Visual & Spatial Reasoning

``

Face A (!)

Face B (@) Face C (#) Face D ($) Face E (%)

Face F (×)

``

Let's re-verify the standard 'L' shaped net:

``

X

Y Z P

Q

R

``

Opposite pairs are (X, Q), (Y, P), and (Z, R). This applies to our net as well if we re-label to fit this structure for easier visualization of the rule.

Let's apply the rule to the given net:

``

$!$

$@$ # $ $\%$

$\ast$

``

× # is the central face. Moving horizontally from #: $@$ and $ are on either side. So, $@$ ($B$) is opposite $ ($D$).

× Moving vertically from #: $!$ and $\%$ are on either side. So, $!$ ($A$) is opposite $\%$ ($E$).

× The remaining two faces $\ast$ ($F$) and # ($C$) must be opposite each other.

So the opposite pairs are:

× ! (A) and $\%(E)$

× @ (B) and $ (D)

× # (C) and $\ast (F)$

Let's check the options:

× A. $!$, & (! (A) and & (E)): These are opposite. (Wait, the question used $\%$ not &. Let's assume $\%$ and & are the same for the example based on my thought process, or adjust the problem to use &.) Re-using my specific symbols: $A{=}^{\prime }{!}^{\prime }$, $B{=}^{\prime }{@}^{\prime }$, C='#', D='$', $E{=}^{\prime }{\%}^{\prime }$, F='^'. Then $A$ opposite $E$, $B$ opposite $D$, $C$ opposite $F$. My question used $@$, #, $, $\%$ and $!$ $\ast$. So let's align the text with the options. If the question uses $!$, $@$, #, $, &, $\ast$ then:

Net:

``

$!$

$@$ # $ &

$\ast$

``

Opposite pairs:

× $!$ is opposite & (separated by # and $).

× $@$ is opposite $ (separated by #).

× # is opposite $\ast$ (separated by $ and &).

Let's re-evaluate options based on these pairs:

× A. $!$, & ($!$ and & are opposite). This would be a correct pair.

× B. $@$, $ ($@$ and $ are opposite). This would also be a correct pair.

This means my original net definition Face B (@) Face C (#) Face D ($) Face E (\%) leads to $B$ opposite $D$ and $C$ opposite $E$. And $A$ opposite $F$.

I need to pick one correct answer. Let's fix the net to be unambiguous for the given choices. A very standard net is the cross shape.

Standard cross net structure:

``

1

2 3 4 5

6

``

Opposite pairs: (1, 6), (2, 4), (3, 5).

Let's apply this to the question's symbols for clarity.

Net:

``

$!$ (Position 1)

$@$ # $ & (Positions 2, 3, 4, 5)

$\ast$ (Position 6)

``

Following the rule for the cross net:

× $!$ (position 1) is opposite $\ast$ (position 6).

× $@$ (position 2) is opposite $ (position 4).

× # (position 3) is opposite & (position 5).

Now, let's re-check the options:

× A. $!$, &: These are $1$ and $5$. Not opposite. (Adjacent to $ and # if folded)

× B. $@$, $ : These are $2$ and $4$. These are opposite. (Correct)

× C. #, $!$: These are $3$ and $1$. Not opposite. (Adjacent)

× D. &, $\ast$: These are $5$ and $6$. Not opposite. (Adjacent)

Therefore, the only pair of opposite faces among the choices is (@, $).

Total number of completely hidden cubes = $4$ (from Layer 1) + $1$ (from Layer 2) + $0$ (from Layer 3) + $0$ (from Layer 4) = $5$ cubes.

• First, consider the cut location (0.75, 0.75) relative to the last diagonal fold line ($y=-x+1.5$). If we plug the coordinates into the line equation: $0.75+0.75=1.5$. This means the cut is made exactly on the fold line. When a cut is made on a fold line, it is not mirrored across that fold; it remains a single cut after that specific fold is undone. So, after unfolding the diagonal fold (reverse Fold 3), there is still only 1 circular hole, located at (0.75, 0.75) in the $0.5\times 0.5$ top-right quadrant.

• Next, unfold the vertical fold (reverse Fold 2, across $x=0.5$). The hole at (0.75, 0.75) will be mirrored across the $x=0.5$ line to create a second hole at $(0.5-(0.75-0.5),0.75)=(0.25,0.75)$. Now there are 2 holes.

• Finally, unfold the horizontal fold (reverse Fold 1, across $y=0.5$). The two holes at (0.75, 0.75) and (0.25, 0.75) will be mirrored across the $y=0.5$ line. This creates two new holes at (0.75, 0.25) and (0.25, 0.25).

In total, $2\times 2=4$ circular holes will be visible when the paper is completely unfolded. These holes are located in each of the four quadrants of the original square, symmetrically placed.