Question

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?

Answer 1

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

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

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

Answer 2

x= 44 ounces y = 66 ounces

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