The digit at the unit’s place of a two-digit number is increased by 50%. And the digit at the ten’s place of the same number is increased by 100%. Now, we find the new numbers is 33 more than original number. Find the original number.Options3562636436

The problem can be solved by understanding the place value of digits in a two-digit number and applying the given conditions.

Let's denote the digit at the unit's place as 'u' and the digit at the ten's place as 't'. So, the original number can be represented as 10t + u.

According to the problem, the digit at the unit's place is increased by 50%. So, the new unit's place digit is 1.5u.

Similarly, the digit at the ten's place is increased by 100%. So, the new ten's place digit is 2t.

The new number, therefore, is 10*(2t) + 1.5u = 20t + 1.5u.

The problem states that this new number is 33 more than the original number. So, we can set up the following equation:

20t + 1.5u = 10t + u + 33.

Solving this equation gives us:

10t + 0.5u = 33.

We can multiply the entire equation by 2 to get rid of the decimal, resulting in:

20t + u = 66.

Since 't' and 'u' are digits, they must be integers between 0 and 9. By checking the options, we can see that the only number that fits this equation is 36. So, the original number is 36.