fish leaves a spot swimming 18kph. A shark leaves the

Hello, OMGSHELA!

Here's a back-door approach to the problem . . .

A fish leaves the reef swimming 18 km/hr.
A shark leaves the same spot on the reef 15 minutes later swimming 23 km/hr.
How long will it take the shark to catch and eat the fish?

The fish has a quarter-hour headstart; it is: \(\displaystyle \,\frac{1}{4}\,\times\,18 \:=\:4.5\) km ahead.


The shark swims \(\displaystyle 23\,-\,18\:=\:5\) km/hr faster than the fish.

It is as if the fish has stopped and the shark approaches at \(\displaystyle 5\) km/hr.


How long does it take the shark to cover 4.5 km?

. . \(\displaystyle \text{Time} \:=\:\frac{\text{Distance}}{\text{Speed}} \:=\:\frac{4.5}{5} \:=\:0.9\text{ hours }\,=\:54\text{ minutes}\)

 

[mammothrob]

But how long does it take for the shark to eat the fish once it catches it? We could be talking about a slow chewer here.
:wink: