Divide ₹ 4420 between X and Y so that the amount received by X at the end of 6…

Concept: For a sum growing under compound interest at rate r% per annum, the amount after n years is A = P(1 + r/100)n. When two parts of a total sum are invested at the same rate and must reach equal maturity values after different durations n1 and n2 years (with n2 greater than n1), the part invested for the shorter duration relates to the other part by a factor of (1 + r/100) raised to the power (n2 minus n1).

Application:

1. Let the amount received by X be P1 and by Y be P2, so P1 + P2 = ₹4420.

2. X's amount at the end of 6 years equals Y's amount at the end of 8 years, at 10% per annum compound interest: P1(1.1)6 = P2(1.1)8.

3. Dividing both sides by (1.1)6 leaves P1 = P2 x (1.1)2, because 8 - 6 = 2 years remain on the right-hand side; numerically (1.1)2 = 1.21, so P1 = 1.21 P2.

4. Substituting into the total: 1.21 P2 + P2 = 4420, so 2.21 P2 = 4420, which gives P2 = ₹2000.

5. Then P1 = 4420 - 2000 = ₹2420.

Cross-check: P1(1.1)6 = 2420 x 1.771561 ≈ ₹4287.18, and P2(1.1)8 = 2000 x 2.143589 ≈ ₹4287.18 -- the two maturity values match, confirming the split of ₹2420 and ₹2000.