Explanation
1. Find possible values of X:
X is a 3-digit binary number with two 1s and one 0 (leading digit cannot be 0).
So, Decimal value of X=5 or 6.
2. Check the options for (Y)s:
Since all options start with 1, let Y=1d0 in base s.
Decimal Value=1×s1+d0×s0=s+d0
(Condition: Digits must be strictly less than the base s)
Option (a) 10: Y=10⟹d0=0⟹Value=s+0=s
If Value=5⟹s=5 (Valid, since digits 1, 0 < 5)
If Value=6⟹s=6 (Valid, since digits 1, 0 < 6)
Option (b) 12: Y=12⟹d0=2⟹Value=s+2
Option (c) 13: Y=13⟹d0=3⟹Value=s+3
If Value=5⟹s=2 (Invalid, digit 3 cannot exist in base 2)
If Value=6⟹s=3 (Invalid, digit 3 cannot exist in base 3)
Option (d) 14: Y=14⟹d0=4⟹Value=s+4
Correct Answer
Option (a) 10 (or b) are mathematically valid. In standard exams, (a) is preferred as (5)10=(10)5 and (6)10=(10)6.