Presentation on theme: "SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS Investment Problems Example 1 An investment counselor invested 75% of a client’s money into a 9% annual simple."— Presentation transcript:
SECTION 2-3 APPLICATIONS OF LINEAR EQUATIONS Investment Problems Example 1 An investment counselor invested 75% of a client’s money into a 9% annual simple interest money market fund. The remainder was invested in 6% annual simple interest government securities. Find the amount invested in each if the total annual interest earned is $3300. Annual Interest is the interest earned EACH YEAR (so t=1). Step 1) What are we being asked to find? The amount invested in each account. Let x be the total amount invested, then 75% of x is the amount invested in the Money Market Fund and 25% of x is the amount invested in the government securities. Given info: Step 2) Form an equation to solve for the unknown quantity..75x(.09) +.25x(.06) = $3300 Step 3) Solve the equation:.0675x +.015x = $3300.0825x = $3300 x = $40,000 Step 4) Check the result..75($40,000)(.09) +.25($40,000)(.06) = $3300? Step 5) State the conclusion. We were asked to find amt in each account. 75% of x is the amount invested in the Money Market Fund =.75($40,000) = $30,000 25% of x is the amount invested in the government securities =.25($40,000) = $10,000 $30,000 invested in Money Market Fund and $10,000 invested in gov. securities. PrtI Money Market Fund.75x.091.75x(.09) Gov. securities.25x.061.25x(.06) TOTAL$3300 This column can be deleted since t just equals 1.
You try this one: An investment of $2500 is made at an annual simple interest rate of 7%. How much additional money must be invested at 10% so that the total interest earned will be 9% of the total investment? Step 1) What are we being asked to find? The additional money to be invested at 10% to earn total interest that is 9% of the total investment. Let x = additional money to be invested. Given info: : Original investment = $2500 annual simple interest rate = 7% x = additional investment 10% = new rate New Interest = 9% of TOTAL investment TOTAL investment = 2500 + x PrI Original Investment at 7% 2500.07.07(2500)= 175 Additional Investment at 10% x.10.10x TOTAL2500 + x.09.09(2500+x) Step 2: Form an equation Step 3: Solve the equation 175 +.10x = 225 +.09x.01x = 50 x = 5000 Step 4: Check the result 175 +.10(5000) =.09(2500 + 5000? Yes Step 5: State the conclusion $5000 should be invested at 10%_ 175 +.10x =.09(2500 + x) d d d d
Value Mixture Problems Strategy: Make a table with a row for each ingredient in the mixture. These ingredients are listed in the first column. For each ingredient, write a numerical or variable expression for the amount of the ingredient used, the unit cost of the bend, and the value of the amount used. The last row of the table will be for the resulting blend (or mixture) of the ingredients. In the last row you ADD the values in the previous rows, EXCEPT for the UNIT COST. The new unit cost will be GIVEN to you or it will be the quantity you are solving for. DO NOT ADD UNIT COSTS. You then use the equation, Amount*Unit Cost = Value to solve for the unknown quantity Amount = how much is used (total ounces, pounds, etc. of each ingredient. Unit Cost = cost per unit of the amount (e.g. cost per ounce $/oz, cost per pound, $/lb) Value = amount in dollars of the total amount Amount * Unit Cost = Value Example: 10 oz * $8/oz = $80
Example: How many ounces of silver alloy that costs $6 per ounce must be mixed with 10 oz of a silver alloy that costs $8 per ounce to make a mixture that costs $6.50 per ounce? What are we being asked to find? How many ounces (AMOUNT) of the $6 alloy. Let x = amount in ounces of $6 alloy AMOUNT (oz)UNIT COST ($/oz)VALUE($) $6 per oz Alloyx66x $8 per oz Alloy10880 $6.50 per Alloy (mixture) x + 106.50$6x + $80 Equation to solve for x: (x + 10)($6.50) = $6x + $80 $6.50x + $65 = $6x + $80 $6.50x - $6x = $80 - $65 $0.50x = $15 x = 30 Check: Does (30 + 10)(6.50) = 6(30) + 80? (40)(6.50) = 180 + 80 260 = 260 Yes State Conclusion: What does x represent? The number of ounces of $6 alloy CONCLUSION: 30 OUNCES NEED TO BE ADDED. dd
PERCENT MIXTURE PROBLEMS The strategy for these problems is virtually the same as for the value mixture problems, but you use a different equation. Recall the Percent Triangle: Amount = Percent * Base In these problems, the amount of each ingredient will actually be the “Base” in our equation. The Amount will be the Quantity of substance in each ingredient. So the new formula is Quantity = Percent * Amount Example A chemist wishes to make 3 L of a 7% acid solution by mixing a 9% acid solution and a 4% acid solution. How many liters of each solution should the chemist use? Step 1: What are we being asked to find? How many liters (AMOUNT) of each solution that the chemist should use to make 3L (amount of mixture) of a 7% (percent acid of mixture) acid solution. Given: The mixture will contain 3L of 7% acid. One ingredient is 9% acid and the other is 4% acid. If the total AMOUNT is 3L and the amount of the ingredients are unknown, we’ll use variables. Let x= amount of 9% acid solution. Since the total amount is 3L 3 – x = amount of 4% solution. Amt (QUANTITY Of SUBSTANCE) Base (AMOUNT of INGREDIENT) Percent
PERCENT (changed to decimal) AMOUNT (L)QUANTITY (L) 9% acid.09x.09x 4% acid.043 – x.04(3-x) 7 % acid (mixture).073.07(3) Notice in this table that if we use the last row for our equation, there is no x to solve for:.07(3) =.07(3) However, if we look at the last column on the left, we can make an equation. We can add the quantities together to get the quantity of acid in the final mixture..09x +.04(3-x) =.07(3).09x +.12 -.04x =.21.09x -.04x =.21 -.12.05x =.09 x =.09/.05 x = 1.8 Check: Does.09(1.8) +.04(3-1.8) =.07(3) ?.162 +.04(1.2) =.21.162 +.048 =.21.21 =.21 YES State the Conclusion. What were we being asked to find? The amount of each acid solution, 9% and 4%. x = amount of 9% acid solution in liters = 1.8 L 3 – x = amount of 4% acid solution in liters = 3 – 1.8 = 1.2 L CONCLUSION: MIX 1.8 L OF 9% ACID SOLUTION AND 1.2 L OF 4% ACID SOLUTION d d d d d d d d d dd
Distance TimeRate (speed) CH. 2.3 UNIFORM MOTION PROBLEMS Just as we did not add unit costs or percentages in Mixture problems, we also DO NOT ADD RATES in uniform motion problems. The only time we add rates is when an object’s rate is affected by another force such as the wind or a current. Also, we do not add TIMES if the objects traveled at the same time, but we do if they traveled at different times. Remember, it sometimes helps to draw pictures with these problems. Example 1 p.223: Two cars, the first traveling 10 mph FASTER THAN the second, start at the same time from the same point and travel in opposite directions. In 3 hours they are 288 miles apart. Find the rate of the second car. Step 1) Analyze the problem. What are we asked to find? The rate of the second car. If we let x= rate of the second car, and the first car is 10 mph faster than the second car, then x + 10 will be the rate of the first car. Draw a picture: Step 2) Form an equation. We know d = rt. Total distance will be Car 1 distance + Car 2 distance, since they went in opposite directions from the same point, and we are given that total distance is 288 miles. Equation to solve for Car 2’s rate: 3(x+10) + 3x = 288 3x + 30 + 3x = 288 6x = 258 x = 43 CONCLUSION: x= Car 1’s RATE (mph). FIRST CAR TRAVELED AT 43 MPH 10 + x mphx mph Total Distance = 288 miles Time =3 hours RateTimeDistance Car 1x + 1033(x + 10) Car 2x33x TOTAL288
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