Mr. Dubey borrows ₹100000 from State Bank of India at 11% per annum compound interest. He repays ₹41000 at the end of first year and ₹47700 at the end of second year. Find the amount outstanding at the beginning of the third year.

Principal for first year = ₹100000, rate = 11%.

Interest for first year = $\dfrac{100000 \times 11 \times 1}{100} = \dfrac{1100000}{100}$ = ₹11000.

Amount after first year = ₹100000 + ₹11000 = ₹111000.

Money refunded at the end of first year = ₹41000.

Principal for second year = ₹111000 - ₹41000 = ₹70000.

Interest for second year = $\dfrac{70000 \times 11 \times 1}{100} = \dfrac{770000}{100}$ = ₹7700.

Amount after second year = ₹70000 + ₹7700 = ₹77700.

Money refunded at the end of second year = ₹47700.

Principal for third year = ₹77700 - ₹47700 = ₹30000.

Hence, the amount outstanding at the beginning of third year = ₹30000.