Question

Serena is making an experiment. For that, she needs 20 grams of a 52% solution of salt. She has two large bottles of salt water: one with 40% and the other with 70% of salt in them. How much of each must she use to make the solution she needs?

I am in need of help

Answer 1

Answer:

Serena needs 12 grams of the 40% solution and 8 grams of the 70% solution.

Step-by-step explanation:

Let x be the weight of 40% salt solution and y be the weight of 70% salt solution.

We have been given that Serena needs 20 grams of solution of salt. We can represent this information in an equation as:

$x+y=20...(1)$

We are also told that she needs 20 grams of a 52% solution of salt by adding  40% and 70% of salt solutions.

We can represent this information in an equation as:

$\frac{40}{100}*x+\frac{70}{100}*y=20*\frac{52}{100}...(2)$

$0.40x+0.70y=20*0.52...(2)$

Now we will use substitution method to solve our system of equations.

From equation (1) we will get,

$x=20-y$

Substituting this value in equation (2) we will get,

$0.40(20-y)+0.70y=20*0.52$

$8-0.40y+0.70y=10.4$

$8-8-0.40y+0.70y=10.4-8$

$0.30y=2.4$

$\frac{0.30y}{0.30}=\frac{2.4}{0.30}$

$y=8$

Therefore, Serena must use 8 grams of 70% salt solution.

Upon substituting y=8 in equation (1) we will get,

$x+8=20$

$x+8-8=20-8$

$x=12$

Therefore, Serena must use 12 grams of 40% salt solution.