There are four different numbers. The average of the first three numbers is three times the fourth number, and the average of all the four numbers is 55. What is the average of the first three numbers?

Options

A. 72
B. 63
C. 60
D. 66

Answer

To solve the problem step by step, we can follow these steps:

1. **Define the Variables**:
Let the four different numbers be \( a, b, c, \) and \( d \).

2. **Set Up the Equations**:
- According to the problem, the average of the first three numbers \( a, b, c \) is three times the fourth number \( d \). This can be expressed as:
\[ \frac{a + b + c}{3} = 3d \]
- The average of all four numbers is given as 55:
\[ \frac{a + b + c + d}{4} = 55 \]

3. **Rearranging the Equations**:
- From the first equation, we can multiply both sides by 3:
\[ a + b + c = 9d \quad \text{(Equation 1)} \]
- From the second equation, we can multiply both sides by 4:
\[ a + b + c + d = 220 \quad \text{(Equation 2)} \]

4. **Substituting Equation 1 into Equation 2**:
- Substitute \( a + b + c \) from Equation 1 into Equation 2:
\[ 9d + d = 220 \]
- Simplifying this gives:
\[ 10d = 220 \]

5. **Solving for \( d \)**:
- Divide both sides by 10:
\[ d = 22 \]

6. **Finding \( a + b + c \)**:
- Substitute \( d \) back into Equation 1:
\[ a + b + c = 9 \times 22 = 198 \]

7. **Calculating the Average of the First Three Numbers**:
- The average of the first three numbers \( a, b, c \) is:
\[ \frac{a + b + c}{3} = \frac{198}{3} = 66 \]

Thus, the average of the first three numbers is **66**.