Let the three parts of the 243 kg weight be \(A\), \(B\), and \(C\).

The total weight is given as 243 kg.

So, we have the equation:

\(A + B + C = 243 \quad (1)\)

The question states that half of the first part, one-third of the second part, and one-fourth of the third part are equal. We can write this relationship as:

\(\frac{A}{2} = \frac{B}{3} = \frac{C}{4}\)

Let's assume this common equal value is \(k\). This means:

• \(\frac{A}{2} = k \implies A = 2k\)
• \(\frac{B}{3} = k \implies B = 3k\)
• \(\frac{C}{4} = k \implies C = 4k\)

Now, substitute these expressions for \(A\), \(B\), and \(C\) into equation (1):

\(2k + 3k + 4k = 243\)

Combine the terms on the left side:

\((2 + 3 + 4)k = 243\)

\(9k = 243\)

To find the value of \(k\), divide both sides by 9:

\(k = \frac{243}{9}\)

\(k = 27\)

Now that we have the value of \(k\), we can find the values of the three parts \(A\), \(B\), and \(C\):

• First part, \(A = 2k = 2 \times 27 = 54\) kg
• Second part, \(B = 3k = 3 \times 27 = 81\) kg
• Third part, \(C = 4k = 4 \times 27 = 108\) kg

So, the three parts are 54 kg, 81 kg, and 108 kg.

Let's verify if these parts satisfy the conditions:

• Do they add up to 243 kg? \(54 + 81 + 108 = 135 + 108 = 243\) kg. Yes.
• Are half of the first, one-third of the second, and one-fourth of the third equal?
  • \(\frac{A}{2} = \frac{54}{2} = 27\)
  • \(\frac{B}{3} = \frac{81}{3} = 27\)
  • \(\frac{C}{4} = \frac{108}{4} = 27\)

  Yes, they are all equal to 27.

The calculated parts satisfy both conditions.

Revision Table: Dividing Weight into Parts

Additional Information on Ratios and Proportion

This problem involves the concept of ratios and proportion. When quantities are in proportion, their relationship can be expressed using a constant factor.

If we have a relationship like \(\frac{a}{x} = \frac{b}{y} = \frac{c}{z}\), it means that the ratio of each quantity (\(a\), \(b\), \(c\)) to its corresponding divisor (\(x\), \(y\), \(z\)) is the same. We can set this common ratio equal to a constant, say \(k\).

\(\frac{a}{x} = k \implies a = xk\)

\(\frac{b}{y} = k \implies b = yk\)

\(\frac{c}{z} = k \implies c = zk\)

This method is useful when dealing with quantities that are proportional to certain numbers, as seen in this problem where the parts \(A, B, C\) are proportional to 2, 3, and 4 respectively.

The sum of the quantities can then be expressed in terms of \(k\):

\(a + b + c = xk + yk + zk = (x+y+z)k\)

If the total sum is known, we can easily find \(k\) and subsequently the individual quantities. This method simplifies solving problems involving division of a total quantity into parts based on given ratios or proportional conditions.