NIMCET 2019 Question Paper with solutions

If $\mathit{A}=1$ then
 $1\mathit{B}1$
 $\times 11$
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 $1\mathit{B}1$
$1\mathit{B}1\times$
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$1\left(\mathit{B}+1\right)\left(\mathit{B}+1\right)1$
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$\therefore \mathit{B}+1=\mathit{C}$
So, different combinations of values of A. B and C
= (1.2, 3), (1, 3. 4), (1. 4. 5). (1, 5, 6). (1. 6, 7), (1, 7, 8). (1, 8. 9)
Hence. 7 d iffe re n t values of the sum of A. B and C arepossible.