A person amortizes a loan of ₹1,50,000 for a new home by obtaining a 10 year mortgage at the rate of 12% compounded monthly. Find (i) The monthly payments (ii) Total interest paid. (Given $a_{\overline{120}|0.01} = 69.6891$)

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> [!TLDR]
> The loan equals the present value of the monthly payments: $R	imes a_{\overline{120}|0.01}=1{,}50{,}000$. With the given factor $a_{\overline{120}|0.01}=69.6891$, the monthly payment is $R=\dfrac{1{,}50{,}000}{69.6891}=₹2{,}152.42$ and the total interest is $₹1{,}08{,}290.40$.

A home loan is the present value of all its future monthly payments, discounted at the monthly rate. Here the annual rate is 12% compounded monthly, so the monthly rate is $i=1\%=0.01$, and there are $n=10	imes12=120$ payments.

1. **Loan parameters.** $P=₹1{,}50{,}000$, $i=0.01$, $n=120$; given $a_{\overline{120}|0.01}=69.6891$.
2. **Monthly payment.** …

Q.A person amortizes a loan of ₹1,50,000 for a new home by obtaining a 10 year mortgage at the rate of 12% compounded monthly. Find

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Q.A person amortizes a loan of ₹1,50,000 for a new home by obtaining a 10 year mortgage at the rate of 12% compounded monthly. Find

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