In compound interest, the interest accrues the initial principal and the accumulated interest from previous periods. Amount = P[1 + (R/100)]t Where, P = Principal R = Rate of interest per annum t = Time in years This article contains the MCQs on 'Compound Interest' that are fetched from previous competitive exams like Bank P.O, M.B.A, M.A.T, etc. The test consists of 13 questions on Simple Interest. No negative marking for this test. No Time limit The pass percentage is 70% The correct answer with a description will be displayed after the answer has been marked. Submit the test to calculate your score once you are done with all the questions. Complexity Level- Moderate Given: The sum = 12000 The time = 2 years The compound interest for two years = 1996.80 Formula used: \({CI}\ =\ {P\ × \ [({1\ +\ {R\over100}})^T\ -\ 1]}\) \({SI}\ =\ {P\ ×\ T\ ×\ R\over 100}\) Where, CI = The compound interest, P = The principle, R = The rate of the interest, SI = The simple interest, and T = The time Calculation: Let us assume the rate of interest be R ⇒ According to the question ⇒ \({1996.8}\ =\ {12000\ × \ [({1\ +\ {R\over100}})^2\ -\ 1]}\) ⇒ \({1996.8}\ =\ {12000\ × \ [({100\ +\ R\over100})^2\ -\ 1]}\) ⇒ \({1996.8}\ =\ {12000\over10000} × \ [({100\ +\ R})^2\ -\ 10000]\) ⇒ \({1996.8}\ =\ {1.2} × \ [({100\ +\ R})^2\ -\ 10000]\) ⇒ \({1996.8}\ =\ {1.2} × \ [({100^2\ +\ R^2\ +\ 200R})\ -\ 10000]\) ⇒ \({1996.8}\ =\ {1.2} × \ [{10000\ +\ R^2\ +\ 200R}\ -\ 10000]\) ⇒ \({1996.8}\ =\ {1.2} × \ [{\ R^2\ +\ 200R}]\) ⇒ 1996.8 = 1.2R2 + 240R ⇒ 19968 = 12R2 + 2400R ⇒ 1664 = R2 + 200R ⇒ By solving the quadratic equation ⇒ R = 8, and -208 ⇒ The value of R = 8 ⇒ The total amount after two years = 12000 + 1996.8 = 13996.8 ⇒ The simple interest for third year on 13996.8 at 8% per annum ⇒ According to the question ⇒ \({SI}\ =\ {13996.8\ ×\ 1\ ×\ 8\over 100}\) ⇒ SI = 139.968 × 8 = 1119.744 ⇒ The simple interest for the third year = 1119.744 ∴ The required result will be 1119.74.
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