Quantity I: The average age of a cricket team of 11 players is x years. After three years, four players leave the team and four new players join in place of them then the current average age becomes (x + 2) years. The current average age of the players who left the team was 18 years then find the average of the current age of the 4 new joiners? Quantity II: In a group of 6 students, the average age of the 5 youngest students is 12 years and the average age of the 5 oldest students is 15 years. What is the average age of all the students? (it is given that the age of them is in arithmetic progression)? Correct Answer Quantity I > Quantity II Quantity I: Sum of the age of 11 players = 11x years After 3 years, sum of their age will become 11x + 11 × 3 = 11x + 33 years The sum of the current age of 7 players after leaving 4 players = (11x + 33) - 18 × 4 = 11x - 39 years Let the sum of the age of the new joiner be m years Then the sum of the current age of 11 players ⇒ 11x - 39 + m = (x + 2) × 11 ⇒ 11x - 39 + m = 11x + 22 ⇒ m = 61 Sum of ages of new joiners is 61 years ∴ Required average = 61 / 4 = 15.25 years Quantity II: Let us assume that the 6 students are p, q, r, s, t and u which are in ascending order. Let the age of the first student be x years and the common difference be d year Then according to the question, p + q + r + s + t = 12 × 5 = 60 years x + (x + d) + (x + 2d) + (x + 3d) + (x + 4d) = 60, ⇒ 5x + 10d = 60 ----(i) q + r + s + t + u = (15) × 5 years = 75 years ⇒ (x + d) + (x + 2d) + (x + 3d) + (x + 4d) + (x + 5d) = 75 ⇒ 5x + 15d = 75 -----(ii) Solving equation (i) and (ii) we get, x = 6, d = 3 Sum of ages of p, q, r, s, t and u = 6 + 9 + 12 + 15 + 18 + 21 = 81 years Required average = 81 / 6 = 13.5 years Quantity I > Quantity II