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. to 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.)

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.

18/5 (or 3.6) liters of solution A should be diluted with solution B.

The answer is 10 bud

38 dollars is your answer.