Out of a group of three numbers, second is thrice of the first and also twice of the third. If the sum of the three numbers is 55 then find the largest number in the group.

Options

A. 26
B. 29
C. 30
D. 32

Answer

Let's solve the problem step by step.

### Step 1: Define the Variables
Let the three numbers be \( A \), \( B \), and \( C \).

### Step 2: Establish Relationships
According to the problem:
- The second number \( B \) is thrice the first number \( A \):
\[ B = 3A \]
- The second number \( B \) is also twice the third number \( C \):
\[ B = 2C \]

### Step 3: Express \( C \) in Terms of \( A \)
From the equation \( B = 2C \), we can express \( C \) in terms of \( B \):
\[ C = \frac{B}{2} \]
Substituting \( B = 3A \) into this equation gives:
\[ C = \frac{3A}{2} \]

### Step 4: Set Up the Sum Equation
The sum of the three numbers is given as 55:
\[ A + B + C = 55 \]
Substituting the expressions for \( B \) and \( C \):
\[ A + 3A + \frac{3A}{2} = 55 \]

### Step 5: Combine Like Terms
Combine the terms on the left side:
\[ 4A + \frac{3A}{2} = 55 \]
To combine these, convert \( 4A \) into a fraction:
\[ 4A = \frac{8A}{2} \]
Now, we can rewrite the equation:
\[ \frac{8A}{2} + \frac{3A}{2} = 55 \]
This simplifies to:
\[ \frac{11A}{2} = 55 \]

### Step 6: Solve for \( A \)
To eliminate the fraction, multiply both sides by 2:
\[ 11A = 110 \]
Now, divide by 11:
\[ A = 10 \]

### Step 7: Find \( B \) and \( C \)
Now that we have \( A \), we can find \( B \) and \( C \):
\[ B = 3A = 3 \times 10 = 30 \]
\[ C = \frac{3A}{2} = \frac{3 \times 10}{2} = 15 \]

### Step 8: Identify the Largest Number
The three numbers are:
- \( A = 10 \)
- \( B = 30 \)
- \( C = 15 \)

The largest number in the group is:
\[ \text{Largest number} = B = 30 \]

### Final Answer
The largest number in the group is **30**.

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