The boys mixed 10 gallons of 20% pure lemon juice mix and 10 gallons of the 70% pure lemon juice mix , because they wanted 20 gallons of lemonade at a 40%lemon juice mixture. They thought that because 40% was almost halfway between 20% and 70%,they should mix equal parts of both, but the lemonade turn out too tart. How much of each should they have used to get a final mixture of 20 gallons at 40% lemon juice?
If they used x gallons of 20% juice, then
.20x + .70(20-x) = .40*20
x = 12
so, 12 of 20% and 8 of 70%
To solve this problem, we can set up a system of equations representing the amount of pure lemon juice in each mixture and the total quantity of lemonade.
Let's denote:
x = the number of gallons of the 20% pure lemon juice mix
y = the number of gallons of the 70% pure lemon juice mix
Now, let's write the equations based on the information given:
Equation 1: The total quantity of lemonade is 20 gallons:
x + y = 20
Equation 2: The desired lemon juice concentration is 40%:
(0.20x + 0.70y) / 20 = 0.40
Let's simplify equation 2 by multiplying both sides by 20:
0.20x + 0.70y = 8
Now, we have a system of equations:
x + y = 20
0.20x + 0.70y = 8
We can solve this system of equations to find the values of x and y, which will give us the amounts of the 20% and 70% lemon juice mixes needed.
One way to solve this system is by substitution. Let's solve equation 1 for x and substitute it into equation 2:
x = 20 - y
0.20(20 - y) + 0.70y = 8
4 - 0.20y + 0.70y = 8
0.50y = 4
y = 8
Now, substitute the value of y back into equation 1 to find x:
x + 8 = 20
x = 12
Therefore, they should have used 12 gallons of the 20% pure lemon juice mix and 8 gallons of the 70% pure lemon juice mix to obtain a 20-gallon lemonade mixture with a 40% lemon juice concentration.