In a two digit number, the digit in the units place is greater than twice the digit in the tens place by 1. If the digits in the units and tens places are interchanged, the difference between the new number and the original number is exactly 1 less than the original number. What is the original number?

Introduction / Context:
This is a two digit number puzzle involving conditions on the digits and the difference between the original and reversed numbers. It uses basic algebra to represent the number and its digits, and then applies the given relationships to find the original number. Such problems are popular in aptitude tests to check number sense and equation forming skills.

Given Data / Assumptions:
- The number is a two digit number with tens digit x and units digit y.
- The digit in the units place is greater than twice the tens digit by 1, so y is 2x plus 1.
- When the digits are interchanged, a new two digit number is formed.
- The difference between the new number and the original number is exactly 1 less than the original number.
- We must find the original number.

Concept / Approach:
Let the tens digit be x and the units digit be y. This makes the original number 10x + y. The units digit condition allows us to express y in terms of x. When the digits are interchanged, the new number becomes 10y + x. We are told that the difference between the new number and the original number equals the original number minus 1. This information gives an equation that can be solved for x, and then y follows from the first condition.

Step-by-Step Solution:
Step 1: Let the tens digit be x and the units digit be y. Then the original number is 10x + y.Step 2: The digit in the units place is greater than twice the tens digit by 1, so y = 2x + 1.Step 3: When the digits are interchanged, the new number becomes 10y + x.Step 4: The problem states that the difference between the new number and the original number is 1 less than the original number, so (10y + x) - (10x + y) = (10x + y) - 1.Step 5: Simplify the left side: 10y + x - 10x - y = 9y - 9x.Step 6: The right side is 10x + y - 1. So we get 9y - 9x = 10x + y - 1.Step 7: Rearrange to 9y - 9x - 10x - y + 1 = 0, which simplifies to 8y - 19x + 1 = 0.Step 8: Substitute y = 2x + 1 into 8y - 19x + 1 = 0 to get 8(2x + 1) - 19x + 1 = 0.Step 9: Expand: 16x + 8 - 19x + 1 = 0, which gives -3x + 9 = 0, so 3x = 9 and x = 3.Step 10: Then y = 2x + 1 = 2 * 3 + 1 = 7. The original number is 10 * 3 + 7 = 37.

Verification / Alternative check:
With the original number 37, the digits are 3 and 7. Twice the tens digit is 6, and the units digit 7 is greater than 6 by 1, which satisfies the first condition. Reversing the digits gives the new number 73. The difference between the new and original numbers is 73 - 37 = 36. The original number minus 1 is 37 - 1 = 36. Thus, the second condition is also satisfied. This confirms that 37 is the correct original number.

Why Other Options Are Wrong:
Numbers such as 35, 36, 39, or 47 do not satisfy the conditions. For example, in 39, twice the tens digit is 6, but the units digit 9 is greater than 6 by 3, not 1. In 47, twice the tens digit is 8, but the units digit 7 is less than 8. Checking the reversed number differences also fails to match the requirement that the difference equals the original number minus 1.

Common Pitfalls:
Some learners may misinterpret the phrase "more than twice" and write y = 2x - 1 instead of y = 2x + 1. Others confuse which number is subtracted from which when forming the difference. It is important to read the problem carefully and translate each phrase correctly into algebraic equations before solving.

Final Answer:
The original two digit number is 37.