To solve this problem, let's break it down step by step:
1. Define the Digits:
- Let the digit in the unit's place be [tex]\(x\)[/tex].
- According to the problem, the digit in the ten's place is twice the digit in the unit's place, so it will be [tex]\(2x\)[/tex].
2. Form the Two-Digit Number:
- A two-digit number can be expressed in terms of the ten's place and the unit's place. Hence, the number formed will be [tex]\(10 \times (2x) + x = 20x + x = 21x\)[/tex].
3. Subtraction from the Number:
- When 27 is subtracted from this number, the digits are reversed. If the original number is 21x, the reversed number will have the digits of the original number swapped. Thus, the ten's digit of the reversed number will be [tex]\(x\)[/tex] and the unit's digit will be [tex]\(2x\)[/tex].
- Hence, the reversed number can be expressed as [tex]\(10 \times x + 2x = 10x + 2x = 12x\)[/tex].
4. Set Up the Equation:
- According to the problem, subtracting 27 from the original number gives us the reversed number. Hence, we can create the equation:
[tex]\[
21x - 27 = 12x
\][/tex]
5. Solve the Equation:
- Move all terms involving [tex]\(x\)[/tex] to one side:
[tex]\[
21x - 12x = 27
\][/tex]
[tex]\[
9x = 27
\][/tex]
- Solve for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{27}{9} = 3
\][/tex]
6. Find the Digits:
- The unit's place digit [tex]\(x\)[/tex] is 3.
- The ten's place digit [tex]\(2x\)[/tex] is [tex]\(2 \times 3 = 6\)[/tex].
7. Form the Original Number:
- Combine the digits for the original number:
[tex]\[
10 \times 6 + 3 = 60 + 3 = 63
\][/tex]
Therefore, the original two-digit number is 63.
To summarize:
- The digit in the unit's place is [tex]\(3\)[/tex].
- The digit in the ten's place is [tex]\(6\)[/tex].
- The original number is [tex]\(63\)[/tex].
