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Q.In the year 2010, Mr. Aggarwal took a home loan of ₹ 30,00,000 from State Bank of India at 7⋅5%7\cdot5\% p.a. compounded monthly for 20 years. Based on the above information, answer the following questions :

(i) Determine the EMI. [1]
(ii) Find the principal paid by Mr. Aggarwal in the 150th150^{th} instalment. [1]
(iii)
(a) Find the total interest paid by Mr. Aggarwal. [2]
(OR)
(iii)
(b) How much was paid by Mr. Aggarwal to repay the entire amount of home loan ? [2] [Use (1⋅00625)240=4⋅4608; (1⋅00625)91=1⋅7629(1\cdot00625)^{240} = 4\cdot4608;\ (1\cdot00625)^{91} = 1\cdot7629]
CBSECBSE Class XII Board 2023Subjective· 4mImportance★★★★★
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EMI ≈\approx ₹24,167.82; 150th150^{\text{th}} principal ≈\approx ₹13,709; total interest ≈\approx ₹28.0 lakh; total repaid ≈\approx ₹58.0 lakh.

EMI=P⋅i (1+i)n(1+i)n−1\text{EMI}=\dfrac{P\cdot i\,(1+i)^{n}}{(1+i)^{n}-1}, with PP = principal, ii = monthly rate, nn = number of months. Principal in the kthk^{\text{th}} instalment =EMI⋅(1+i)−(n−k+1)=\text{EMI}\cdot(1+i)^{-(n-k+1)}. Total paid =EMI×n=\text{EMI}\times n.

  1. P=₹30,00,000P=\text{₹}30{,}00{,}000; i=7.5%12=0.625%=0.00625i=\dfrac{7.5\%}{12}=0.625\%=0.00625; n=20×12=240.n=20\times12=240. Given (1.00625)240=4.4608.(1.00625)^{240}=4.4608.

(i) EMI

2. EMI=3000000×0.00625×4.46084.4608−1=18750×4.46083.4608=836403.4608≈₹24,167.82.\text{EMI}=\dfrac{3000000\times0.00625\times4.4608}{4.4608-1}=\dfrac{18750\times4.4608}{3.4608}=\dfrac{83640}{3.4608}\approx \text{₹}24{,}167.82.

(ii) Principal in the 150th150^{\text{th}} instalment

3. Remaining instalments after the 149th149^{\text{th}}: n−k+1=240−150+1=91.n-k+1=240-150+1=91.

4. Principal part =EMI⋅(1.00625)−91=24167.821.7629≈₹13,709.=\text{EMI}\cdot(1.00625)^{-91}=\dfrac{24167.82}{1.7629}\approx \text{₹}13{,}709.

(iii)(a) Total interest …

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