Figure 1 shows a plate and Figure 2 shows two stacks of identical plates. There are 3 plates in a shorter stack and 7 plates in a taller stack. The height of the shorter stack is 20 cm and the height of the taller stack is 44 cm.

1. Identify the difference in height and the difference in the number of plates between the two stacks.

Difference in height $= 44 - 20 = 24$ cm
Difference in number of plates $= 7 - 3 = 4$ plates

2. Calculate the height of one plate gap (the thickness added by each additional plate).

Height of 1 gap $= 24 \div 4 = 6$ cm

3. Calculate the height of a single plate by subtracting the gaps from the total height of a stack.

Height of 3 plates stack $= 20$ cm
This stack consists of 1 base plate and 2 gaps.
Height of 2 gaps $= 2 \times 6 = 12$ cm
Height of 1 plate $= 20 - 12 = 8$ cm

1. Find the height of a single plate and the height added by each subsequent plate in a normal stack.

From the given stacks:
Height of 7 plates = $44$ cm
Height of 3 plates = $20$ cm
Difference in height for 4 plates = $44 - 20 = 24$ cm
Height added by each additional plate in a stack = $24 \div 4 = 6$ cm
Height of the first plate = $20 - 2 \times 6 = 8$ cm

2. Calculate the maximum number of plates that can be packed in Arrangement A.

In Arrangement A, the plates are stacked normally.
Let $n$ be the number of plates.
Total height = $8 + 6(n - 1) = 6n + 2$ cm
We need the height to be at most $1$ m = $100$ cm.
$6n + 2 \le 100$
$6n \le 98$
$n \le 16.33$
So, a maximum of $16$ plates can be packed in Arrangement A.

3. Calculate the maximum number of plates that can be packed in Arrangement B and find the difference.

In Arrangement B, the plates are in an alternating orientation (e.g., top-to-top, base-to-base), meaning they do not overlap and each plate contributes its full height of $8$ cm.
Number of plates in Arrangement B = $100 \div 8 = 12.5$
So, a maximum of $12$ plates can be packed in Arrangement B.
Difference = $16 - 12 = 4$
Arrangement A allows Matthew to pack $4$ more plates.