SOLUTION: 1. A rectangular garden has dimensions of 18 feet by 13 feet. A gravel path of uniform width is to be built around the garden. How wide can the path be if there is enough gravel fo

Question 172102: 1. A rectangular garden has dimensions of 18 feet by 13 feet. A gravel path of uniform width is to be built around the garden. How wide can the path be if there is enough gravel for 516 square feet?
2. A business invests $10,000 in a savings account for two years. At the beginning of the second year, an additional $3500 is invested. At the end of the second year, the account balance is $15,569.75. What was the annual interest rate?
3. Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
4. The Hudson River flows at a rate of 3 miles per hour. A patrol boat travels 60 miles upriver, and returns in a total time of 9 hours. What is the speed of the boat in still water?

Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
#1: Let the width of the path be , then the overall dimensions of the garden plus the path must be and because there is width of path on each of the four sides. Since the area of the garden by itself is , the overall area of the garden plus the path must be

But the overall area must also be , so we can write:

Once you multiply the binomials and put the constant term on the left, you will have a quadratic equation in x, which is the value we want to find. You will get two roots to this equation, one of which will be negative. You can exclude the negative root as extraneous introduced by the process of squaring the variable (negative dimensions are absurd anyway). The remain root will be your answer.

2. The IRR function in Excel with the arguments 10,000, 3,500, and -15569.75 produces a result of 8.5% Checking this result:

10,000 * 1.085 = 10,850
(10,850 + 3,500) * 1.085 = 15,569.75 Answer checks

IRR is an iterative calculation based on a collection of pairs of cash flows and time. Use the Net Present Value formula set to zero.

In this case:

and solve for r.

3. Distance = 200 miles. where is the distance, is the speed for the given situation, and is the time for the given situation, hence . In the "what if" situation, the speed is and the time is , hence .

Solve both equations for

and

Since ,

Applying the lowest common denominator of we get:

Simplifying and setting the numerators equal:

Collecting like terms and putting in standard form:

Note that and , so the quadratic factors:

Hence or . Since, presumably, he wasn't going backwards at any time, the -50 root is extraneous. Therefore the speed of the given trip was 40 mph.

Check:
200/40 = 5 hours for the trip

200/50 = 4 hours for the 'what if' trip

50 is 10 more than 40 and 4 is 1 less than 5, Answer checks.

4. This is really the same problem as the previous one. The speed of the current subtracts from the speed in still water on the upstream trip and adds on the downstream trip. Call the speed in still water , and the time to go upstream . Since the time to go upstream plus the time to go downstream is 9 hours, the time to go downstream can be expressed as .

So:

and

Solve both equations for and set the resulting right-hand expressions equal to each other. When you simplify by using the LCD, collecting like terms, and putting it in standard form, you will have a quadratic in . Solve by factoring or using the quadratic formula as is appropriate. You will most likely have an extraneous root because of squaring the variable in the process of setting up the equation. The positive root will be the value for that answers the question.
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