DIPTI got 2 plots which also means Chitra gets (12 - 4 - 4 - 2 = 2) plots .Now let Abha gets “a” tree. So the number of trees with Chitra = a – 20And the number of trees with Dipti = a + 6Now if the number of teak trees in column 2 was x then:Total number of teak trees = x + 2x + 4x + 21 = 7x + 21From statement (9) number of mango trees = 2(7x + 21) = 14x + 42Now as we can see minimum number of trees in Row Z – column 1 and column 2 can be 7 (3, 4 or 4,3)Which means:7x + 21 + 14x + 42 ≥ 205 - (7 + 9 + 28) = 161 or 21x ≥ 98As x is an integer and multiple of 3 or 4 so x can be either 3 or 4 only if x = 5 , 21x = 105 > 98)If x = 3, not possible because if x = 3, 4x = 12, two plots (Row X column 1 and Row Y column 4) will have the same number of trees, which is not possible.Thus x = 4⇒ Number of teak trees = 7x + 21 = 7 × 4 + 21 = 49⇒ Number of mango trees = 14x + 42 = 14 × 4 + 42 = 98∴ Number of pine trees = 205 – 98 – 49 = 58Sum of number of trees in (Row Z) – column 1 and column 2 = 58 – 28 – 9 = 21\therfore the possible PAIR are (3,18) or (6,15)From the statement (5) and (6), Dipti must get 2 plots one each in column 3 and column 4. So she can get a plot only in Row X or row Z.Now from point (8), Dipti didn’t get a plot adjacent to Chitra so Dipti should get plots in row X.Thus we can make the following arrangements :Number of trees with A = 9 + 21 + 16 + 4 = 50So, a = 50Now number of trees with C = a - 20 = 50 – 20 = 30Number of trees with D = a + 6 = 50 + 6 = 56Number of trees with B = 205 - 56 - 50 - 30 = 69So number of trees in Row Z column 2 = 30 -12 = 18number of trees in Row Z column 1 = 21 - 18 = 3Number of trees in Row X column 2 = 69 - 3 - 8 - 28 = 30D has 32 tree plots∴ Final trees – plot diagram will be as below –