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Read the following passage carefully:
In year 2000, Mr. Talwar took a home loan of (3000000 from State Bank of India at 7.5% p.a. compounded monthly for 20 years. Based on the above information, answer the following questions The equated monthly install
Question:
Interest paid by Mr. Talwar in 150th payment was
$\begin{array}{l} [Use (1.00625)^{240} = 4.4608, (1.00625)^{91} \\ = 1.7629, (1.00625)^{48} = 1.1187] \end{array}$

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Detailed Solution

Interest paid in 150th payment

$\begin{array}{l} = \frac{EMI [(1 + i)^{240 - 150 + 1} - 1]}{(1 + i)^{240 - 150 + 1}} \\ = \frac{24167.82 [(1.00625)^{91} - 1]}{(1.00625)^{91}} \\ = \frac{24167.82 \times 0.7629}{1.7629} \\ = ₹ 10458.69 \end{array}$

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