Which of the following is the correct sequence of trees received by Abha, Bina, Chitra and Dipti in that order?

	1	2	3	4
X	C		D	D
Y	A			A
Z	B	C

Let first find plots of each daughter.

Clue 6 says,

Dipti got two adjoining plots in the same row.

Also, clue 8 says,

Chitra and Dipti did not get plots which were adjacent to each other (either in row / column / diagonal).

=> two adjoining plots of Dipti can't be at Y2 , Y3, Z3, and X2(as they are adjacent plots to C)

Therefore, only place for D's plot is X3 and X4.

	1	2	3	4
X	C	B	D	D
Y	A			A
Z	B	C		B

Clue 7 says,

Bina was the only one who got a plot in each row and each column.

=> X3 will be Bina as this is the only place in row X.

Similarly, In column 4 we only have D and A

Therefore, Z3 will be of Bina.

	1	2	3	4
X	C	B	D	D
Y	A			A
Z	B	C		B

Now, we know,

Each daughter got an even number of plots

1. For D :

D can't be at any other place (as they are adjacent to C)

therefore, D have 2 plots in total.

2. For C:

If we put C in any of the blank column, it will become adjacent to D

(therefore, clue 6 and 8 will get contradicted.)

Therefore, C will also have 2 plots.

3. For B,

We want atleast one B in column 3 and row Y each.

Currently, we have 3 plots of B and 3 blank plots.

Therefore, to make B's plot an even number, B will have either 1 more or 3 more plots.

But, if B have 3 more plots , then will be left with only 2 plots

But, clue 4 says, Both Abha and Bina got a higher number of plots than Dipti, which will get contradicted.

Therefore, B will have one more plot, i.e., total of 4 plots

And A will have 4 plots.

	1	2	3	4
X	C	B	D	D
Y	A	A	B	A
Z	B	C	A	B

We know, Bina was the only one who got a plot in each row and each column.

Therefore, one more plot of Bina will be in column 3 and row Y

And other 2 plots will be of A.

	1	2	3	4
X (Mango)	12- C	B	D	D
Y (Teak)	21 - A	A	B	A
Z (Pine)	B	C	9 - A	28 - B

Now, lets find the number of trees in each plot.

(Numbers filled in the table is already given to us in the set)

Also, we know, plots in rows X, Y, Z contained mango, teak and pine trees respectively.

	1	2	3	4
X (Mango)	12- C	B	D	D
Y (Teak)	21 - A	a - A	2a - B	4a - A
Z (Pine)	B	C	9 - A	28 - B

Clue 3 says,

The number of teak trees in Column 3 was double of that in Column 2 but was half of that in Column 4

=> Y3 = 2 (Y2)

and, Y3 = 1/2 (Y4)

Let Y3 = 2a

=> Y2 = a

and, Y4 = 4a

	1	2	3	4
X (Mango)	12- C	B	D	D
Y (Teak)	21 - A	a - A	2a - B	4a - A
Z (Pine)	B	C	9 - A	28 - B

Now, let's try to put X values,

We know, Each plot had trees in non-zero multiples of 3 or 4 and none of the plots had the same number of trees

If a = 3

=> 2a = 6

and 4a will be 12

but, we already have X1 = 12

Therefore, x can't be 3

Now, let's try for a = 6

then 2a will be 12

but, again we have X1 = 12

Therefore, case rejected

If we put a = 9

then , 4a = 36

But, according to clue 2,

The largest number of trees in a plot was 32,

Therefore, case is rejected

Now, we can say for sure that a won't be a multiple of 3

	1	2	3	4
X (Mango)	12- C	B	D	D
Y (Teak)	21 - A	4 - A	8 - B	16 - A
Z (Pine)	B	C	9 - A	28 - B

Now, let's try for a to be multiple of 4

If a = 4

then 2a = 8 and 4a = 16

It satisfies all conditions.

If a = 8

then 4a = 32 => Y3 or Abha's plot = will have 32 teak trees.

But according to clue 2,

The largest number of trees in a plot was 32, but it was not with Abha.

Therefore, this case is also rejected

And, anything more then 8 will make 4a greater then 32 which is also not possible.

Therefore, only possibility of a is 4.

	1	2	3	4
X (Mango)	12- C	B	D	D
Y (Teak)	21 - A	4 - A	8 - B	16 - A
Z (Pine)	B	C	9 - A	28 - B

Now, clue 1 says,

Abha got 20 trees more than Chitra but 6 trees less than Dipti.

We know, Total number of trees with Abha = $21 + 4 + 16 + 9 = 50$

=> total trees with Chitra = $50 - 20 = 30$

and, total number of trees with Dipti = $50 + 6=56$

=> total number of trees with Bina = total number of trees - trees with Abha, Chitra and dipti

Therefore, total number of trees with Bina = $205 - 50- 30 - 56 = 69$

	1	2	3	4
X (Mango)	12- C	B	D	D
Y (Teak)	21 - A	4 - A	8 - B	16 - A
Z (Pine)	B	18 - C	9 - A	28 - B

We know, C has only 2 plots (X1 & Z2)

and total of 30 trees

Where, X1 have 12 trees

=> Z3 = $30 - 12 = 18$ trees.

	1	2	3	4
X (Mango)	12- C	B	32/24 - D	24/32 - D
Y (Teak)	21 - A	4 - A	8 - B	16 - A
Z (Pine)	B	18 - C	9 - A	28 - B

Now, according to clue 2 ,

largest number of trees in a plot was 32, but it was not with Abha.

=> It will be with either Bina or Dipti

1. For Bina,

We know total trees with Bina = $69$

Also, we already know, Y3 & Z4 plots are with B

Total = $8 + 28 = 36$

=> $69 - 36 = 33$

If Bina had 32 trees in one of the plot then she will be left with just one more tree in another plot.

Therefore, Bina does not have 32 trees

Therefore, Dipti will have 32 trees

So, Other plot with Dipti will have = $56 - 32 = 24$ trees

	1	2	3	4
X (Mango)	12- C	30 - B	32/24 - D	24/32 - D
Y (Teak)	21 - A	4 - A	8 - B	16 - A
Z (Pine)	B	18 - C	9 - A	28 - B

Now, Using clue 9,

number of mango trees was double the number of teak trees.

We know, number of teak trees = $21+4+8+16 = 49$

Therefore, number of Mango trees = $2 * 49 = 98$

=> X1 + X2 + X3 + X4 = $98$

=> X2 = $98 - 12-32-24$

=> X2= $30$

	1	2	3	4
X (Mango)	12- C	30 - B	32/24 - D	24/32 - D
Y (Teak)	21 - A	4 - A	8 - B	16 - A
Z (Pine)	3 - B	18 - C	9 - A	28 - B

We know, total number of trees with B = $69$

=> Z1 + X2 + Y3 + Z4 = $69$

=> Z1 = $69 - 30 - 8 - 28$

=> Z1 = $3$

	1	2	3	4
X (Mango)	12- C	30 - B	32/24 - D	24/32 - D
Y (Teak)	21 - A	4 - A	8 - B	16 - A
Z (Pine)	3 - B	18 - C	9 - A	28 - B

50, 69, 30, 56 is the correct sequence of trees received by Abha, Bina, Chitra and Dipti.