Question 132924: A spice mixture is 25% thyme. How many grams of thyme must be added to 12 g. of the misture to increase the thyme content to 40%? Found 2 solutions by ptaylor, nycsharkman:Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of pure thyme that needs to be added
Now we know that the amount of pure thyme in the 12 g mixture (0.25*12) plus the amount of pure thyme that is added (x) has to equal the amount of pure thyme in the final mixture (0.40(12+x)). So our equation to solve is:
0.25*12+x=0.40(x+12) get rid of parens and simplify
3+x=0.40x+4.8 subtract 3 and also 0.40x from each side
3-3+x-0.40x=0.40x-0.40x+4.8-3 collect like terms
0.60x=1.8 divide both sides by 0.60
x=3 gms--------------------amount of pure thyme needed
CK
0.25*12+3=0.40(3+12)
3+3=0.40(15)
6=6
Hope this helps---ptaylor
You can put this solution on YOUR website! A spice mixture is 25% thyme. How many grams of thyme must be added to 12 grams of the mixture to increase the thyme content to 40%?
Let x = amount of grams of thyme that must be added
Let's pretend that we have three cups.
In the cup 1, we have 12 grams times 25%.
In cup 2, we have x grams.
In cup 3, we have (0.40)(x + 12) grams.
We now have this equation:
12(0.25) + x = 0.40x + 4.80
3 + x = .40x + 4.80
x - .40x = 4.80 - 3
.60x = 1.80
x = 1.80/.60
x = 3 grams MUST BE ADDED
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