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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 :

Answer:

[tex] A +B = 40[/tex]  (1)

[tex] A*0.65 + B*0.90 = 40*0.85[/tex]  (2)

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

[tex] A = 40 -B[/tex]   (3)

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

[tex] (40-B)*0.65 +0.9 B =34[/tex]

And solving the last equation for B we got:

[tex] 26 -0.65 B +0.9 B= 34[/tex]

[tex] B = \frac{8}{0.25}= 32[/tex]

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

[tex] A = 40-32 = 8[/tex]

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:

[tex] A +B = 40[/tex]  (1)

[tex] A*0.65 + B*0.90 = 40*0.85[/tex]  (2)

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

[tex] A = 40 -B[/tex]   (3)

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

[tex] (40-B)*0.65 +0.9 B =34[/tex]

And solving the last equation for B we got:

[tex] 26 -0.65 B +0.9 B= 34[/tex]

We subtract in both sides 36:

[tex] 0.25 B =8[/tex]

And dividing both sides by 0.25 we got:

[tex] B = \frac{8}{0.25}= 32[/tex]

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

[tex] A = 40-32 = 8[/tex]

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

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