Question

Can anyone solve this im not sure how to ???

How many litres of pure antifreeze must be added to 77 litres of a 20​% antifreeze solution to obtain a 60​% antifreeze​ solution?

Answer 1

231 L

Explanation:

We can put it into fractions is that’s easier

So `(77L) / (20%) = (xL)/(60%)`

X represents the unknown L of solution.

So there’s two ways of doing this:

1) To get 60% from 20%, you would need to multiply that by 3

So `((77L)*3)/((20%)*3)=(231L)/(60%)`, and therefore x = 231 L

OR

2) Cross multiply to solve for x

`(77L)*(60%)=(20%)*x`

`4620 = 20x` (I removed the units for better visual)

You want to isolate x, so divide both sides by 20 to get:

`4620/20 = 20x/20`

`231 = x`

So in the end, there would be 231 L of pure antifreeze to get 60% solution