How many ounces of 1% antifreeze solution must be added to 100 ounces of 40%
antifreeze solution to obtain a 16% antifreeze solution?
how do I set this equation up?
amount of the 1% stuff to be added ---- x ounces
.1x + .4(100) = .16(x + 100)
solve for x
but first change .1x to .01x
...
Thanks Steve, how sloppy of me
To set up an equation for this problem, we need to determine the amount of ounces of 1% antifreeze solution that must be added. Let's call this unknown quantity "x".
First, let's determine how much antifreeze is in the original 100 ounces of 40% solution, which is 40% of 100 ounces or 0.4 * 100 = 40 ounces of antifreeze.
Next, we need to determine the amount of antifreeze in the desired 16% solution. Since the total solution will be 100 + x ounces, the amount of antifreeze in the final solution will be 16% of (100 + x) ounces or 0.16 * (100 + x) ounces.
Now, we can set up an equation: the amount of antifreeze in the original solution plus the amount of antifreeze in the added 1% solution should equal the amount of antifreeze in the desired 16% solution.
So, the equation can be written as:
40 ounces of antifreeze + x ounces of antifreeze (from the added 1% solution) = 0.16 * (100 + x) ounces of antifreeze.
Now that we've set up the equation, we can solve it to find the value of 'x', which represents the number of ounces of 1% antifreeze solution that must be added.