In a city, the water company increased the monthly water bill by . The initial monthly cost is . What is the cost after the increase?
Table of contents
- 1. Review of Real Numbers2h 39m
- 2. Linear Equations and Inequalities3h 38m
- 3. Solving Word Problems2h 43m
- 4. Graphing Linear Equations in Two Variables3h 17m
- 5. Systems of Linear Equations1h 43m
- 6. Exponents and Polynomials1h 27m
- 7. Factoring2h 42m
- 8. Rational Expressions and Equations2h 18m
- 9. Inequalities and Absolute Value2h 52m
- 10. Relations and Functions1h 10m
- 11. Roots, Radicals, and Complex Numbers2h 33m
- 12. Quadratic Equations and Functions1h 23m
- 13. Inverse, Exponential, & Logarithmic Functions1h 5m
- 14. Conic Sections & Systems of Nonlinear Equations58m
- 15. Sequences, Series, and the Binomial Theorem1h 20m
3. Solving Word Problems
Percent Problem Solving
Multiple Choice
Sofia plans to buy a car. Her loan statement indicates she will pay in interest for a -year loan at simple interest per year. How much did Sofia borrow for the car?
A
\$5,000
B
\$4,500
C
\$15,000
D
\$6,350
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Verified step by step guidance1
Identify the formula for simple interest: \(I = P \times r \times t\), where \(I\) is the interest paid, \(P\) is the principal (amount borrowed), \(r\) is the annual interest rate (in decimal form), and \(t\) is the time in years.
From the problem, note the values: interest \(I = 1350\), time \(t = 3\) years, and interest rate \(r = 9\% = 0.09\) (convert percent to decimal by dividing by 100).
Substitute the known values into the simple interest formula: \(1350 = P \times 0.09 \times 3\).
Simplify the right side by multiplying the rate and time: \(1350 = P \times 0.27\).
Solve for \(P\) by dividing both sides of the equation by \$0.27$: \(P = \frac{1350}{0.27}\). This will give the amount Sofia borrowed.
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