Instructions

A farmer had a rectangular land containing 205 trees. He distributed that land among his fourÂ daughters - Abha, Bina, Chitra and Dipti by dividing the land into twelve plots along threeÂ 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 hadÂ trees in non-zero multiples of 3 or 4 and none of the plots had the same number ofÂ trees. Each daughter got an even number of plots. In the figure, the number mentioned in topÂ left corner of a plot is the number of trees in that plot, while the letter in the bottom right cornerÂ is the first letter of the name of the daughter who got that plot (For example, Abha got the plotÂ in row Y and column 1 containing 21 trees). Some information in the figure got erased, but theÂ 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 thatÂ 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 /Â diagonal).

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

Solution

There are 12 plots and each of them got even number of plots. So, possible cases are 4,4,2,2 or 6,2,2,2.

From 4, A and B got more plots than D. So, the only possible case is A, B each got 4 and C,D each got 2.

From 6, D has to get two adjacent plots and From 8, plots of C, D are nit adjacent to each other => D must have got plots in X3, X4.

C already has two plots in X1, Z2. So, the corner plot Z4 should belong to B.

From 7, B has a plot in each row and each column. So, X2 should belong to B.

Now, out of the remaining Y2, Y3, Y4 and Z3 three plots belong to A and one belongs to B.

Till now B hasn't got any plot in Third column and 2nd row.

So, Y3 belongs to B and Y2, Y4, Z3 belongs to A.

Let the number of trees in Y4 be 4x from 3, number of trees in Y3, Y2 will be 2x, x respectively.

The number of teak trees=7x+21

.'. Number of mango trees=14x+42

The table now looks like:

Each plot had trees in non-zero multiples of 3 or 4 and none of the plots had the same number of trees and from 2, B didn't have the largest number of trees in a plot => x<8.

x can't be 7,5,3,2,1 as for these cases at least one of x,2x,4x is neither multiple of 3 or 4.

x can be 6 or 4.

If x=6, number of Teak trees will be 63 and Mango trees will be 126 => Number of Pine trees= 205-126-63=16 but number of trees in Z3+Z4>16 so, x$$\ne\ $$6.

If x=4, Number of Teak trees=49 and Mango trees=98 => Number of Pine trees=58. Valid case.

Number of trees with A= 30+5x=50.

From 1, number of trees with C, D= 30, 56 respectively.

So, number of trees in Z2= 18.

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

From 2, largest number of trees in a plot is 32. They can be in the plot of either B or D. If they are from B, they have to be from X2 but in that case number of trees in Z1=1 which is neither a multiple of 3 or 4.

So, highest number of trees in a plot are with D and it is 32 -=> number of trees in X3, X4 are 32, 24 in any order.

So, number of trees in X2= 98-56-12=30

.'. Number of trees in Z1=69-30-28-8=3.

The final table will look like:

Number of PIne trees received by Chitra = 18.

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