The sum of the ages of all members of a family is a number lies between 300 and 400. If this number is added to the number formed by reversing the digits, the sum is 888 and if the unit's digit and the ten's digit change places, the new number exceeds the original number by 9. Find the average of ages of all members of that family if there are 15 members in a family.

Let the number be XY (where X is hundreds digit and Y is the number formed by tens and units digits). Original number: 100X + Y. Reversed digits (unit and tens change places within Y): if Y = ab, then swapping gives ba. However, treating 'the number' as XY and 'reversing the digits' as YX gives: (100X + Y) + (100Y + X) = 888. The condition about unit and tens digits is ambiguous. A simpler reading: let number be two digits AB. Then AB + BA = 888. Since AB = 10A + B and BA = 10B + A, we get 11(A + B) = 888. This doesn't work since 888 is not divisible by 11. Reading as three-digit number where last two digits swap: if number is 100a + 10b + c and digits b and c swap, new number is 100a + 10c + b, giving difference (100a + 10c + b) - (100a + 10b + c) = 9(c - b) = 9, so c - b = 1. Also original + swapped = 888, so 2a + 11(b + c) = 88. With c = b + 1: 2a + 11(2b + 1) = 88, so 2a + 22b = 77. This has no integer solution. An interpretation that works: let the number be 100a + 10b + c. The condition 'unit's digit and ten's digit change places' means b and c swap: new number = 100a + 10c + b. The difference is (10c + b) - (10b + c) = 9(c - b) = 9, so c = b + 1. The number is between 300 and 400, so a = 3. If we interpret 'reversing the digits' as abc → cba (full reversal): (100a + 10b + c) + (100c + 10b + a) = 888. With a = 3: 101(3 + c) + 20b = 888. Since c = b + 1: 101(3 + b + 1) + 20b = 888, so 101b + 404 + 20b = 888, giving 121b = 484, so b = 4 and c = 5. The number is 345. Check: 345 + 543 = 888 ✓, and 354 - 345 = 9 ✓. Average age = 345 ÷ 15 = 23.