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4 Answers

The first thing to do is to set up a table... hopefully you will see the work below as a table. The first two rows are the two things you are combining, the last row is the total. The hardest part of this problem is expressing the amounts you have of the two solutions... the first is x, and since you want a total of 945ml, the second is... 945-x. Tricky. | percent | amount | calculation of amount acid solution 1 | 75 | x | 75x solution 2 | 30 | 945-x | 30(945-x) total | 50 | 945 | 50(945) The last column gives you the math problem: the acid in solution 1 + acid in solution 2 = total acid. 75x + 30(845-x) = 50(945) which you can now use to solve for x.

Let x and y, be the quantities in milliliters, for the two acid solutions.

Mixing them, the percent acidity of the mixed acid

(75x + 30y) / (x + y) = 50.

We can simplify this

75x + 30y = 50x + 50y

25x - 20y = 0.

5x - 4y = 0 

Also, the combined amount of the solution is x+y = 945.

So we have a system of two linear equations 5x - 4y = 0 and x+y = 945.

Adding four times the second equation to the first gives 9x = 3780.

Thus x = 3780/9 = 420, and y = 945 - x = 945 - 420 = 525.

So you need 420 ml of the 75% solution and 525 ml of the 30% solution.

 

 

Method I. Standard way:

x = the amount of 75% acid solution

Thus, 945-x = the amount of 30% acid solution

Balance: .75x + .3(945-x) = .5*945

Solve for x,

x = 420 mL

Answer: 420 mL of 75% acid solution is required.

 

Method II. Compare to the final equilibrium.

75-50 = 25

50-30 = 20

By inverse proportion, the amount of 75% acid solution = (20/(20+25))(945) = 420 mL

What quantity of 75 per cent acid solution must be mixed with a 30 per cent solution to produce 945 mL of a 50 per cent solution?

Let x = amount (mL) of 75% solution

then

945-x = amount (mL) of 30% solution

.75x + .30(945-x) = .50(945)

.75x + 283.5-.30x = 472.5

.45x + 283.5 = 472.5

.45x = 189

x = 420 mL (of 75% solution)

30% solution:

945-x = 945-420 = 525 mL (of 30% solution)

 

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