In the following question out of the four options only one is correct, write the correct answer.

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) 11b + 3

(d) 10b + 3

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Solution

Let digit at unit's place be b.
Then digit at ten's place = (b+3)
$∴$ The number is 10(b + 3) + b
=10b + 30 + b
=11b + 30

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In the following question out of the four options only one is correct, write the correct answer. 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) 11b + 3 (d) 10b + 3

Let digit at unit's place be b. Then digit at ten's place = (b+3) ∴ The number is 10(b + 3) + b =10b + 30 + b =11b + 30

Let digit at unit's place be b.
Then digit at ten's place = (b+3)
$∴$ The number is 10(b + 3) + b
=10b + 30 + b
=11b + 30