A, B, C, D, and E are five friends who went on a picnic. Their average age is 43 years. The present age ratio of D to E is 4:5, and the present age ratio of A to B is also 4:5. Four years from now, the ratio of A's age to C's age six years from now will be 4:5. Additionally, four years from now, A’s age will be 10 years less than E's current age. What is the average age of D and E?

Let the present age of A and B = 4x: 5x Four years from now the age of A and Six years from now the age of C = 4y : 5y Equate A in both cases 4x = 4y – 4 X = y – 1, y = x+1 Substitute the value of Y Age of C=5y – 6 => 5(x+1) – 6 = 5x+5-6=5x-1 So present ages of A, B and C = 4x : 5x : 5x – 1 Four years from now the age of A is 10 years less than E 4x +4 = E - 10 => E = 4x + 14 5 parts of E = 4x + 14 1 parts of E= 0.8x+2.8  4 parts of D = 3.2x + 11.2 So present ages of A, B, C, D and E = 4x : 5x : 5x – 1 : 3.2x+11.2 : 4x+14 Average age of all friends = 43 Total age of all friends = 43*5 = 215 years 4x + 5x + 5x – 1 + 3.2x+11.2 + 4x+14 = 215 21.2x = 190.8 X = 9 Substitute x value in D and E Average age of D and E =7.2 * 9 + 25.2 / 2 = 45 years.