Explanation:
l = large pipe
s = small pipe
s needs x hours alone to fill the pool.
l needs x-4 hours alone to fill the pool.
s does 1/x of the work in 1 hour.
l does 1/(x-4) of the work in 1 hour.
3 hours 45 minutes = 3.75 hours
3.75 × 1/x + 3.75 × 1/(x-4) = 1 (= the whole work)
3.75 + 3.75 × x/(x-4) = x
3.75×(x-4) + 3.75×x = x(x-4) = x² - 4x
3.75x - 15 + 3.75x = x² - 4x
7.5x - 15 = x² - 4x
-15 = x² - 11.5x
0 = x² - 11.5x + 15
the general solution to such a quadratic equation is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = -11.5
c = 15
x = (11.5 ± sqrt(132.25 - 4×1×15))/(2×1) =
= (11.5 ± sqrt(72.25))/2 = (11.5 ± 8.5)/2
x1 = (11.5 + 8.5)/2 = 20/2 = 10
x2 = (11.5 - 8.5)/2 = 3/2 = 1.5
x = 1.5 would turn l = 1/(x-4) negative, which does not make sense.
so, x = 10 is our solution.
that means
s does 1/10 of the work in 1 hour.
so, the small pipe alone fills the pool in 10 hours.
l does 1/(10-4) = 1/6 of the work in 1 hour.
so, the large pipe alone fills the pool in 6 hours.