Question
What is the simple interest of a loan for $1,000 with 5 percent interest after 3 years?
$1,150
$1,050
$150
$50
$1,150
$1,050
$150
$50
Answers
GPT-4o mini
To calculate the simple interest on a loan, you can use the formula:
\[ \text{Simple Interest} = P \times r \times t \]
where:
- \( P \) is the principal amount (the initial loan amount),
- \( r \) is the annual interest rate (as a decimal),
- \( t \) is the time in years.
Given:
- \( P = 1000 \) (the amount of the loan),
- \( r = 0.05 \) (5 percent interest as a decimal),
- \( t = 3 \) (the number of years).
Now, substituting the values into the formula:
\[ \text{Simple Interest} = 1000 \times 0.05 \times 3 \]
\[ \text{Simple Interest} = 1000 \times 0.15 \]
\[ \text{Simple Interest} = 150 \]
Now, to find the total amount to be paid back after 3 years:
\[ \text{Total Amount} = P + \text{Simple Interest} \]
\[ \text{Total Amount} = 1000 + 150 = 1150 \]
The simple interest after 3 years is **$150**. So the options are:
- $1,150 (Total Amount)
- $1,050 (Principal + Part of Interest)
- **$150 (Correct Answer)**
- $50 (Incorrect)
Therefore, the correct answer is **$150**.
\[ \text{Simple Interest} = P \times r \times t \]
where:
- \( P \) is the principal amount (the initial loan amount),
- \( r \) is the annual interest rate (as a decimal),
- \( t \) is the time in years.
Given:
- \( P = 1000 \) (the amount of the loan),
- \( r = 0.05 \) (5 percent interest as a decimal),
- \( t = 3 \) (the number of years).
Now, substituting the values into the formula:
\[ \text{Simple Interest} = 1000 \times 0.05 \times 3 \]
\[ \text{Simple Interest} = 1000 \times 0.15 \]
\[ \text{Simple Interest} = 150 \]
Now, to find the total amount to be paid back after 3 years:
\[ \text{Total Amount} = P + \text{Simple Interest} \]
\[ \text{Total Amount} = 1000 + 150 = 1150 \]
The simple interest after 3 years is **$150**. So the options are:
- $1,150 (Total Amount)
- $1,050 (Principal + Part of Interest)
- **$150 (Correct Answer)**
- $50 (Incorrect)
Therefore, the correct answer is **$150**.