A farmer had a rectangular land containing 205 trees. He distributed that land among his four daughters – Abha, Bina, Chitra and Dipti by dividing the land into twelve plots along three rows (X,Y,Z) and four Columns (1,2,3,4) as shown in the figure below:

The plots in rows X, Y, Z contained mango, teak and pine trees respectively. Each plot had trees in non-zero multiples of 3 or 4 and none of the plots had the same number of trees. Each daughter got an even number of plots. In the figure, the number mentioned in top left corner of a plot is the number of trees in that plot, while the letter in the bottom right corner is the first letter of the name of the daughter who got that plot (For example, Abha got the plot in row Y and column 1 containing 21 trees). Some information in the figure got erased, but the following is known:

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

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

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

4. Both Abha and Bina got a higher number of plots than Dipti.

5. Only Bina, Chitra and Dipti got corner plots.

6. Dipti got two adjoining plots in the same row.

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

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

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

- 49
- 84
- 126
- 98

- 60, 39, 40, 66
- 50, 69, 30, 56
- 44, 87, 24, 50
- 54, 57, 34, 60

- 15
- 18
- 30
- 21

- Bina, 3 trees
- Abha, 4 trees
- Dipti, 6 trees
- Bina, 4 trees

- Bina got 32 pine trees.
- Chitra got 12 mango trees.
- Abha got 41 teak trees.
- Dipti got 56 mango trees

- 4
- 2
- Cannot be determined
- 3

From the statement point 4 and 6 , we can say both Abha and Bina will get 4 plots each (as each daughter got an even number of plots) .

As Dipti got 2 plots which also means Chitra gets (12- 4-4-2 = 2) plots .

Now let Abha gets “a” trees. So number of trees with Chitra = a – 20

And number of trees with Dipti = a + 6

Now if the number of teak trees in column 2 was x then

Total number of teak trees = x+2x+4x +21 = 7x+21

From statement point 9) number of mango trees = 2(7x+21 ) = 14x +42

Now 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 <= 98

As x is an integer and multiple of 3 or 4 so x can be ether 3 or 4 only ( if x = 5 , 21x = 105 > 98)

If x = 3, not possible because if x = 3 , 4x = 12 which means two plots ( Row X column 1 and Row Y column 4) will have 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

Thus Number of pine trees = 205 – 98 – 49 = 58

Sum of number of trees in Row Z – column 1 and column 2 = 58 – 28 – 9 = 21

Possible pair (3,18) or (6,15)

From the statement point 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 = 50

So a = 50

Now number of trees with C = a -20 = 50 – 20 = 30

Number of trees with D= a+6 = 50+6 =56

Number of trees with B = 205 -56-50-30 = 69

So number of trees in Row Z column 2 = 30 – 12 = 18

number of trees in Row Z column 1 = 21 – 18 = 3

Number of trees in Row X column 2 = 69 – 3 – 8 – 28 = 30

D has 32 tree plots

Thus final trees – plot diagram will be as below –

All the questions can be answered now.

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