Question 817747
R1 is the rate of pipe 1
R2 is the rate of pipe 2
when the pipes are both filling the tank together, their rates are additive.
pipe 1 fills the tank i  3 hours.
this means that pipe 1 fills 1/3 of the tank per hour.
R1 is therefore equals to 1/3.
pipe 1 and pipe2, working together, fill the tank in 3/4 of an hour.
this means that pipe 1 and pipe 2, working together, fill 4/3 of the tank per hour.
since, when pipe 1 and pipe 2 are working together, their rates are additive, then this means that R1 + R2 = 4/3
you now have 2 equations.
R1 = 1/3
R1 + R2 = 4/3
substitute 1/3 for R1 in the second equation and you get:
1/3 + R2 = 4/3
subtract 1/3 from both sides of this equation to get:
R2 = 4/3 - 1/3 which is equal to (4-1) / 3 which is equal to 3/3 which is equal to 1.
this means that R2 is equal to 1.
this means that R2, working alone, would fill the rank in 1 hour.
let's see if that holds up.
we are told that pipe 2 and pipe 2, working together fill the tank in 3/4 of an hour.
the formula is R * T = Q
R is the rate of work.
T is the time
Q is the quantity.
for pipe 1 alone, R = 1/3 and Q = 1, so the formula becomes:
1/3 * T = 1
solve for T to get T = 3 hours.
for pipe 2 alone, R = 1 and Q = 1, so the formula becomes:
1 * T = 1
solve for T to get T = 1
when they work together, their rates are additive.
formula becomes:
(R1 + R2) * T = Q
R1 + R2 = 1 + 1/3 = 4/3
Q = 1
formula becomes:
4/3 * T = 1
multiply both sides of this equation by 3/4 and you get:
3/4 * 4/3 * T = 3/4 * 1
simplify to get:
T = 3/4
solution to this problem is:
pipe 2 alone can fill the tank in 1 hour.