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EMI, CAGR & Depreciation · Question 27 of 66

Question 27

HardOriginal question

A loan of ₹1,00,000 at 10% a year is repaid in 3 equal yearly instalments by the reducing-balance method. To the nearest rupee, what is each instalment? (Take (1.1)3=1.331(1.1)^{3} = 1.331.)

Options for question 27

Show the answer and explanation
CorrectNot quiteAnswer B

Correct answer: B

Explanation

  1. E=P i (1+i)n(1+i)n−1=1,00,000×0.1×1.3310.331E = \frac{P\,i\,(1 + i)^{n}}{(1 + i)^{n} - 1} = \frac{1{,}00{,}000 \times 0.1 \times 1.331}{0.331}.

  2. E=13,3100.331≈40,211E = \frac{13{,}310}{0.331} \approx 40{,}211 rupees.

Worth noting

₹43,333 is the flat-rate instalment; ₹33,333 leaves out interest altogether.

Concept
A yearly instalment by reducing balance
Typical time
120 s
Your time
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The rule this testsEMI by the reducing-balance method
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Ten questions from EMI, CAGR & Depreciation, at your own pace.