Question

The digit in the tens place of a two digit number is more than the digit in the unit place.

Let the digit at units place be b then the number is​

Answer 1

Given: The digit in the tens place of a two digit number is 3 more than the digit in the unit place. Let the digit at units place be b.

To find: The number.

Solution:

Let the digit in the tens place be equal to x. So the number can be written as follows.

$10x + b$

According to the question, the digit in the tens place is 3 more than the digit in the units place. This can be represented as follows.

$x = 3b$

Thus, the number can now be written as follows.

$10(3b) + b = 30 b + b$

                $= 31 b$

So the number is the product of 31 and b.

Therefore, the number is 31b.

Although part of your question is missing, you might be referring to this full question:

The digit in the ten's place of a two-digit number is 3 more than the digit in the units place. Let the digit at the unit's place be b. Then the number is

Answer 2

Number is 11b + 30 if The digit in the tens place of a two digit number is 3 more than the digit in the unit place and digit at units place is b

Given:

• A two digit number
• The digit in the tens place  is 3 more than the digit in the unit place.
• Digit at unit place is b

To Find:

• Number

Solution:

Step 1:

Add 3 in b to find digit at tens place

3 + b

Step 2:

Find the number by multiplying tens digit with 10 and Adding unit digit.

10 (3 + b) + b

Step 3:

Simplify the expression:

30 + 10b + b

= 11b + 30

Number is 11b + 30 if The digit in the tens place of a two digit number is 3 more than the digit in the unit place and digit at units place is b

( Question is Missing information that how much more is tens digit than unit digit but Probably Correct Question is tens place of a two digit number is 3 more than the digit in the unit place )

Refer this : https://brainly.in/question/16965561