Solution:Let y be the amount of salt added.Amount of saltPercentconcentrationResult Original 2020%=0.200.2x20Addedy100%= 11xyFinal Solution20 + y25% = 0.250.25(20+y)

Original + Added= Result(0.2x20)+(1xy)= 0.25(20+y)4+y= 5+0.25yY-0.25y= 5-40.75y=10.75y = 1y= 4/3 or 1 1/3 Answer: 1 1/3 of salt should be added.Problem 4( Removing from solution)4.John has 20 ounces of a 20% salt solution. How much water should beevaporated to make it a 30% solution?Solution:20 % of salt solution = 80 % of Water30 % of salt solution = 70 % of WaterLet y be the amount of water evaporated.Amount of saltWaterResult Original 2080%=0.8020x0.80Addedy100%= 11xyFinal Solution20 - y70% = 0.700.70(20-y)Original-Removed= Result(0.8x20)-(1xy)= 0.70(20-y)16-y=14-0.7yY-0.7y=16-140.3y=2Y=2/0.3~ 6.67Answer: 6.67 ounces of water should be evaporated.Problem 5(Replace Solutions)5.A tank has a capacity of 10 galloons. When it is full, it contains 15% alcohol.How many galloons must be replaced by an 80% alcohol solution to give 10galloons of 70% solution?Solution:Let y be the amount of alcohol replaced.

Amount of saltWaterResult Original 1015%=0.150.15x10Removedy15%=0.150.15yAddedy80%=0.80.8yFinal Solution1080%=0.70.7x10Exercises:1.How many gallons of a 50% salt solution must be mixed with 60 gallonsof a 15% solution to obtain a solution that is 40% salt?2.A chemist has one solution that is 60% acid and another that is 30% acid.How much of each solution is needed to make a 750 mL solution that is50% acid?3.How many pounds of coffee beans selling for $2.20 per pound should bemixed with 2 pounds of coffee beans selling for $1.40 per pound to obtaina mixture selling for $2.04 per pound?4.John wants to make a 100 ml of 5% alcohol solution mixing a quantity of a2% alcohol solution with a 7% alcohol solution. What are the quantities ofeach of the two solutions (2% and 7%) he has to use?5.Sterling Silver is 92.5% pure silver. How many grams of Sterling Silvermust be mixed to a 90% Silver alloy to obtain a 500g of a 91% Silveralloy?AGE PROBLEMS-are algebra wordproblemsthat deal with theagesof people currently, in thepast or in the future. If theprobleminvolves a single person, then it is similarto an Integerproblem.Examples:Problem 11.A mother is ten times as old as her daughter; in 24 years, she will be twiceas old as her daughter. Find their present age.

Solution:Let x = the daughter’s present age10x = the mother’s present age. If the mother will be twice as old as her daughter in 24 years, the equationis 2 times the daughter’s age in 24 years = the mother’s age in 24 years or2(x + 24) = 10x + 242(x + 24) = 10x + 242x + 48 = 10x + 242x – 10x + 48 =24-8x + 48 = 24-8x= 24 – 48-8x = -24-8x = -24-8-8x = 3 (daughter’s age)10x = 10 ( 3 ) = 30 (mother’s age)Problem 22.Bill is 8 years older than his brother. In 3 years, bill will be twice as old as hisbrother. Find their present ages.Solution:Let x = Bill’s brother’s agex + 8 = Bill’s age. In 3 years, their ages will be x + 3= Bill’s brother’s age (x + 8) + 3 = Bill’s age. Now in 3 years, Bill will be twiceas old. This means theequation is 2 times Bill’s brother’s age in 3 years = Bill’s age in 3 years or2(x + 3) = (x + 8) + 3.

#### You've reached the end of your free preview.

Want to read all 23 pages?

- Fall '19