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Q.A man took a home loan of ₹ 40,00,000 from a bank at the interest of 6⋅75%6\cdot75\% per annum compounded monthly which is to be amortized by equal payments at the end of each month for 10 years. Based on the above information, answer the following questions :

(i) Find the monthly instalment. [Use (1⋅005625)−120=0⋅510120(1\cdot005625)^{-120} = 0\cdot510120]
(ii) Find the principal outstanding at the beginning of 61st61^{st} month. [Use (1⋅005625)60=1⋅400115(1\cdot005625)^{60} = 1\cdot400115]
(iii)
(a) Find the interest amount paid in the 61st61^{st} instalment.
(OR)
(iii)
(b) Find the principal amount paid in the 61st61^{st} instalment.
CBSECBSE Class XII Board 2025Subjective· 4mImportance★★★★★est
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EMI =Pi1−(1+i)−n≈=\dfrac{Pi}{1-(1+i)^{-n}}\approx ₹ 45,930; outstanding before the 61st61^{st} payment (PV of the last 6060 instalments) ≈\approx ₹ 23,33,431; its interest part ≈\approx ₹ 13,126 and principal part ≈\approx ₹ 32,804.

EMI: EMI=P i1−(1+i)−n\text{EMI}=\dfrac{P\,i}{1-(1+i)^{-n}}. Outstanding balance before the (k+1)th(k{+}1)^{th} payment (prospective) =EMI⋅1−(1+i)−(n−k)i=\text{EMI}\cdot\dfrac{1-(1+i)^{-(n-k)}}{i}. Interest in a payment =i×(outstanding)=i\times(\text{outstanding}); principal =EMI−interest=\text{EMI}-\text{interest}.

Given P=₹40,00,000P=₹40{,}00{,}000, i=6.751200=0.005625i=\dfrac{6.75}{1200}=0.005625, n=120n=120 months.

(i) Monthly instalment

  1. EMI=P i1−(1+i)−n=4000000×0.0056251−(1.005625)−120\text{EMI}=\dfrac{P\,i}{1-(1+i)^{-n}}=\dfrac{4000000\times0.005625}{1-(1.005625)^{-120}}.
  2. Numerator =4000000×0.005625=22500=4000000\times0.005625=22500.
  3. Using (1.005625)−120=0.510120(1.005625)^{-120}=0.510120: denominator =1−0.510120=0.489880=1-0.510120=0.489880.
  4. EMI=225000.489880≈45929.6≈₹45,930\text{EMI}=\dfrac{22500}{0.489880}\approx45929.6\approx₹45{,}930.

(ii) Principal outstanding at the beginning of the 61st61^{st} month

  1. After 6060 payments, 6060 instalments remain; the outstanding equals their present value: B=EMI⋅(1+i)60−1i (1+i)60B=\text{EMI}\cdot\dfrac{(1+i)^{60}-1}{i\,(1+i)^{60}}.
  2. Using (1.005625)60=1.400115(1.005625)^{60}=1.400115: B=45930⋅1.400115−10.005625×1.400115=45930⋅0.4001150.007875647B=45930\cdot\dfrac{1.400115-1}{0.005625\times1.400115}=45930\cdot\dfrac{0.400115}{0.007875647}. …

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