SOLUTION: Hello,
Please help me. The question:
How many ounces from a 15% solution must be mixed with 4 ounces of a 20% solution to make a 17% solution?
Your help is much appreciate
Algebra.Com
Question 183490: Hello,
Please help me. The question:
How many ounces from a 15% solution must be mixed with 4 ounces of a 20% solution to make a 17% solution?
Your help is much appreciated. Thank you.
Found 2 solutions by ptaylor, josmiceli:
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=amount of 15% solution needed
Now we know that the amount of pure stuff in the 15% solution (0.15x) plus the amount of pure stuff in the 20% solution (0.20*4) has to equal the amount of pure stuff in the final mixture (0.17(4+x)). So:
0.15x+0.20*4=0.17(4+x) get rid of parens
0.15x+0.80=0.68+0.17x subtract 0.68 and also 0.15x from each side
0.15x-0.15x+0.80-0.68=0.68-0.68+0.17x-0.15x collect like terms
0.12=0.02x divide each side by 0.02
x=6 oz----------------------amont of 15% solution needed
CK
0.15*6+0.20*4=0.17*10
0.90+0.80=1.70
1.70=1.70
Hope this helps---ptaylor
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = ounces of the 15% solution needed
Let = ounces of the 17% solution which will result
given:
(1)
---------------
(2)
I have 2 equations and 2 unknowns, so it's solvable
Multiply both sides of (1) by and subtract
from (2)
(2)
(1)
(3)
Plug this value back into (1)
(1)
(1)
(1)
6 ounces of the 15% solution must be mixed with
4 ounces of the 20% solution to make 10 ounces
of the 17% solution
check answer:
(2)
(2)
(2)
(2)
OK
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