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 24m
- 2. Linear Equations and Inequalities3h 42m
- 3. Solving Word Problems2h 48m
- 4. Graphing4h 42m
- 5. Systems of Linear Equations2h 6m
- 6. Exponents and Polynomials3h 25m
- 7. Factoring2h 36m
- 8. Rational Expressions and Equations3h 51m
- Simplifying Rational Expressions39m
- Multiplying and Dividing Rational Expressions25m
- Adding and Subtracting Rational Expressions with Common Denominators24m
- Least Common Denominators32m
- Adding and Subtracting Rational Expressions with Different Denominators39m
- Rational Equations44m
- Direct & Inverse Variation27m
- 9. Roots and Radicals2h 46m
- 10. Quadratic Equations3h 2m
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, \(P\) is the principal (amount borrowed), \(r\) is the annual interest rate (in decimal), and \(t\) is the time in years.
From the problem, note the values: interest \(I = 1350\), time \(t = 3\) years, and interest rate \(r = 9\% = 0.09\).
Substitute the known values into the formula: \(1350 = P \times 0.09 \times 3\).
Simplify the right side: \(1350 = P \times 0.27\).
Solve for \(P\) by dividing both sides by \(0.27\): \(P = \frac{1350}{0.27}\).
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