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.

Sure, let's solve this step by step.

Step 1: Let's assume the original two-digit number is 10X + Y, where X is the digit at the tens place and Y is the digit at the units place.

Step 2: According to the problem, the digit at the unit’s place (Y) is increased by 50%. So, the new digit at the unit's place is Y + 0.5Y = 1.5Y.

Step 3: The digit at the ten’s place (X) is increased by 100%. So, the new digit at the tens place is X + X = 2X.

Step 4: The new number is therefore 10*(2X) + 1.5Y.

Step 5: According to the problem, the new number is 33 more than the original number. So, we can set up the following equation: 10*(2X) + 1.5Y = 10X + Y + 33.

Step 6: Simplifying the equation, we get 10X + 1.5Y = X + Y + 33.

Step 7: Further simplifying, we get 9X + 0.5Y = 33.

Step 8: Since X and Y are digits, they must be integers. The only way for 9X + 0.5Y to be an integer is if Y is even. So, Y can be 0, 2, 4, 6, or 8.

Step 9: Substituting these values into the equation, we find that the only solution is X = 3 and Y = 6.

So, the original number is 36.