SOLUTION: Could you please help me with this How many ounces of 1% antifreeze solution must be added to 100 oz of 40% antifreeze solution to obtain a 16% antifreeze solution? Thank you

Question 1185429: Could you please help me with this
How many ounces of 1% antifreeze solution must be added to 100 oz of 40% antifreeze solution to obtain a 16% antifreeze solution?
Thank you so much

Found 4 solutions by josgarithmetic, Theo, MathTherapy, greenestamps:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
x ounces of 1%
added to 100 ounces of 40%
to make 16% solution

, fairly easy to compute the value
.
.
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
let x = the number of ounces of 1% solution.
let y = total number of ounces of 16% solution.

your equations are:

.01x + .4*100 = .16y
x + 100 = y

simplify to get:

.01x + 40 = .16y
x + 100 = y

multiply both sides of the first equation by 100 and leave the second equation as is to get:

x + 4000 = 16y
x + 100 = y

subtract the second equation from the first to get:

3900 = 15y

solve for y to get:

y = 3900 / 15 = 260

confirm that your original equations are correct when you replace y with 260.

the original equations are:

.01x + .4*100 = .16y
x + 100 = y

simplify to get:

.01x + 40 = .16y
x + 100 = y

replace y with 260 to get:

.01x + 40 = .16*260
x + 100 = 260

simplify to get:

.01x + 40 = 41.6
x + 100 = 260

subtract 40 from both sides of the first equation and subtract 100 from both sides of the second equation to get:

.01x = 41.6 - 40
x = 260 - 100

simplify to get:

.01x = 1.6
x = 160

go back to your original equations and replace y with 260 and x with 160 to get:

.01x + 40 = .16y becomes 1.6 + 40 = .16 * 260
x + 100 = y becomes 160 + 100 = 260

simplify to get:

41.6 = 41.6
260 = 260

this confirms the values of x and y are correct.

your solution is that:

x = 160 and y = 260

let x = the number of ounces of 1% solution.
let y = total number of ounces of 16% solution.

this means that 160 ounces of 1% solution are added to 100 ounces of 40% solution to get 260 ounces of 16% solution.

that's your solution.

normally you would want to move all the variables to the left side of the equal sign and all the constants to the right side of the equal sign, but this is not necessary, and in this case, it was preferable to leave them where they were.

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Could you please help me with this
How many ounces of 1% antifreeze solution must be added to 100 oz of 40% antifreeze solution to obtain a 16% antifreeze solution?
Thank you so much

Let amount of 1% solution to mix, be S
Then we get: .01S + .4(100) = .16(S + 100)
.01S + .4(100) = .16S + .16(100)
.4(100) - .16(100) = .16S - .01S
.24(100) = .15S
Amount of 1% solution needed, or
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

You have three responses showing formal algebraic solutions to your problem. All are valid....

If a formal algebraic solution is not required, here is a quick and easy way to solve this kind of problem (if the numbers are "nice") informally.

You are starting with a 40% antifreeze solution and adding a 1% antifreeze solution until the mixture is 16% antifreeze.

To model the problem, picture starting at 40% on a number line and "walking" towards 1%, stopping when you reach 16%.

In doing that, you have walked 24/39 = 8/13 of the total distance. (40 to 1 is a difference of 39; 40 to 16 is a difference of 24; hence the fraction 24/39=8/13.)

That means 8/13 of the mixture is the 1% antifreeze that you are adding.

So the 100 ounces of 40% antifreeze you started with was 5/13 of the mixture; that means the 8/13 of the mixture that you added was 160 ounces.

ANSWER: 160 ounces

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