Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
M098 Carson Elementary and Intermediate Algebra 3e Section 3.5 Objectives 1. 2. Use a table to solve problems involving two investments. Use a table to solve problems involving mixtures. Prior Knowledge Converting from a percent to a decimal: To convert from a percent to a decimal, move the decimal point two places to the left. Clearing decimals: To clear decimals in an equation, multiply each term by an appropriate power of 10 (e.g. 10, 100, 1000, etc.) Vocabulary Principal APR Concentration An amount of money invested into (or borrowed from) an account Annual Percentage Rate Amount of a substance per unit of volume New Concepts Use a table to solve problems involving two investments When solving problems involving investments, it is helpful to remember the formula for simple interest: I = Prt In this formula, “I” is the interest earned on account where a principal “P” is invested at a given rate “r” for a certain time (in years) “t.” Example 1 You invest $5000 into an account that earns a 4% APR. How much interest will you earn after 1 year? First, identify the key information about the principal, rate and time, then use the formula: Notice, the interest rate of 4% was converted to a decimal P 5000 I Pr t r 0.04 I (5000)(0.04)(1) t 1 I 200 This means that after 1 year, the account earns $200 in interest. In this section, we will be dealing with interest earned between two accounts for one year. Since the “t” will always be 1, our formula for interest will be just I = Pr We can use a table to organize the information like we did when we solved problems involving distances. When solving problems involving investments: Identify the key words indicating the principal, interest rate and total interest earned. Organize information in a table. Write the equation and solve. Example 2 Suppose you invested $8000 into two separate accounts. The APR for the first account was 4% and the APR for the second account was 8%. After one year the total amount of interest earned was $600. How much did you invest in each account? C.Stewart 2011 page 1 M098 Carson Elementary and Intermediate Algebra 3e Section 3.5 Before we fill in the table, let’s break this down into its components: Principal – the total principal invested into the two accounts is $8000. To figure out how much is invested into the two accounts, look at the following table: Amount Invested First Account 1000 1500 3700 p Now, we can fill in the table: Accounts Interest Rate First Account 0.04 Second Account 0.08 (totals) Amount Invested Second Account 8000 – 1000 = 7000 8000 – 1500 = 6500 8000 – 3700 = 4300 8000 – p Principal p 8000 - p In general terms Since we do not know the interest for each account, we will use the formula R•P = I Interest 0.04p 0.08(8000 – p) 600 0.04 p 0.08(8000 p) 600 In general, you will need to add the interest from the first account (I1) and the interest from the second account (I2) to find the total interest (It). Equation to solve: I1 + I2 = It Clear decimals as necessary. (100)[0.04 p 0.08(8000 p)] (100)600 4 p 8(8000 p) 60000 4 p 64000 8 p 60000 4 p 64000 60000 4 p 4000 4 p 4000 4 4 p 1000 Looking back at the table, the answer p = 1000 represents the amount invested into the first account, therefore the amount invested into the second account is 8000 – p or $7000. Example 3 Marvin invests $12,000 in two plans. Plan 1 is at an APR of 6% and Plan 2 is at an APR of 9%. If the total interest earned after one year is $828, what principal was invested in each plan? Accounts Plan 1 Plan 2 (totals) Interest Rate 0.06 0.09 Principal p 12000 - p 0.06 p 0.09(12000 p) 828 6 p 9(12000 p) 82800 Equation to solve: I1+I2 = It Interest 0.06p 0.09(12000 – p) 828 Clear decimals by multiplying by 100 6 p 72000 9 p 82800 3p 7200 82800 3p 10800 p 3600 Therefore, $3600 was invested into the first plan and $8400 was invested into the second account. C.Stewart 2011 page 2 M098 Carson Elementary and Intermediate Algebra 3e Section 3.5 Use a table to solve problems involving mixtures Sometimes, it is necessary to create a mixture. The process for solving problems involving mixtures is the same as solving investment problems; the only difference is the terminology: Investments: Rate x Principal = Total Interest (PR = I) Chemical Mixtures: Concentration of Chemical x Volume of Solution = Total Amount of Chemical (CV = A) Other Mixtures: Price x Quantity = Total Cost Notice how similar all of these formulas are. IN GENERAL: Value x Quantity = Total Example 4 Suppose you buy a 250 mL bottle of hydrogen peroxide at your local pharmacy with a concentration of 3%. How much of that bottle is pure hydrogen peroxide? First, identify the key information about the concentration and volume, then use the formula: CV = A The concentration of 3% was converted to a decimal. V 250 250(0.03) A 7.5 A C 0.03 This means that there are 7.5 mL of pure hydrogen peroxide in a 250 mL bottle. Again, the process for solving a mixture problem is much like solving an investment problem. You can still use a table to organize your information. The main difference is that usually, two solutions are mixed together to achieve a third solution. Example 5 Ken has 80 milliliters of 15% acid solution. How much of a 20% acid solution must be added to create a solution that is 18% acid? What will the total volume of the mixture be? Solutions First Solution Second Solution Mixture Concentration 0.15 0.20 0.18 Volume of Sol 80 x x + 80 Amount of Sol 0.15(80) 0.20x 0.18(x + 80) Since we do not know the amount for each solution, we will use the formula C•V = A The problem tells us that he is using 80 mL of the first solution but asks “how much” of the second solution, so we will call that “x.” Since we are adding these two solutions together to get the third solution we will call the volume of the mixture “x + 80.” In general, you will need to add the amount from the first solution and the amount from the second solution to find the total amount. Equation to solve: A1 + A2 = At Clear decimals as necessary. 0.15(80) 0.20 x 0.18(x 80) 15(80) 20 x 18(x 80) 1200 20 x 18 x 1440 2 x 240 x 120 Therefore, he must use 120 mL of the second 20% solution. The total amount of the new mixture is “x + 80” or 200 mL. C.Stewart 2011 page 3 M098 Carson Elementary and Intermediate Algebra 3e Section 3.5 Example 6 The Candy Shoppe wants to mix 115 pounds of candy to sell for $0.80 per pound. How many pounds of $0.60 candy must be mixed with a candy costing $1.20 per pound to make the desired mix? It’s helpful to keep the mixture on the bottom row. Types $0.60 candy $1.20 candy $0.80 Mixture Price 0.60 → 0.6 1.20 → 1.2 0.80 → 0.8 0.6 x 1.2(115 x) 0.8(115) 6 x 12(115 x) 8(115) 6 x 1380 12 x 920 Quantity x 115 - x 115 Total Cost 0.6x 1.2(115 – x) 0.8(115) Finding the quantity here is much like example 2 6 x 1380 920 6 x 460 x 76.6 6 115 x 38.34 This means that 76.66 pounds of the $0.60 candy should be mixed with 38.34 pounds of the $1.20 candy to make 115 pound mixture that sells for $0.80 per pound. C.Stewart 2011 page 4