Question

Albert and fernandes have two leg swimming race. both start from opposite ends of the pool. on the first leg, the boys pass each other at 18 m from the deep end of the pool. during the second leg they pass at 10 m from the shallow end of the pool. both go at constant speed but one of them is faster. each boy rests for 4 seconds at the end of the first leg. what is the length of the pool?

Answer 1

Answer:

Step-by-step explanation:

Let the length of swimming pool be d meters

let their speed be x and y.

So according to question the fast swimmer (let x) would start from shallow end.  

Thus , Let they first meet after time, t seconds

x × t = d - 18 .......(1)

y × t = 18 .....(2)

Divide equation (2) by equation (1)

$\bf\implies \frac{y}{x}=\frac{18}{d-18}..........(3)$

Let T be the time after which they meet 2nd time (the 4 sec delay is cancelled as both wait for 4 sec)

Now, since x traveled one length complete to deep end + length from deep end to 10 m before shallow end

⇒ x × T = 2d - 10 ........(4)

Also, y traveled one length complete to shallow end + 10 m from shallow end

⇒ 4y × T = d + 10........(5)

Now, divide equation (5) by (4)

$\bf\implies\frac{y}{x}=\frac{d+10}{2d-10}......(6)$

On comparing equations (5) and equation (6) , We get,

$\frac{18}{d-18}=\frac{d+10}{2d-10}\\\\\implies 36d-180=d^2+10d-18d-180\\\\\implies d^2-44d=0\\\\\implies d(d-44)=0\\\\\implies d= 0\text{ or }d = 44$

Since, distance cannot be 0 so d = 0 is rejected.

Hence, d = 44 meters.

Therefore, Length of the pool = 44 meters