A) 6
B) 8
C) 4
D) None of the above
Correct Answer: B
Solution :
Let the digit at unit's place be x and the digit at ten's place = y such that x < y. Two digit number = 10y + x Product of the digit = xy = 24 Difference between digits = \[y-x=5\] ... (i) \[\therefore \,\,\,\,\,x<y\therefore {{\left( x+y \right)}^{2}}={{\left( x-y \right)}^{2}}+4xy\] \[=\,\,\,\,\,\,{{5}^{2}}+4\times 24=121\] \[\Rightarrow \,\,\,\,\,\,x+y=11\] ... (ii) Adding (i) & (ii), we have \[2y=16\Rightarrow y=8\] Shortcut method:- Product of digits \[=24=8\times 3\] Difference between digits = 5 Which satisfies the question Hence, digit at ten's place = [8]You need to login to perform this action.
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