Question

Of the three numbers, the first is one-third of the second and twice the third. The average of these numbers is 27. The largest of these numbers is

• A. 18
• B. 36
• C. 54
• D. 108

Hint:

When numbers are given by relations, express all in one variable and use the average/sum to solve quickly.

Answer 1

The Correct Option is C

Let the three numbers be \( x, y, z \). According to the problem, we know:

• The first number ( \( x \) ) is one-third of the second number ( \( y \) ), so: \( x = \frac{y}{3} \)
• The first number ( \( x \) ) is twice the third number ( \( z \) ), so: \( x = 2z \)
• The average of the three numbers is 27, which means: \(\frac{x+y+z}{3} = 27\)

From the average condition, we have:

\(x+y+z=81\)

Substitute \( x = \frac{y}{3} \) in the equation \( x = 2z \):

\(\frac{y}{3} = 2z \implies y = 6z\)

Now substitute \( x \) and \( y \) in the equation \( x+y+z=81 \):

\(\frac{y}{3} + y + z = 81\)

Using \( y = 6z \):

\(\frac{6z}{3} + 6z + z = 81 \implies 2z + 6z + z = 81\)

\(9z = 81\)

\(z = 9\)

Now find \( y \):

\(y = 6z = 6 \times 9 = 54\)

Now find \( x \):

\(x = \frac{y}{3} = \frac{54}{3} = 18\)

The largest of these numbers is clearly \( y \), which is:

54