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Q.Mahesh purchased a house from a company for ₹ 70,00,000 and made a down payment of ₹ 15,00,000. He repays the balance in 25 years by monthly instalments at 9% p.a. compounded monthly.

(i) What is the amount of monthly payment ?
(ii) What is the total interest payment ? [Given : (1.0075)−300=0.1062878338(1.0075)^{-300} = 0.1062878338]
CBSECBSE Class XII Board 2024Subjective· 5mImportance★★★★★est
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Loan principal =₹55,00,000=\text{₹}55{,}00{,}000, n=300n=300 months, i=0.0075i=0.0075; the EMI formula gives ₹46,155.8046{,}155.80 per month and total interest ₹83,46,74083{,}46{,}740.

EMI E=P i1−(1+i)−nE=\dfrac{P\,i}{1-(1+i)^{-n}}, where P=P= loan, i=i= monthly rate, n=n= number of months. Total interest =nE−P.=nE-P.

  1. Loan amount (after down payment): P=70,00,000−15,00,000=₹55,00,000.P=70{,}00{,}000-15{,}00{,}000=\text{₹}55{,}00{,}000.
  2. Number of months: n=25×12=300.n=25\times12=300. Monthly rate: i=9%12=0.0912=0.0075.i=\dfrac{9\%}{12}=\dfrac{0.09}{12}=0.0075.
  3. (i) Instalment: E=P i1−(1+i)−n=55,00,000×0.00751−(1.0075)−300.E=\dfrac{P\,i}{1-(1+i)^{-n}}=\dfrac{55{,}00{,}000\times0.0075}{1-(1.0075)^{-300}}.
  4. Numerator =55,00,000×0.0075=41,250.=55{,}00{,}000\times0.0075=41{,}250. Using (1.0075)−300=0.1062878338(1.0075)^{-300}=0.1062878338, denominator =1−0.1062878338=0.8937121662.=1-0.1062878338=0.8937121662. …

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