NTSE Stage 2 MAT 2018 Exam Question Paper

Given that B=6 and E = 8
‌‌‌‌‌A‌‌6‌‌‌C
×‌‌‌‌‌‌D‌‌‌8
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A‌‌C‌‌‌F‌‌6
8‌‌A‌‌‌G‌‌‌0
––––––––––
F‌‌H‌‌‌F‌‌‌6
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As at the Thousand’s place A+8=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 A=1 and thus that of F = 9. Also, in the first multiplication, E×C=B, i.e., 8×C=6, So ‘C’ can either be 2 or 7. By hit and trial, we find that C=2. Therefore,
‌‌‌‌‌‌‌‌1‌‌6‌‌‌2
‌‌×‌‌‌‌‌‌D‌‌‌8
––––––––––
‌‌1‌‌‌2‌‌‌‌9‌‌‌6
+8‌‌1‌‌‌G‌‌‌0
––––––––––
‌‌‌9‌‌H‌‌‌9‌‌‌6
––––––––––
As 9+G=9, therefore, G should be ‘0’ which can be obtained when D=5, thus the difference between D & F isF–D=9–5=4