The cost of gold varies directly as the cube of its weight. A gold piece weighing 20 decigram costs ₹1,000. If it is broken into two pieces whose weights are in the ratio 2 : 3, then what is the profit or loss incurred?

Correct — D, ₹720 loss. Cost varies as the cube of weight, so cost = k × weight³. From the whole piece, 1000 = k × 20³ = 8000k, giving k = 0.125. Broken in the ratio 2 : 3, the twenty decigrams split into 8 and 12. The two pieces are then worth 0.125 × 8³ = 0.125 × 512 = ₹64 and 0.125 × 12³ = 0.125 × 1728 = ₹216, a total of ₹280. The whole piece was worth ₹1,000, so breaking it destroys 1000 − 280 = ₹720 of value, which is a loss. There is a shortcut that skips k entirely: the pieces are 2/5 and 3/5 of the weight, so their values are (2/5)³ + (3/5)³ = (8 + 27)/125 = 35/125 = 7/25 of the original, and 7/25 of ₹1,000 is ₹280.

• (a)₹280 profit — ₹280 is the value of the two pieces put together, not the change in value. Reporting it as a profit reverses the direction as well.
• (b)₹280 loss — The direction is right but the figure is the wrong one. The loss is the original value minus the new value, that is 1000 − 280 = ₹720.
• (c)₹720 profit — The right figure with the wrong sign. Because the cube function grows faster than the quantity itself, splitting a piece always reduces total value; there is no way to gain by breaking it.

When a quantity varies as the cube of another, splitting the second quantity always reduces the total of the first. The reason is that for positive a and b, (a + b)³ is greater than a³ + b³, since the expansion carries the extra terms 3a²b and 3ab². This is why a diamond or a gold piece is worth less when broken, and why questions of this type always end in a loss unless the pieces are being joined rather than split.

The fraction method is worth adopting because it dispenses with the constant altogether. If a piece splits in the ratio m : n, the surviving fraction of the value is (m³ + n³) divided by (m + n)³. For 2 : 3 that is 35/125, or 7/25, so 72 per cent of the value is destroyed. The same formula answers every version of this item — for a 1 : 1 split it gives 2/8, a quarter of the original, which is the classic result that halving a diamond leaves three-quarters of its value gone.

• Cost = k × weight³; from 1000 = k × 8000, k = 0.125.
• A 20 decigram piece split in the ratio 2 : 3 gives pieces of 8 and 12 decigrams.
• The pieces are worth 0.125 × 512 = ₹64 and 0.125 × 1728 = ₹216, totalling ₹280.
• The loss is 1000 − 280 = ₹720.
• For a split in the ratio m : n the surviving fraction of value is (m³ + n³) ÷ (m + n)³ — here 35/125 = 7/25.

• Reporting ₹280, the value of the pieces, as the answer instead of the change in value.
• Getting the sign wrong; a split under cube variation is always a loss.
• Dividing 20 in the ratio 2 : 3 as 10 and 10, or as 2 and 3.

Asked as a variation problem wearing a profit-and-loss costume. Both the correct figure with the wrong sign and the intermediate figure are offered as options.

• practice — not a real PYQ

  A diamond whose cost varies as the square of its weight is broken into two equal halves. What percentage of its value is lost?

  • (a)25%
  • (b)50%
  • (c)60%
  • (d)75%

  Answer(b) 50% — the halves are worth (1² + 1²) ÷ 2² = 1/2 of the original, so half the value goes.

• practice — not a real PYQ

  The cost of a gem varies as the cube of its weight. If a gem is broken into two pieces in the ratio 1 : 2, what fraction of its original value survives?

  • (a)1/3
  • (b)1/2
  • (c)1/9
  • (d)2/9

  Answer(a) 1/3 — (1³ + 2³) ÷ (1 + 2)³ = 9/27 = 1/3.