The cost of a diamond varies directly as the square of its weight. A diamond broke into four pieces with weights in the ratio 4 : 3 : 2 : 1. When the pieces were sold separately, the merchant got ₹63,000 less than the price of the original diamond. Find the original price of the diamond.

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Accepted Answer

Let the original weight of the diamond be W and the original price be P.

Since cost varies directly as the square of the weight, we have:
P = k * W^2 for some constant k.

The diamond breaks into four pieces with weights in the ratio 4:3:2:1.
Let the common factor be x, so the weights are 4x, 3x, 2x, and 1x.

Total weight: 4x + 3x + 2x + 1x = 10x.

Original weight W = 10x.

Original price P = k * (10x)^2 = k * 100x^2.

Price of pieces sold separately:
Sum = k*(4x)^2 + k*(3x)^2 + k*(2x)^2 + k*(1x)^2
= k*(16x^2 + 9x^2 + 4x^2 + 1x^2) = k * 30x^2.

Difference between original price and sum of pieces = P - Sum = k*100x^2 - k*30x^2 = k*70x^2.

Given difference = ₹63,000
=> k*70x^2 = 63,000
=> k*x^2 = 63,000 / 70 = 900.

Original price P = k*100x^2 = 100 * (k*x^2) = 100 * 900 = ₹90,000.

But the problem states the merchant got ₹63,000 less than the price of one piece, which means the difference is ₹63,000 less than the original price, so the original price is ₹90,000.

However, the options show ₹86,000 as correct answer in the problem statement, so let's re-examine:

Re-examining the problem statement: "When the pieces were sold, the merchant got ₹63,000 less than that of one piece."

This means the merchant got ₹63,000 less than the price of the original diamond (one piece).

So difference = ₹63,000

From above, difference = k*70x^2 = 63,000

k*x^2 = 900

Original price = k*100x^2 = 100 * 900 = ₹90,000

Therefore, the original price is ₹90,000.

Hence, the correct answer is ₹90,000.

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