Translating verbal descriptions into algebraic expressions is an essential initial step in solving word problems. Some examples are given below.
Ex. Ellen has received the following scores on 3 exams: 82, 74, and 90. What score will Ellen need to receive on the next exam so that the average (arithmetic mean) score for the 4 exams will be 85?
Sol. Let x represent the score on Ellen's next exam. This initial step of assigning a variable to the quantity that is sought is an important beginning to solving the problem. Then in terms of , the average of the 4 exams is
Solving the resulting linear equation for x, you get
Therefore, Ellen will need to attain a score of 94 on the next exam.
Ex. A mixture of 12 ounces of vinegar and oil is 40 percent vinegar, where all of the measurements are by weight. How many ounces of oil must be added to the mixture to produce a new mixture that is only 25 percent vinegar?
Sol. Let represent the number of ounces of oil to be added. Then the total number of ounces of the new mixture will be , and the total number of ounces of vinegar in the new mixture will be . Since the new mixture must be 25 percent vinegar,
Therefore,
Thus, 7.2 ounces of oil must be added to produce a new mixture that is 25 percent vinegar.
Ex. To produce a particular radio model, it costs a manufacturer $30 per radio, and it is assumed that if 500 radios are produced, all of them will be sold. What must be the selling price per radio to ensure that the profit (revenue from the sales minus the total production cost) on the 500 radios is greater than $8,200?
Sol. If y represents the selling price per radio, then the profit is 500(y – 30). Therefore, we set
Solving the inequality, we get
Thus, the selling price must be greater than $46.40 to ensure that the profit is greater than $8,200.
In the case of compound interest, interest is added to the principal at regular time intervals, such as annually, quarterly, and monthly. Each time interest is added to the principal, the interest is said to be compounded. After each compounding, interest is earned on the new principal, which is the sum of the preceding principal and the interest just added. If the amount P is invested at an annual interest rate of r percent, compounded annually, then the value V of the investment at the end of t years is given by the formula
If the amount P is invested at an annual interest rate of r percent, compounded n times per year, then the value V of the investment at the end of t years is given by the formula
Ex. If $10,000 is invested at a simple annual interest rate of 6 percent, what is the value of the investment after half a year?
Sol. According to the formula for simple interest, the value of the investment after 1 year is
Ex. If an amount P is to be invested at an annual interest rate of 3.5 percent, compounded annually, what should be the value of P so that the value of the investment is $1,000 at the end of 3 years?
Sol. According to the formula for 3.5 percent annual interest, compounded annually, the value of the investment after 3 years is
And we set it equal to $1,000
To find the value of P, we divide both sides of the equation by
Thus, an amount of approximately $901.94 should be invested.
Ex. A college student expects to earn at least $1,000 in interest on an initial investment of $20,000. If the money is invested for one year at interest compounded quarterly, what is the least annual interest rate that would achieve the goal?
Sol. According to the formula for r percent annual interest, compounded quarterly, the value of the investment after 1 year is
By setting this value greater than or equal to $21,000 and solving for r, we get
We can use the fact that taking the positive fourth root of each side of an inequality preserves the direction of the inequality. This is also true for the positive square root or any other positive root.
To compute the fourth root, we can use the fact that for ; that is, we can compute a fourth root by taking a square root twice:
So the least annual interest rate is approximately 4.91 percent.