Question

If the Red side is exchanged with the Green side and the Blue is swapped with Black, then which of the following is false?

• A. Red is opposite to Black
• B. White is adjacent to Brown
• C. Green is opposite to Brown
• D. White is opposite to Blue

Hint:

When questions involve multiple swaps on a cube, freeze the \emph{positions} (top, bottom, the four side slots) and just relabel them. Then check opposites/adjacencies by positions, not colours.

Answer 1

The Correct Option is B

Step 1: Record the \emph{initial layout (from Q53).}
Top \(=\) Black, Bottom \(=\) Red. Side-ring (in order) \(=\) White \(\to\) Blue \(\to\) Brown \(\to\) Green (so that Blue is adjacent to both White and Brown). Hence opposites initially: \(\text{White}\leftrightarrow\text{Brown}\), \(\text{Blue}\leftrightarrow\text{Green}\). Step 2: Perform the swaps as "exchange positions of colours''.
Swap \( \text{Red} \leftrightarrow \text{Green}\): Green moves to Bottom; Red moves to Green's side position.
Swap \( \text{Blue} \leftrightarrow \text{Black}\): Blue moves to Top; Black moves to Blue's side position.
The side-ring order (by positions) is unchanged; only the \emph{labels} on two side positions change. New layout: Top \(=\) Blue, Bottom \(=\) Green; side ring \(=\) White \(\to\) Black \(\to\) Brown \(\to\) Red. Step 3: Evaluate each option on the \emph{new cube.}
(a) Red opposite Black? — On the ring, Black and Red occupy the two positions that were opposite (they replaced Blue and Green), so \textit{True}.
(b) White adjacent to Brown? — On the ring, White and Brown remain opposite (unchanged), not adjacent \(\Rightarrow\) \textit{False}.
(c) Green opposite Brown? — Green is Bottom; its opposite is Top (Blue), not Brown. This also reads as \textit{False}.
(d) White opposite Blue? — White is a side; Blue is Top; opposites are side–side and top–bottom only, so this is \textit{False}. Step 4: Important note on the question.
With the natural "swap positions" interpretation, \emph{(b), (c), and (d) all come out false}. Most exam keys expect a single choice; in that case, (b) is the intended standout because the White–Brown pair stays opposite (hence never adjacent) regardless of the swaps. If your paper insists on exactly one answer, choose \(\boxed{\text{(b)}}\) and annotate the ambiguity.