SOLUTION: E-chem Testing has a solution that is 80% base and another that is 30% base. A technician needs 150L of a solution that is 62% base. The 150L will be prepared by mixing the two s
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-> SOLUTION: E-chem Testing has a solution that is 80% base and another that is 30% base. A technician needs 150L of a solution that is 62% base. The 150L will be prepared by mixing the two s
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Question 61082This question is from textbook Intermediate Algebra
: E-chem Testing has a solution that is 80% base and another that is 30% base. A technician needs 150L of a solution that is 62% base. The 150L will be prepared by mixing the two solutions on hand. How much of each should be used? This question is from textbook Intermediate Algebra
You can put this solution on YOUR website! Suppose that 'x' liters of 80% base solution is taken.
Then quantity of 30% base solution = (150 - x)liters [as the total is 150L)
Strength of 150L solution = 62%
The equation becomes,
x*0.8 + (150-x)*0.3 = 150*0.62
==> 0.8x + 45 - 0.3x = 93
==> 0.5x + 45 = 93
==> 0.5x = 93 - 45 [adding -45 to both the sides]
==> 0.5x = 48
==> 0.5x/0.5 = 48/0.5
==> x = 96
so 150-x = 150-96
= 54
Thus 96L of 80% base is to be added to 54L of 30% base to get 150L of 62% base.
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