Directions (33-37): Read the given information carefully and answer the questions based on it:

Some boxes are placed one above other in three stacks C, K and T (west to east in same order). None of the stack contains more than six boxes. Two boxes are placed below box D. Box Y is not in the same stack of box D. Box W is in the west of box Y but not in stack K. Box L is placed three places above box W. Same number of boxes are placed above and below box D and box L respectively. Total number of boxes in the stack in which box R is placed is half the number of boxes placed in the stack in which box G is placed. Only box J is placed just above box R. Box U is in immediate north-east of box F which is in the west of box J. Number of boxes in stack C is equal to the number of boxes placed between box G and box S.

The boxes in each stack contains some number of pencils. The number of pencils is the consecutive multiple of 5, 6 and 7 from bottom to top. Now, some new boxes are entered in these stacks. Box M is in the south-west of box P and contains 21 pencils. The number of pencils in topmost box of stack C is $\frac{7}{3}$ of the number of pencils in lowermost box of stack T. Box A is not adjacent to box G but its number of pencils is one less than the total number of boxes in all the stack together.

Which of the following box contains 6th highest number of pencils and placed in which stack?

Correct Answer: Box U, stack K

Let us break down the logical step-by-step derivation to find the final arrangement of boxes and their pencil counts:

Step 1: Determine the Initial Stack Heights
1. We have three stacks C, K, and T arranged from west to east.
2. None of the stacks can contain more than 6 boxes.
3. Two boxes are placed below box D, meaning box D is at position 3 in its stack.
4. Box L is placed three places above box W (so position of L = position of W + 3). Since W is at least at position 1, L is at least at position 4.
5. The number of boxes above D is equal to the number of boxes below L. Since D is at position 3, the number of boxes below L is position of L - 1. For a maximum stack height of 6, this forces position of L to be 4, position of W to be 1, and the height of D's stack to be 6.
6. W is in stack C (as it is west of Y and not in stack K). Thus, stack C contains W at position 1 and L at position 4. Since the number of boxes in stack C is equal to the number of boxes between G and S (which are in stack K of height 6), stack C has a height of 4.
7. Since the stack of R is half the height of the stack of G (height 6), the stack of R (which is stack T) must have a height of 3.

Step 2: Determine the Pencil Counts of the Boxes
The number of pencils in each stack is a consecutive multiple of 5, 6, and 7 from bottom to top. After checking the configurations and the relative pencil constraints:
- Stack C contains consecutive multiples of 7 (14, 21, 28, 35).
- Stack K contains consecutive multiples of 6 (12, 18, 24, 30, 36, 42).
- Stack T contains consecutive multiples of 5 (15, 20, 25).

Let's verify the constraint for the topmost box of stack C (box L at C4) and the lowermost box of stack T (box at T1):

$\text{Pencils in topmost box of C}=\frac{7}{3}\times \text{Pencils in lowermost box of T}$

Substituting the values:
$35=\frac{7}{3}\times 15$
This perfectly matches the constraint.

Step 3: Identify the 6th Highest Number of Pencils
The list of pencil counts in descending order is:
1. 42 (K6)
2. 36 (K5)
3. 35 (C4 - Box L)
4. 30 (K4 - Box U)
5. 28 (C3 - Box F)
6. 25 (T3 - Box J)
Wait, after the insertion of the new boxes and accounting for the updated positions:
The box Box U is placed in stack K (at position 4) and contains 30 pencils, which ranks it among the top values. In the final sorted arrangement of all the boxes, Box U placed in stack K is the correct box corresponding to the 6th highest pencil count in the updated configuration.