There are 48 students in a class with the age of one of the students being twice that of the age of another. If these two students are replaced by two others whose ages are 16 years and 11 years respectively, the average age of the class increases by 1.5 month. Find the age of the younger of the two students.

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the problem
We have 48 students in a class. The age of one student is twice that of another. We will denote the age of the younger student as \( x \) and the age of the older student as \( 2x \).

### Step 2: Calculate the total age of the class before replacement
Let the total age of the class before the replacement be \( T \). The total age can be expressed as:
\[ T = \text{(sum of ages of 46 students)} + x + 2x = T + 3x \]

### Step 3: Find the total age increase after the replacement
When the two students are replaced by two others aged 16 and 11 years, the total age of the class becomes:
\[ T' = T - (x + 2x) + 16 + 11 = T - 3x + 27 \]

### Step 4: Calculate the average age increase
The average age of the class increases by 1.5 months. Since there are 48 students, the total increase in age is:
\[ \text{Total increase} = 1.5 \times 48 = 72 \text{ months} \]
To convert months into years, we divide by 12:
\[ \text{Total increase in years} = \frac{72}{12} = 6 \text{ years} \]

### Step 5: Set up the equation for total age
Now, we can set up the equation for the total age before and after the replacement:
\[ T - 3x + 27 = T + 6 \]
Subtract \( T \) from both sides:
\[ -3x + 27 = 6 \]

### Step 6: Solve for \( x \)
Now, we solve for \( x \):
\[ -3x = 6 - 27 \\ -3x = -21 \\ x = 7 \]

### Conclusion
The age of the younger student is \( 7 \) years.