Question 398521
It takes pump A 2 hours less time than pump B to empty a certain swimming pool.
 Pump A is started at 8:00 A.M., and pump B is started at 11:00 A.M.
 If the pool is still half full at 5:00 P.M., then how long would it take pump A working alone?
:
Let t = time required by pump A working alone
then
(t+2) = time required by pump B working alone
:
Pump A: 8 AM to 5 PM: 9 hrs,
Pump B: 11Am to 5 PM: 6 hrs
:
Let the empty pool = 1, then a half full pool = {{{1/2}}}
:
{{{9/t}}} + {{{6/((t+2))}}} = {{{1/2}}}
Multiply by 2t(t+2), results:
9(2(t+2)) + 6(2t) = t(t+2)
18(t+2) + 12t = t^2 + 2t
18t + 36 + 12t = t^2 + 2t
30t + 36 = t^2 + 2t
A quadratic equation
t^2 + 2t - 30t - 36 = 0
t^2 - 28t - 36 = 0
Use the quadratic formula to find t
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this equation: x=t; a=1; b=-28; c=-36
{{{x = (-(-28) +- sqrt(-28^2-4*a*-36 ))/(2*1) }}}
Do the math here, should get a positive solution of:
t = 29.23 hrs, Pump A will empty the pool alone
:
:
Check solution
{{{9/29.23}}} + {{{6/31.23}}} = 
.308 + .192 = .500 (a half full pool)