MATH SOLVE

5 months ago

Q:
# A scientist has two solutions, which she has labeled Solution A and Solution b. Each contains salt. She knows that Solution A is 55% salt and Solution B 80% is salt. She wants to obtain 110 ounces of a mixture that is 75% salt. How many ounces of each solution should she use?

Accepted Solution

A:

x oz -solution A , 55% salt

y oz -solution B, 80 % salt

x+y=110

0.55x -oz salt in A solution

0.8y -oz salt in B solution

0.75*110 =82.5 oz salt in 75% solution

0.55x +0.8y=82.5

System of equations:

x+y=110 /*-0.55 ----> -0.55x-0.55y=110*(-0.55)

0.55x +0.8y=82.5

-0.55x-0.55y=-60.5, add both equations

0.55x +0.8y=82.5

-0.55x-0.55y+0.55x +0.8y=-60.5+82.5

-0.55y+0.8y= 22

0.25y=22

y=88 oz -solution B, 80 % salt

x+y=110

x=110-88=22 oz -solution A , 55% salt

88 oz -solution B, 80 % salt and 22 oz -solution A , 55% salt

y oz -solution B, 80 % salt

x+y=110

0.55x -oz salt in A solution

0.8y -oz salt in B solution

0.75*110 =82.5 oz salt in 75% solution

0.55x +0.8y=82.5

System of equations:

x+y=110 /*-0.55 ----> -0.55x-0.55y=110*(-0.55)

0.55x +0.8y=82.5

-0.55x-0.55y=-60.5, add both equations

0.55x +0.8y=82.5

-0.55x-0.55y+0.55x +0.8y=-60.5+82.5

-0.55y+0.8y= 22

0.25y=22

y=88 oz -solution B, 80 % salt

x+y=110

x=110-88=22 oz -solution A , 55% salt

88 oz -solution B, 80 % salt and 22 oz -solution A , 55% salt