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opticaltempest
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Hello, I am stuck on the following problem.
How many milliliters of a 30% acid solution and a 55% acid solution must be mixed to obtain 100 milliliters of a 50% acid solution?
My solution:
Let X = The unknown milliliters of 30% acid.
Let Y = The unknown milliliters of 55% acid.
(1) X + Y = 100
(2) 0.3X + 0.55Y = 0.5
Multiplying (2) by 100 in order to work with integers,
(2) 30X + 55Y = 50
Multiplying (1) by -30 in order to eliminate X
(1) -30X - 30Y = -3000
So our new equations are,
(1) -30X - 30Y = -3000
(2) 30X + 55Y = 50
Adding the two equations in order to eliminate X,
25Y = -2950 => Y = -118
Back-substituting Y into (1) in order to solve for X,
-30X - 30(-118) = -3000
-30X + 3540 = -3000
-30X = -6540
X = 218
The answer in the book says:
20 milliliters of 30% acid solution and 80 milliliters of 55% acid solution.Any hints as to where I am going wrong?
How many milliliters of a 30% acid solution and a 55% acid solution must be mixed to obtain 100 milliliters of a 50% acid solution?
My solution:
Let X = The unknown milliliters of 30% acid.
Let Y = The unknown milliliters of 55% acid.
(1) X + Y = 100
(2) 0.3X + 0.55Y = 0.5
Multiplying (2) by 100 in order to work with integers,
(2) 30X + 55Y = 50
Multiplying (1) by -30 in order to eliminate X
(1) -30X - 30Y = -3000
So our new equations are,
(1) -30X - 30Y = -3000
(2) 30X + 55Y = 50
Adding the two equations in order to eliminate X,
25Y = -2950 => Y = -118
Back-substituting Y into (1) in order to solve for X,
-30X - 30(-118) = -3000
-30X + 3540 = -3000
-30X = -6540
X = 218
The answer in the book says:
20 milliliters of 30% acid solution and 80 milliliters of 55% acid solution.Any hints as to where I am going wrong?
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