NTSE Stage 2 MAT 2018 Exam Question Paper

Given that $\mathit{B}=6$ and E = 8
$\mathrm{A}6\mathrm{C}$
$\times \mathit{D}8$
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$\mathrm{A}\mathrm{C}\mathrm{F}6$
$8\mathrm{A}\mathrm{G}0$
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$\mathrm{F}\mathrm{H}\mathrm{F}6$
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As at the Thousand’s place $\mathit{A}+8=\mathit{F}$, there is no carry over, thus the value of A can either be 0 or 1, but 0 will make it a 3-digit number. Hence, the value of $\mathit{A}=1$ and thus that of F = 9. Also, in the first multiplication, $\mathit{E}\times \mathit{C}=\mathit{B}$, i.e., $8\times \mathit{C}=6$, So ‘C’ can either be 2 or 7. By hit and trial, we find that $\mathit{C}=2.$ Therefore,
$\mathrm{1}6\mathrm{2}$
$\times \mathit{D}8$
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$1296$
$+81\mathrm{G}0$
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$9\mathrm{H}\mathrm{9}6$
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As $9+\mathit{G}=9$, therefore, G should be ‘0’ which can be obtained when $\mathit{D}=5$, thus the difference between D & F is$\mathit{F}\mathit{–}\mathit{D}=9\mathit{–}5=4$