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 :243672108

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, or x = 1.5z.

The problem also tells us that the average of the three numbers is 44, so:

(x + y + z) / 3 = 44 x + y + z = 132

Substituting y = 2x and x = 1.5z into the equation, we get:

1.5z + 2(1.5z) + z = 132 4.5z = 132 z = 132 / 4.5 z = 29.33 (approximately)

Substituting z = 29.33 into y = 3z, we get:

y = 3 * 29.33 y = 88 (approximately)

So, the largest number is approximately 88. However, this is not one of the options given. It seems there might be a mistake in the problem or the options provided.