Question 218873: A certain solution of salt water is 15% salt and weighs 35 pounds. How much salt must be added to produce a solution that is 25% salt?
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! Let x = number of pounds of pure salt that must be added.
35 = number of pounds of solution at the beginning
.15*35 = number of pounds of salt at the beginning
x+35 = total number of pounds of solution at the end
.25(x+35) = total number of pounds of salt at the end
The equation is based upon the fact that the amount of salt in the solution + the amount of salt added = total amount of salt in the solution.
.15(35) + 1.00*x = .25(x+35)
5.25+ 1.00x = .25x + 8.75
Subtract .25x from each side:
5.25 +.75x = 8.75
Subtract 5.25 from each side:
.75x = 8.75-5.25=3.50
Divide both sides by .75:
x= 3.50/.75
x= 350/75= 14/3 = 4 2/3 = 4.66666. . . pounds of salt
I know this is an ugly answer, and the check is even uglier! I did check this, and it seems to work, although I will spare the details!!
For additional explanation on this topic, please go to my website. Do a "Bing" search for my last name "Rapalje". Look for "Rapalje Homepage" at the top of this search. Look near the top of my Homepage for "Basic, Intermediate and College Algebra: One Step at a Time", click on "Intermediate Algebra" and look in Chapter 1 for Section 1.05 Applications.
R^2
Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Altamonte Springs Campus
|
|
|
