Problem 8
Question
If the price of a pound of coffee is \(\$ 3.20\) and the projected rate of inflation is \(5 \%\) per year, how much per pound should we expect coffee to cost in 5 years? Express your answer to the nearest cent. $4.08
Step-by-Step Solution
Verified Answer
The expected cost is \( \$4.08 \) per pound.
1Step 1: Understand the problem
We need to determine how much the price of coffee will be in 5 years, given a current price of \( \$3.20 \) per pound and an annual inflation rate of \( 5\% \).
2Step 2: Determine the formula to use
To calculate the future price of an item considering inflation, we use the formula: \( \text{future price} = \text{current price} \times (1 + \text{inflation rate})^n \), where \( n \) is the number of years into the future.
3Step 3: Insert known values into the formula
Here, the current price is \( 3.20 \), the inflation rate is \( 5\% \), which is equivalent to \( 0.05 \) as a decimal, and \( n = 5 \), so we get: \[ 3.20 \times (1 + 0.05)^5 \]
4Step 4: Calculate the future price
Evaluate the expression by calculating \( 1 + 0.05 = 1.05 \) and raising it to the power of 5: \( 1.05^5 \approx 1.2762815625 \). Then, multiply by \( 3.20 \):\[ 3.20 \times 1.2762815625 \approx 4.0841 \]
5Step 5: Round the result
Round the result to the nearest cent, which gives us \( \$4.08 \).
Key Concepts
Algebraic FormulasPercentage IncreaseFuture Value Calculations
Algebraic Formulas
Algebraic formulas are expressions composed of variables and constants, using operations like addition, subtraction, multiplication, and division. They are used to describe relationships between quantities and to calculate unknown values.
In the exercise, we used an algebraic formula to determine the future price of coffee considering inflation. The formula applied is:
By inserting given values into our formula, we can solve for the future cost, reflecting how algebraic manipulation is key to solving real-world problems like inflation adjustments.
In the exercise, we used an algebraic formula to determine the future price of coffee considering inflation. The formula applied is:
- \( \text{future price} = \text{current price} \times (1 + \text{inflation rate})^n \)
By inserting given values into our formula, we can solve for the future cost, reflecting how algebraic manipulation is key to solving real-world problems like inflation adjustments.
Percentage Increase
Percentage increase measures how much a quantity grows compared to its original size. It's commonly used to express growth, often in prices or populations. The inflation rate is a typical example of a percentage increase, signifying how prices are expected to rise over time.
In our scenario, inflation is projected to increase the coffee price by 5% annually. Understanding what this means is crucial:
In our scenario, inflation is projected to increase the coffee price by 5% annually. Understanding what this means is crucial:
- A 5% increase means for every dollar, an extra 5 cents is added each year.
- This rate is compounded over multiple years, adding to the previous year’s price.
Future Value Calculations
Future value calculations help determine the value of an investment or item at a future date, accounting for interest rates or inflation. This is particularly valuable in financial planning, allowing individuals to anticipate the cost of items over time.
The calculation used in the exercise predicts the future price of coffee after 5 years of inflation:
The calculation used in the exercise predicts the future price of coffee after 5 years of inflation:
- Current price: $3.20
- Inflation rate: 5%, or 0.05 in decimal form
- Time period: 5 years
Other exercises in this chapter
Problem 7
A tank contains 16,000 liters of water. Each day one-half of the water in the tank is removed and not replaced. How much water remains in the tank at the end of
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7\. Solve \(i=P r t\) for \(P\), given that \(r=9 \%, t=3\) years, and \(i=\$ 216\).
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8\. Solve \(i=P r t\) for \(P\), given that \(r=8 \frac{1}{2} \%, t=2\) years, and \(i=\$ 204\).
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A tank contains 5832 gallons of water. Each day onethird of the water in the tank is removed and not replaced. How much water remains in the tank at the end of
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