Two pails of different sizes contain 34.5 liters of water altogether. When 0.68 liter of water is poured from the bigger pail into the smaller pail, the amount of water in the bigger pail is 9 times that in the smaller pail. How much water was there in each pail at first?
S+L=34.5
S+.68 =1/9(L-.68)
S+L=34.5
9S-L=9*.68-.68/9
add the equations..
10S=9*.68-.68/9 +34.5
solve for S first.
Tfh66ml
31.68 in big pail
2.82 in small pail
To solve this problem, we can use a system of equations.
Let's assume the capacity of the smaller pail is x liters, and the capacity of the bigger pail is y liters.
From the problem, we know that the total amount of water in both pails is 34.5 liters:
x + y = 34.5 -- Equation 1
We also know that when 0.68 liters of water is poured from the bigger pail into the smaller pail, the amount of water in the bigger pail becomes 9 times that in the smaller pail:
y - 0.68 = 9(x + 0.68) -- Equation 2
Now we can solve these equations simultaneously to find the values of x and y.
First, let's solve Equation 1 for x:
x = 34.5 - y
Now substitute this value of x into Equation 2:
y - 0.68 = 9((34.5 - y) + 0.68)
Simplify the equation:
y - 0.68 = 9(34.5 - y + 0.68)
y - 0.68 = 9(35.18 - y)
y - 0.68 = 316.62 - 9y
10y - 0.68 = 316.62
10y = 317.3
y = 31.73
Now substitute this value of y back into Equation 1 to find the value of x:
x + 31.73 = 34.5
x = 34.5 - 31.73
x = 2.77
So, the initial amount of water in the smaller pail was 2.77 liters, and the initial amount of water in the bigger pail was 31.73 liters.