The digit in the ten’s place of a two-digit number

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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