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SS3: MATHEMATICS - 2ND TERM

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  1. Matrices I | Week 1
    6 Topics
  2. Matrices II | Week 2
    1 Topic
    |
    1 Quiz
  3. Commercial Arithmetic | Week 3
    7 Topics
    |
    1 Quiz
  4. Coordinate Geometry | Week 4
    8 Topics
    |
    1 Quiz
  5. Differentiation of Algebraic Expressions | Week 5 & 6
    7 Topics
  6. Application of Differentiation | Week 7
    4 Topics
    |
    1 Quiz
  7. Integration | Week 8
    8 Topics
    |
    1 Quiz



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Interest is paid on money saved in a bank or borrowed from a lender. The original amount saved or lent is known as Principal. The period (either in years, months, weeks, etc.) which the money borrowed is to be used by the borrower is called Time. The amount of compensation in terms of a certain percentage of the amount lent out is called Rent.

In general, if the principal, P is lent or borrowed for T years at a rate of R% per annum, therefore the simple interest is given by

I = \( \frac{P \; \times \; T \; \times \; R }{100}\)

Thus P = \( \frac{100I }{TR}\)

R= \( \frac{100I }{PT}\)

T= \( \frac{100I }{PR}\)

With these, simple interest problems can be solved by substituting the appropriate values into any of these formulae.

Example 1:

Find the simple interest on a loan of #2700 for \(\scriptsize \; 6 \frac{1}{4}\) years at 3% per annum

Solution

:- \(\scriptsize 6 \frac {1}{4} = \frac{25}{4} \)

I = \( \frac{PTR }{100}\)

= \( \frac{2700 \; \times \; \frac{25}{4} \; \times \; 3 }{100}\)

= \( \frac{2700 \; \times \; 25 \; \times \; 3 }{400}\)

= #506.25k

Example 2:

A sum of #3500 invested at 2½% per annum for a number of years earns #612.50k interest. Find the time.

Solution

T = \( \frac{100I }{PR}\)

= \( \frac{100 \; \times \; 612.5 }{3500 \; \times \; \scriptsize 2 \frac{1}{2}}\)

= \( \frac{61,250 }{8750}\)

= 6.99years ≅ 7years

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