Mr. Robinson saves rupees 10000 per month in SBI ay 11% interest per annum. What amount will be get after 6 years ?

Dear Student,

Please find below the solution to the asked query:

Given : A man deposit Rs 10000 per month in SBI .

Interest rate =  11 % per annum

We know
Simple interest  = $\frac{P \times r \times t }{100}$ , Here P  =  principle amount , r  = Rate of interest and t  = time ( in years )

Money he deposit in a year  =  12 ( 10000 ) = 120000 , So

Simple interest  = $\frac{120000 \times 11 \times 1 }{100}= 1200 \times 11 = \mathbf{13200}$ , So 

Money at starting of 2^nd year = 120000 + 13200 = 133200

And

Money at end of 2^nd year = 120000 + 133200 = 253200

Simple interest  = $\frac{253200 \times 11 \times 1 }{100}= 2532 \times 11 = \mathbf{27852}$ , So 

Money at starting of 3^rd year = 253200 + 27852 = 281052

And

Money at end of 3^rd year = 120000 + 281052 = 401052

Simple interest  = $\frac{401052 \times 11 \times 1 }{100}= 4010.52 \times 11 = \mathbf{44115}\mathbf{.}\mathbf{72}$ , So 

Money at starting of 4^th year = 401052 + 44115.72= 445167.72

And

Money at end of 4^th year = 120000 + 445167.72 = 565167.72

Simple interest  = $\frac{565167.72 \times 11 \times 1 }{100}= 5651.6772\times 11 = \mathbf{62168}\mathbf{.}\mathbf{4492}$ , So 

Money at starting of 5^th year = 565167.72 + 62168.4492= 627336.1692

And

Money at end of 5^th year = 120000 + 627336.1692 = 747336.1692

Simple interest  = $\frac{747336.1692 \times 11 \times 1 }{100}= 7473.361692\times 11 = \mathbf{82206}\mathbf{.}\mathbf{978612}$ , So 

Money at starting of 6^th year = 747336.1692 + 82206.978612= 829543.147812

And

Money at end of 6^th year = 120000 + 829543.147812 = 949543.147812

Simple interest  = $\frac{949543.147812 \times 11 \times 1 }{100}= 9495.43147812\times 11 = \mathbf{104449}\mathbf{.}\mathbf{74625932}$ , So 
Therefore,

Amount will be get after 6 years = 949543.147812 + 104449.74625932 = 1053992.89407132 $\approx$Rs.1053992.90     ( Ans )

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