Question 211163: A hose can fill a swimming pool in 6 hours. Another hose needs 3 more hours to fill the pool than the two hoses combined. How long would it take the second hose to fill the pool.
Found 2 solutions by Alan3354, Theo:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A hose can fill a swimming pool in 6 hours. Another hose needs 3 more hours to fill the pool than the two hoses combined. How long would it take the second hose to fill the pool.
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h1 takes 6 hours, so it fills 1/6 pool per hour.
h2 takes x hours, so it fills 1/x pool per hour.
Together, they do 1/6 + 1/x per hour.
1/6 + 1/x = (x+6)/6x per hour (together), so it takes 6x/(x+6) hours.
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The time the 2nd hose takes, x hours, is 3 hours more than the time together.
x-3 = 6x/(x+6)
(x-3)*(x+6) = 6x
(x-6)*(x+3) = 0
x = 6 hours (Ignore the -3 hours)
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Each hose takes 6 hours. Together they take 3 hours, so the time for the 2nd hose is 3 hours more than the 3 hours together.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! let X = the rate hose 1 takes to fill the swimming pool.
let Y = the rate hose 2 takes to fill the swimming pool.
let T1 = time it takes for hose 1 to fill the swimming pool.
let T2 = time it takes for hose 2 to fill the swimming pool.
let T3 = time it takes for hose 1 and hose 2 to fill the swimming pool.
let G = number of gallons in the swimming pool.
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formula for hose 1 is:
X * T1 = G
formula for hose 2 is:
Y * T2 = G
formula for hose 1 and 2 is:
(X+Y)*T3 = G
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hose 1 takes 6 hours to fill the pool.
T1 = 6
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formula for hose 1 becomes:
X * 6 = G
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hose 2 takes 3 more hours to fill the pool than the two hoses combined.
T2 = T3 + 3
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formula for both hoses working together is:
(X+Y)*T3 = G
X = G/6
formula becomes:
(G/6 + Y) * T3 = G
Y = G/T2
formula becomes:
(G/6 + G/T2) * T3 = G
T3 = T2 - 3
formula becomes:
(G/6 + G/T2) * (T2-3) = G
factor out G on left hand side of equation to get:
G * (1/6 + 1/T2) * (T2-3) = G
divide both sides by G to get:
(1/6 + 1/T2) * (T2-3) = 1
multiply both sides by 6*T2 to get:
(T2 + 6) * (T2-3) = 6*T2
simplify to get:
(T2)^2 - 3*T2 + 6*T2 - 18 = 6*T2
subtract 6*T2 from both sides and combine like terms to get:
(T2)^2 - 3*T2 - 18 = 0
factor equation on the left to get:
(T2-6)*(T2+3) = 0
T2 = 6
or:
T2 = -3 not good because can't be negative.
Answer is:
T2 = 6
T3 = T2 - 3 = 6 - 3 = 3
We have:
X*6 = G
Y*6 = G
This makes:
X*6 = Y*6
which makes:
X = Y
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Formulas become:
X*6 = G
Y*6 = G
(X+Y)*3 = G
G can be any number of gallons and the equation will work out.
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The question was:
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A hose can fill a swimming pool in 6 hours. Another hose needs 3 more hours to fill the pool than the two hoses combined. How long would it take the second hose to fill the pool.
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The answer is:
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It would take the second hose 6 hours to fill the pool.
The 2 hoses working together take 3 hours to fill the pool.
The second hose takes 3 more hours than both hoses combined to fill the pool.
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What was not stated was that the first hose takes 3 more hours than both hoses combined to fill the pool as well.
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G is unknown, but we could get the answer we needed because it canceled out in the equations when solving for T2. G can be anything and the equations will work. To understand this better, substitute any value for G and solve the equation.
Example:
Let G = 1200 gallons
X * 6 = 1200
X = 1200/6 = 200 gallons per hour.
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Y * 6 = 1200
Y = 1200/6 = 200 gallons per hour.
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(X+Y)*3 = 1200
X = 200
Y = 200
(200+200)*3 = 1200
(400)*3 = 1200
1200 = 1200 proving that the values for X and Y are good and that hose 1 and hose 2 working together take 3 hours to fill the pond.
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