243 has been divided into three parts such that half of the first part, one - third of the second part and one - fourth of the third part are equal. The largest part is

We are given that 243 has been divided into three parts, say $x$, $y$, and $z$. According to the problem:

• Half of the first part: $\frac{x}{2}$
• One-third of the second part: $\frac{y}{3}$
• One-fourth of the third part: $\frac{z}{4}$

We are told that these three expressions are equal:

2x​=3y​=4z​

Let this common value be $k$. Therefore:

$$\frac{y}{3} = k \quad \Rightarrow \quad y = 3k$$ $$\frac{z}{4} = k \quad \Rightarrow \quad z = 4k$$

Now, the sum of the parts is given as 243:

x+y+z=243

Substituting the values of $x$, $y$, and $z$ in terms of $k$:

2k+3k+4k=243

Simplifying the equation:

9k=243

Solving for $k$:

k=9243​=27

Now, we can find the values of $x$, $y$, and $z$:

$$y = 3k = 3 \times 27 = 81$$ $$z = 4k = 4 \times 27 = 108$$

Thus, the largest part is $z = 108$.