Saline solution A is 10% strength and saline solution B is 5% strength. I have an order for 6 litres of an 8% saline solution. T o fill the order how many litres of solution A should be diluted with solution B? (Hint. Assume a 1% solution contains r grams of salt per litre.)
2 answers:
Let x = volume of saline solution A, y = volume of saline solution B. We're going to set up two equations. Firstly, x + y = 6. That gives us y = 6 – x (1) Second, 0.1x + 0.05y = 0.08•6, or 0.1x + 0.05y = 0.48 (2) Substitute (1) into (2): 0.1x + 0.05(6 – x) = 0.48 0.1x + 0.3 – 0.05x = 0.48 0.05x + 0.3 = 0.48 Subtract 0.3 from both sides: 0.05x = 0.18 x = 3.6 Then y = 6 – 3.6 = 2.4 So we have to use 3.6 litres of solution A and 2.4 litres of solution B.
<span>18/5 (or 3.6) liters of solution A should be diluted with solution B.</span>
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