Question 157698
not terribly sure because ran out of time but i think this is the answer.
let p1 = rate of the larger pump.
let p2 = rate of the smaller pump.
both pumps working for 5 hours can drain the tank.
so 5*p1 + 5*p2 = x where x is the amount of fuel in the tank.
which means p1 is pumping for 5 hours and p2 is pumping for 5 hours.
problem states that pump 1 can drain the tank by itself in 4 hours less than pump 2 can drain the tank by itself.
so p1 * (t-4) = x
and p2 * (t) = x
if they both work by themselves than they both can drain 2 * the tank.
so p1 * (t-4) + p2 * (t) = 2x = 10*p1 + 10*p2 since in 10 hours they can both drain the tank twice working together.
so, .....
p1 * (t-4) = 10 * p1
and p2 * (t) = 10 * p2
solving for p1 we get
p1 * t - 4 * p1 = 10 * p1
which becomes
p1 * t = 6 * p1
which becomes
t = 6
it takes p1 6 hours to drain the tank.
solving for p2 we get
p2 * t = 10 * p2
which becomes

t = 10
it takes p2 10 hours to drain the tank.
answer is it takes p2 10 hours to drain the tank.
p1 takes 6 hours which is 4 hours less so the equation looks sound.