how many liters of a 90% acid solution must be mixed with a 15% acid solution to get 675 L of an 80% acid solution?
will give medal and fan plz help
Let M equal the gallons of the 90% mixing solution. Let F equal the gallons of the final solution. So: 90 + M = F. Also, the number of gallons of pure antifreeze in the final will equal the sum of the gallons of antifreeze in the two mixing parts: Original solution: ( 90 gal )*(0.15) = 13.5 gal [0.15 represents 15%]Mixing solution: M*0.90Final solution: F*0.80 So 13.5 + M*0.90 = F*0.80 Now you have 2 linear equations and 2 unknowns, you can solve for M & F, using your favorite method: M = 585 and F = 675. Add 585 gallons of the 90% to get 675 gallons of 80% solution.
help
@pphalke In what way are you confused?
aveline the problem in general
A will represent the liters of the 90% acid solution, and B will represent the liters of the 15% acid solution A+B=675 Both add up to a total of 675 liters We only want one unknown variable, so let's rearrange the equation: B=675-A Now for the equation. The coefficients represent the percent acid per liter. 0.90A + 0.15B = (675)(0.80) Let's substitute (675-A) for B: 0.90A+0.15(675-A)=(675)(0.80) Use the distributive property: 0.90A+101.25-0.15A=540 You should be able to solve from here.
when I collect like terms I get 0.75? I subtracted 0.90-0.15 is that right
@Aveline
That's the correct coefficient for A. What about the rest of the equation?
I did 0.75A+101.25=540 then I subtract 101.25 from both sides but I get a negative answer its -9585?? is it right
Aveline
Ok, so: 0.90A+101.25-0.15A=540 Combine A's 0.75A+101.25=540 Next, subtract 101.25
@Aveline I get a negative answer
What is 540-101.25?
438.75
Now divide that by 0.75A
585 @Aveline
Yes, you will use 585 liters of A, the 90% acid solution. It was mentioned that we needed to make 675 liters, and that A+B=675. Can you find out how many liters of B we need?
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