help with this math advanced algeb problem?

Question

a health clinic uses a solution of bleach to sterilize petri dishes in w/c cultures are grown. the sterilization tank contains100 gal of a solution of 2% ordinary householdbleach mixed with pure distilled water. New research indicates that the concentration of bleach should be 5% for complete sterilization. how much of the solution should be drained and replaced with bleach to increase the bleach content to the recommended?

Answer 1

Take N to be the amount of the original solution kept.
Since the original is 2% bleach (a concentration of 2/100 or 1/50), there will be N*(1/50) amount of bleach contributed from the initial mixture to the final.

The new added is 100% bleach so the amount of bleach it contributes to the final mixture will be (100 - N).

Since 5% is wanted there must be 5 gals of bleach in the final mixture. So:

5 = N/50 + 100 - N
49N = 95*50 = 4750 and N = 96.9388 gallons

So 100 - 96.9388 or 3.0613 gallons of the original mixture should be replaced.

Answer 2

'Going to drain x gallons of the .02 sol. Have left (100 - x) gallons of a 2% solution. Need to end up with 5 gal. of bleach in the 100 gal. tank. So 2%(100-x gals) + x gallons of pure bleach = 5 gallons of bleach.
.02(100-x) = 5; 2 - .02x + x = 5; 2 + .98x = 5; .98x = 3; x = 3.06 gallons should be drained and replaced by pure household bleach.

Answer 3

tricky

100 gal of 2% solution is 98 gal water + 2 gal bleach

100 gal of 5% solution is 95 gal water + 5 gal bleach

if you don't like math, drain half the tank into another 100 gal sterilization tank and then add 46 gallons of water and 4 gallons of bleach to each tank giving you two 100 gallon tanks of 5% solution

but, assuming you don't have another 100 gal sterilization tank,

you want to subtract x amount of (2% water plus bleach) from 100 gallons of (2% water plus bleach) and replace it with the same amount: x of pure bleach so that the resulting 100 gallons is 5% bleach.
100(.98w+.02b)-(.98w +.02b)x+ 1bx=100(.95w+.05b)
98w+2b-.98wx-.02bx+1bx=95w+5b
98w-95w+2b-5b-.98wx+1.00bx-.02bx=0
3w-3b-.98wx+.98bx=0
3(w-b)-.98x(w-b)=0
3(w-b)=.98(w-b)x
3(w-b)/.98(w-b)=x
x=3/.98 = 3.0612245