Question

- 8th Grade Math -

How many grams each of a 10% salt solution and a 5% salt solution must be mixed in order to obtilain 400 grams of an 8% solution? (10% = 10/100)

Answer 1

240g and 10% of sugar

Answer 2

240g of 10% salt solution and 160g of 5% salt solution

Explanation:

Simpler way:

Let X be the amount of 10% salt solution and Y be the amount of 5% salt solution.

Then X:Y = 3:2 as you need more 10% salt solution than 5% salt solution to create a 8% salt solution. The ratio 3:2 is given as 10 - 8 = 2 and 8-5 = 3.

If X:Y = 3:2 = 60:40, then 60% of the 400g of 8% salt solution must be the 10% solution and 40% of the 400g of 8% salt solution must be the 5% solution. Therefore, 240g of the 10% salt solution and 160g of the 5% salt solution is needed to create 400g of the 8% salt solution.

This is correct as the amount of salt in the solution is equal to (10% X 240g = 24g) + (5% of 160g = 8g). So the total amount (24+8) 32g should be equal to the amount of salt in the 400g of 8% salt solution.

8% of 400g = 32g which means our answer is correct.

More complex way:

A more complex way is to use algebra.

Let X be the amount of 10% salt solution and Y be the amount of 5% salt solution.

X + Y = 400g

and 10% of X + 5% of Y = 8% of 400

so 1/10 X + 1/20 Y = 32

2X + Y = 640 (multiply the previous equation by 20)

Now from the first line we know that X + Y = 400

So if we subtract (2X+Y) by (X+Y) we can get the value of X:

(2X+Y) - (X+Y) = 640 - 400

2X+Y - X - Y = 240

Therefore X = 240

And since X + Y = 400, Y = 160.