A scientist has two solutions, which she has labeled Solution A and Solution B. Each contains salt. She knows that Solution A is 70%salt and Solution B is 95% salt. She wants to obtain 110 ounces of a mixture that is 80% salt. How many ounces of each solution should she use?

x= 44 ounces y = 66 ounces  

Step-by-step explanation:

Distribution would be an increase to find 80% salt.

Where we use 4/5 = 80 but use here 4/5.

We find 19/20 but just say 95 for solution B =

Let x = the number of ounces of Solution A

Let y = the number of ounces of Solution B

x + y = 110        y = 110 - x

(70x + 95y) = 4/5 x 110

70x + 95y = 88    

70x + 95 y = 88

70x + 95 (88 - x) = 8800 make units equal or for start convert decimalization = same units as 2nd multiplied value.

70x + 8360 - 88x = 8800

-10x - 8360 - 8800 = y

-10x = 440

x = 44 ounces

y = 110- 44

y =  66 ounces  

66 ounces 70% salt solution and of 44 ounces of 95% salt solution

Step-by-step explanation:

Let x = volume needed of 70% salt solution.

Let y = volume needed of 95% salt solution.

Total needed: 110 ounces of 80% salt solution.

Equation of total volumes:

x + y = 110    Equation 1

Equation of percentages of salt:

0.7x + 0.95y = 0.8(110)

0.7x + 0.95y = 88     Equation 2

Multiply both sides of equation 1 by -0.7; write equation 2 below it and add the equations.

       -0.7x - 0.7y = -77

(+)   0.7x + 0.95y = 88

                  0.25y = 11

y = 44

x + y = 110

x + 44 = 110

x = 66

66 ounces 70% salt solution and of 44 ounces of 95% salt solution

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00