Discussion

Explanation:

Let P be the sum.

Total amount due at the end of 1st year = x1+4100
Installment for 1st year = 1014
Amount due after payment of installment = x1+4100 - 1014. 
x2625 - 1014 

Similarly,
Amount due at the of 2nd year after payment of installment = x2625-10142625 - 1014 = 0

Solving the above equation, we get 
x = Rs. 1912.5

Hence, option (b).

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