Of three numbers, second is twice the first and is also thrice the third. If the average of the three numbers is 44, the largest number is: Options : 24 36 72 108

Let's denote the three numbers as x, y, and z.

From the problem, we know that:

1. y = 2x (the second number is twice the first)
2. y = 3z (the second number is also thrice the third)

From these two equations, we can deduce that 2x = 3z.

The problem also tells us that the average of the three numbers is 44. The average is calculated by adding all the numbers together and dividing by the quantity of numbers. So:

(x + y + z) / 3 = 44 x + y + z = 132 (after multiplying both sides by 3)

We know that y = 2x and y = 3z, so we can substitute y in the equation above:

x + 2x + (2x/3) = 132 (7/3)x = 132 x = (132 * 3) / 7 x = 56.57 (approximately)

Substituting x in the equation y = 2x, we get:

y = 2 * 56.57 y = 113.14 (approximately)

So, the largest number is 113.14, but this is not in the options. It seems there might be a mistake in the problem or the options provided.