Divide 243 into three parts, so that half of the I part, 1/3rd of the II part…

Concept: A chain of equal ratios like A/2 = B/3 = C/4 means every fraction equals the same constant k, so each part can be written as a multiple of k (here A = 2k, B = 3k, C = 4k). Once the parts are expressed this way, any other condition -- such as a fixed total -- turns into a single equation in k.

Applying this to the given parts A, B and C:

1. Let A/2 = B/3 = C/4 = k, so A = 2k, B = 3k and C = 4k.

2. Since the three parts must add up to 243: A + B + C = 243, i.e. 2k + 3k + 4k = 243.

3. Combine the terms: 9k = 243, so k = 243/9 = 27.

4. Substitute k = 27 back: A = 2(27) = 54, B = 3(27) = 81, C = 4(27) = 108.

Cross-check: 54 + 81 + 108 = 243, and half of 54, one-third of 81 and one-fourth of 108 are each 27, so both conditions given in the question hold.