Question:
The digit in the ten’s place of a two-digit number is 3 more than the digit in the unit’s place. Let the digit at unit’s place be b. Then, the number is
(a) 11b+30
(b) 10b+ 30
(c) 11 b + 3
(d) 10b + 3
Solution:
(a) Let digit at unit’s place be b.
Then, digit at ten’s place = (3 + b)
Number = 10 (3 + b) + b – 30 + 10b + b = 11b + 30
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