Question:medium
\[\begin{array}{|c|c|c|} \hline 4 & 7 & 9 \\ \hline 8 & 6 & 8 \\ \hline 3 & 7 & 9 \\ \hline 35 & 49 & ? \\ \hline \end{array} \]
\[\begin{array}{|c|c|c|} \hline 4 & 7 & 9 \\ \hline 8 & 6 & 8 \\ \hline 3 & 7 & 9 \\ \hline 35 & 49 & ? \\ \hline \end{array} \]
Show Hint
When a table has a "result" row, try applying a simple operation column-wise such as \(a\times b + c\).
Updated On: Aug 18, 2026
- 63
- 89
- 81
- 64
Show Solution
The Correct Option is C
Solution and Explanation
Step 1: Let each column's top, middle and third-row entries be \(x,y,z\); check the rule \(x\times y+z\) on the first two columns: \(4\times8+3=35\) and \(7\times6+7=49\), both matching.
Step 2: Apply the same rule to the third column, where \(x=9,\ y=8,\ z=9\): \(9\times8=72\).
Step 3: Add the third-row value: \(72+9=81\).
\[ \boxed{81} \]
Step 2: Apply the same rule to the third column, where \(x=9,\ y=8,\ z=9\): \(9\times8=72\).
Step 3: Add the third-row value: \(72+9=81\).
\[ \boxed{81} \]
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