1 liter of solution has .31 liter antimicrobial and .69 liter water.
Now suppose we add x
liter water to bring down the concentration
solve for x
So add .107 liter of water in 1 liter of solution to get cncentration of 28%
We have just received 25 litres of an antimicrobial solution that is to be added to the hens drinking water, but the solution that we have is 31% of the concentrate, whereaswe require it to be 28%.
My question is this... If I have 1 litre of concentrate at 31%, how much water must I add to this in order to bring this down to the specified 28% concentration?
This was an error on the manufacturers side, and as I am not mathematically equipped any help from you guys would be much appreciated!
Kind regards,
Adam H
Normally I wouldn't post an (essentially) identical working of a problem that was already solved, but since errors can have significant consequences in real-life applications like these, I thought you might like confirmation of arpitagarwal82's solution:
In one liter of the solution, the volume of the concentrate is 0.31 L and the volume of the water is 0.69 L. Let represent the amount of water that must be added. The volume of the solution will then become To get the appropriate mixture, we need
Solve for
So that is about 107 ml, which does agree with the above answer.
Yet another confirmation - perhaps this one is really for the budding mathematicians. Can I urge you to try to understand what we used to call (about 200 years ago, when I was taught arithmetic) 'multiplying factors'? These are basically fractions that increase or decrease things in a certain ratio. The numbers you are given to play with here are 28 and 31. And the multiplying factors you can make with these are simply and .
To make something bigger in this ratio, then, you multiply by , and to make something smaller you multiply by .
We have 1 litre of solution at 31%; if we increase this to 1 litre litres the concentration will be 28%. It really is that easy!
Grandad