Of the three numbers, second is twice the first and also thrice the third. If the average of the three numbers is 44, the largest number Is :

Options

A. 24
B. 72
C. 36
D. 108

Answer

To solve the problem step by step, let's denote the three numbers as follows:

1. Let the first number be \( A \).
2. The second number \( B \) is twice the first number, so \( B = 2A \).
3. The third number \( C \) is such that the second number is thrice the third, so \( B = 3C \). This means \( C = \frac{B}{3} = \frac{2A}{3} \).

Now, we know that the average of the three numbers is 44. The average is calculated as:

\[ \text{Average} = \frac{A + B + C}{3} \]

Substituting the expressions for \( B \) and \( C \) in terms of \( A \):

\[ \text{Average} = \frac{A + 2A + \frac{2A}{3}}{3} \]

Combining the terms in the numerator:

\[ = \frac{3A + 2A/3}{3} \]

To combine \( 3A \) and \( \frac{2A}{3} \), we can convert \( 3A \) into a fraction with a common denominator:

\[ 3A = \frac{9A}{3} \]

Now, we can add the fractions:

\[ = \frac{\frac{9A}{3} + \frac{2A}{3}}{3} = \frac{\frac{11A}{3}}{3} = \frac{11A}{9} \]

Setting the average equal to 44:

\[ \frac{11A}{9} = 44 \]

To solve for \( A \), multiply both sides by 9:

\[ 11A = 44 \times 9 \]

Calculating \( 44 \times 9 \):

\[ 44 \times 9 = 396 \]

So we have:

\[ 11A = 396 \]

Now, divide both sides by 11:

\[ A = \frac{396}{11} = 36 \]

Now that we have \( A \), we can find \( B \) and \( C \):

1. \( B = 2A = 2 \times 36 = 72 \)
2. \( C = \frac{2A}{3} = \frac{2 \times 36}{3} = 24 \)

Now, we have:

- First number \( A = 36 \)
- Second number \( B = 72 \)
- Third number \( C = 24 \)

The largest number among \( A \), \( B \), and \( C \) is \( B \), which is 72.

Thus, the largest number is:

\[ \text{Largest number} = 72 \]