Question

11. A man borrows 6000 for 2 years at the rate of 12% pa, compounded annually. He repays ₹3000 at the end of each year, How much does he still owe after the second repayment? ​

Answer 1

Answer:

Amount owe after 2nd repayment  = Rs 1166.4

Step-by-step explanation:

Given,

Principal ( P ) = Rs 6000

Time ( T ) = 2 years

Rate ( R ) = 12 %

Case 1: For 1st Year

Principal ( P ) = Rs 6000

Time ( T ) = 1 year

Rate ( R ) = 12 %

    SI = $\frac{P*T*R}{100}$

⇒  SI = $\frac{6000*12*1}{100}$

⇒  SI = Rs 720

Total Amount ( A ) after 1st year = P + SI

                     = Rs 6000 + Rs 720

                     = Rs 6720

Case 2: For 2nd Year

Principal for 2nd year = Amount of 1st Year - Amount Repays after 1st year

                                    = Rs 6720 - Rs 3000

                                    = Rs 3720

Time ( T ) = 1 year

Rate ( R ) = 12 %

  SI = $\frac{P*T*R}{100}$

      = $\frac{3720*12*1}{100}$

      = Rs 446.4

Total Amount ( A ) after 2nd year = Rs 3720 + Rs 446.4

                     = Rs 4166.4

Now,

Amount owe after 2nd repayment = Rs 4166.4 - Rs 3000

                                                         = Rs 1166.4