Question

A scientist has two solutions, which she had labeled solution A and solution B. each contains salt. she knows that solution A is 65% salt and Solution B is 90% salt. She wants to obtain 40 ounces of a mixture that is 85% salt. How many ounces of each solution should she use?

Answer 1

Answer:

$A +B = 40$  (1)

$A*0.65 + B*0.90 = 40*0.85$  (2)

From equation (1) we can solve for A and we got:

$A = 40 -B$   (3)

And replacing equation (3) into equation (2) we got:

$(40-B)*0.65 +0.9 B =34$

And solving the last equation for B we got:

$26 -0.65 B +0.9 B= 34$

$B = \frac{8}{0.25}= 32$

And solving for A from equation (3) we got:

$A = 40-32 = 8$

So then we need 8 ounces of solution A with concentration of 65% of salt and 32 ounces of solution B with 90% of salt

Step-by-step explanation:

Let A represent the amount of solution A and B the amount of solution B. We know that the concentration of A is 65% and the concentration for B is 90%.

We want to obtain a solution of 40 ounces with a concentration of 85% of salt.

Based on this we can set up the following equations:

$A +B = 40$  (1)

$A*0.65 + B*0.90 = 40*0.85$  (2)

From equation (1) we can solve for A and we got:

$A = 40 -B$   (3)

And replacing equation (3) into equation (2) we got:

$(40-B)*0.65 +0.9 B =34$

And solving the last equation for B we got:

$26 -0.65 B +0.9 B= 34$

We subtract in both sides 36:

$0.25 B =8$

And dividing both sides by 0.25 we got:

$B = \frac{8}{0.25}= 32$

And solving for A from equation (3) we got:

$A = 40-32 = 8$

So then we need 8 ounces of solution A with concentration of 65% of salt and 32 ounces of solution B with 90% of salt