Please, some help on this one
Some of 50%-intensity red paint is replaced with 25% solution of red paint such that the new paint intensity is 30%. What fraction of the original paint was replaced?
Let H = amt of high intensity paint, L = amt of low intensity paint
We want H/L --- What is the ratio of H to L in the final mixture?
(0.5H + 0.25L)/(H + L) = 0.3 ---> 0.5H + 0.25L = 0.3H + 0.3L
0.2H = 0.05L ----> H/L = 1/4 ... so for every 1 part H we have 4 part L in the mixture
We now have 4 L for every H left in the can, Thus 4/5 of the H was replaced by L
(We could have solved for L/H = 4, thus 4 L for every 1, same thing)