IPMAT Indore 2025 · Quantitative Ability · Question 13

Step 1 — compute {A}^{2}.

{A}^{2}=\left[ \begin{matrix}2 & n \\ 4 & 1\end{matrix}\right]\left[ \begin{matrix}2 & n \\ 4 & 1\end{matrix}\right]=\left[ \begin{matrix}4+4n & 3n \\ 12 & 4n+1\end{matrix}\right]

Step 2 — multiply once more to get {A}^{3}={A}^{2}A.

{A}^{3}=\left[ \begin{matrix}20n+8 & 4{n}^{2}+7n \\ 16n+28 & 16n+1\end{matrix}\right]

Step 3 — match the top-left entry first (it is the only one with no unknown letter on the right):

20n+8=27\times 4=108 \Rightarrow n=5

Step 4 — substitute n=5 and read off p, q, r.

{A}^{3}=\left[ \begin{matrix}108 & 135 \\ 108 & 81\end{matrix}\right]=27\left[ \begin{matrix}4 & 5 \\ 4 & 3\end{matrix}\right]

So q=5, p=4, r=3 (watch the positions: p is bottom-left, q is top-right).

p+q+r=4+5+3=12

Answer: 12