SBI Junior Associate 2015 · Question 127 of 200
arithmetic
$$C.I. = P [(1 + \frac{R}{100})^T - 1]$$ => $$5632 = P [(1 + \frac{20}{100})^2 - 1]$$ => $$5632 = P [(\frac{6}{5})^2 - 1]$$ => $$5632 = \frac{11 P}{25}$$ => $$P = \frac{5632 \times 25}{11} = 12,800$$ $$\therefore$$ $$S.I. = \frac{12800 \times 12 \times 3}{100}$$ = $$128 \times 36$$ = Rs. $$4,608$$
Source: SBI Clerk 25 Jan 2015 Paper with Solutions — Cracku · reliable-secondary
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