Solving a 2-Digit Number Word Problem

This problem asks us to find a 2-digit number based on two conditions related to its digits and the result of adding 27 to it.

Understanding the Digits of a 2-Digit Number

A 2-digit number can be represented using its tens digit and its units digit. Let's use variables for these:

• Let the tens digit be \(t\).
• Let the units digit be \(u\).

The value of the 2-digit number is \(10 \times \text{tens digit} + 1 \times \text{units digit}\). So, the number is \(10t + u\).

When the digits are interchanged, the new number has \(u\) as the tens digit and \(t\) as the units digit. The value of the number with interchanged digits is \(10u + t\).

Setting Up Equations from the Conditions

We are given two pieces of information, which we can translate into two algebraic equations:

Condition 1: The sum of the digits of the 2-digit number is 9.

This means the sum of the tens digit and the units digit is 9.

\(t + u = 9\) (Equation 1)

Condition 2: When 27 is added to the number, the digits get interchanged.

This means the original number plus 27 equals the number with interchanged digits.

\((10t + u) + 27 = 10u + t\) (Equation 2)

Solving the System of Linear Equations

Now we have a system of two linear equations with two variables (\(t\) and \(u\)). Let's simplify Equation 2:

\(10t + u + 27 = 10u + t\)

Subtract \(t\) and \(10u\) from both sides to group the variables:

\(10t - t + u - 10u + 27 = 0\)

\(9t - 9u + 27 = 0\)

Divide the entire equation by 9 to simplify:

\(\frac{9t}{9} - \frac{9u}{9} + \frac{27}{9} = \frac{0}{9}\)

\(t - u + 3 = 0\)

Rearrange this to isolate the constant term:

\(t - u = -3\) (Equation 3)

Now we have the simplified system:

• Equation 1: \(t + u = 9\)
• Equation 3: \(t - u = -3\)

We can solve this system using the elimination method. Add Equation 1 and Equation 3:

\((t + u) + (t - u) = 9 + (-3)\)

\(t + u + t - u = 9 - 3\)

\(2t = 6\)

Now, solve for \(t\):

\(t = \frac{6}{2}\)

\(t = 3\)

Now substitute the value of \(t\) (which is 3) back into Equation 1 to find \(u\):

\(t + u = 9\)

\(3 + u = 9\)

Subtract 3 from both sides:

\(u = 9 - 3\)

\(u = 6\)

So, the tens digit \(t\) is 3 and the units digit \(u\) is 6.

Finding the Original 2-Digit Number

The original number is \(10t + u\).

Substitute the values \(t=3\) and \(u=6\):

\text{Number} = 10(3) + 6

\text{Number} = 30 + 6

\text{Number} = 36

Verification of the 2-Digit Number

Let's check if the number 36 satisfies both original conditions:

• Condition 1: Sum of digits is 9.
  Digits of 36 are 3 and 6. \(3 + 6 = 9\). This condition is satisfied.
• Condition 2: Adding 27 interchanges the digits.
  Original number = 36. Add 27: \(36 + 27 = 63\).
  The digits of 36 are 3 and 6. Interchanging them gives the number 63. This condition is satisfied.

Both conditions are met by the number 36.

Comparing with the Options

The options provided were:

• A. 45
• B. 36
• C. 18
• D. 27

Our calculated number is 36, which matches option B.

As shown in the table, only the number 36 satisfies both conditions of the problem.

Revision Table: Key Concepts

Additional Information: Number Properties

Problems involving the digits of a number are common in basic algebra. Understanding how to represent numbers based on their place values (tens, units, hundreds, etc.) is crucial. For a 3-digit number with digits \(h\), \(t\), and \(u\), the value is \(100h + 10t + u\). Interchanging digits or adding/subtracting values often leads to systems of linear equations or sometimes quadratic equations depending on the complexity.

The problem we solved is a classic example of setting up and solving linear equations derived from a word problem about number properties.