How to Solve Simple Interest Problems Part 3

This is the third and last part of the Solving Simple Interest Problems Series. In the first and second part of this series, we have learned how to solve simple interest problems. In this post, we continue with two more problems, one with the rate missing and the other one with the principal missing.

Example 5

Mr. Wong deposited $15,000 in a bank for a certain rate of interest per year. After two years, the interest was $900. What was the rate of interest in percent?

Solution and Explanation

We have the following given:

Principal (P) = $15,000
Interest (I) = 900
Time = 2 (years)
Rate (R) = ?

As we have learned in the previous posts, simple interest is the product of the principal, the rate, and time or

I = PRT.

Substituting the given we have

900 = (15000)(R)(2)
900 = 30000R

Dividing both sides by 30000, we have

R = 900/30000
R = 0.03

The rate is 0.03 which is we need to convert to percent by multiplying it by 100. Therefore, the rate is 3%.

Example 6

Mrs. Lansangan invested a certain amount of money in a bank that gives 4% interest per year. She got an interest of Php2400 after 3 years.

Solution and Explanation

I = 2400
R = 4%
T = 3 years

Using the simple interest formula mentioned above, we have

2400 = (P)(4%)(3)

Converting 4 percent to decimal, we have

2400 = (P)(0.04)(3)
2400 = 0.12P

Dividing both sides by 0.12, we have

20000 = P

Therefore, Mrs. Lansangan invested Php20000.

That’s it for our series on how to solve simple interest problems. I hope you have learned well. Good luck for the exams.

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2 Responses

  1. August 20, 2015

    […] How to Solve Simple Interest Problems Part 3 is a discussion of simple interest problems where the unknowns are the principal and the rate. […]

  2. August 20, 2015

    […] the next post, we will be solving simple interest problems whose uknowns are rate and […]

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