How many gallons of a 80% alcohol solution and a 30% alcohol solution must be mixed to get 10 gallons of a 70% alcohol solution?
80% solution is= to how many gallons
30% solution is = to how many gallons
0.80x is the amount of alcohol in x gallons of 80% mix.
0.30(10-x) is the amount in the 30% mix
if you add them up, the amount of alcohol must be the same as that in 10 gallons of 70%.
I just multiplied everything by 100 to get rid of the pesky decimals.
So, if you solve that, you get x=8
to check, note that 80% of 8 gals is 6.4 gals of alcohol
30% of 2 gals is 0.6 gals
70% of 10 gals is 7 gals, which is 6.4 + 0.6
so the equation works.
look at just the alcohol content. If there are x gallons of 80%, then the rest (10-x) must be 30%. So,
80x + 30(10-x) = 70*10
I don't understand your answer
Thank you
To determine the number of gallons of the 80% alcohol solution and the 30% alcohol solution required to obtain a 10-gallon mixture with a concentration of 70% alcohol, we can use a mixture equation.
Let's assume x represents the number of gallons of the 80% alcohol solution and y represents the number of gallons of the 30% alcohol solution.
The total volume of the mixture is given as 10 gallons:
x + y = 10 ...(Equation 1)
To find the concentration of alcohol in the mixture, we need to consider the amount of alcohol in each solution. The 80% alcohol solution will have 80% of x gallons as pure alcohol, which is 0.8x gallons. Similarly, the 30% alcohol solution will have 30% of y gallons as pure alcohol, which is 0.3y gallons.
The total alcohol in the mixture is given as 70% of the total 10 gallons:
(0.8x + 0.3y)/(x + y) = 0.7 ...(Equation 2)
Now we have a system of two equations (Equation 1 and Equation 2) that we can solve to find the values of x and y.
First, let's rearrange Equation 1 to express y in terms of x:
x + y = 10
y = 10 - x
Substitute this expression for y into Equation 2:
(0.8x + 0.3(10 - x))/(x + 10 - x) = 0.7
(0.8x + 3 - 0.3x)/10 = 0.7
0.5x + 3 = 0.7(10)
0.5x + 3 = 7
0.5x = 7 - 3
0.5x = 4
x = 4/0.5
x = 8
Now that we have the value of x, substitute it back into Equation 1 to find y:
8 + y = 10
y = 10 - 8
y = 2
Therefore, to obtain a 10-gallon mixture with a concentration of 70% alcohol, you would need 8 gallons of the 80% alcohol solution and 2 gallons of the 30% alcohol solution.