In a two-digit number, the digitat the unit's place is 1 less than twice the digit at the ten's place. If the digits at unit's and ten's place are interchanged, the difference between the new and the original number is less than the original number by 20. The original number is

Options

A. 59
B. 23
C. 35
D. 47

Answer

To solve the problem step-by-step, let's denote the digits of the two-digit number as follows:

Let:
- \( X \) = digit at the ten's place
- \( Y \) = digit at the unit's place

### Step 1: Set up the equations based on the problem statement
From the problem, we have two key pieces of information:

1. The digit at the unit's place is 1 less than twice the digit at the ten's place:
\[ Y = 2X - 1 \]

2. If the digits are interchanged, the difference between the new number and the original number is less than the original number by 20:
- The original number can be expressed as \( 10X + Y \).
- The new number after interchanging the digits is \( 10Y + X \).
- The difference is given by:
\[ (10Y + X) - (10X + Y) = (10Y + X - 10X - Y) = 9Y - 9X \]
According to the problem, this difference is equal to the original number minus 20:
\[ 9Y - 9X = (10X + Y) - 20 \]

### Step 2: Simplify the second equation
Rearranging the second equation:
\[ 9Y - 9X = 10X + Y - 20 \]
This simplifies to:
\[ 9Y - Y = 10X + 9X - 20 \]
\[ 8Y = 19X - 20 \]

### Step 3: Substitute the value of \( Y \) from the first equation into the second equation
Now, substitute \( Y = 2X - 1 \) into the equation \( 8Y = 19X - 20 \):
\[ 8(2X - 1) = 19X - 20 \]
Expanding this gives:
\[ 16X - 8 = 19X - 20 \]

### Step 4: Solve for \( X \)
Rearranging the equation to isolate \( X \):
\[ 16X - 19X = -20 + 8 \]
\[ -3X = -12 \]
\[ X = 4 \]

### Step 5: Find \( Y \)
Now that we have \( X \), we can find \( Y \):
\[ Y = 2X - 1 = 2(4) - 1 = 8 - 1 = 7 \]

### Step 6: Construct the original number
The original number is:
\[ 10X + Y = 10(4) + 7 = 40 + 7 = 47 \]

### Final Answer
The original number is **47**.