Question
The value of a diamond varies directly as the square of its weight. A diamond falls and breaks into 3 pieces with their weights in the ratio 1 : 2 : 3. Had the three pieces been equal in weight, the loss in value would have been a further Rs. 4,500. Find the value of the original unbroken diamond.

Consider the weights in the three cases …

Case 1: Weight of the three pieces as they broke

Case 2: Weight of the unbroken diamond (there is no loss of weight when diamond breaks)

Case 3: Weight of the three pieces, had they been equal.

Next square the weights and work in value in the three cases.

First working on the weights in three cases …

Case 1: The pieces as they broke …. let weights be k, 2k and 3k

Case 2: The unbroken diamond … the weight would have been k + 2k + 3k i.e. 6k

Case 3: Had the the three pieces been equal, teach piece’s weight would have been 2k, 2k and 2k.

Now, it is given that value ∝ (weight)2 i.e. value = n × (weight)2

Thus, the value in the three respective cases would be …

Case 1: Value = n × k2 + n × (2k)2 + n × (3k)2 = 14nk2

Case 2: Value = n × (6k)2 = 36nk2

Case 3: Value = n × (2k)2 + n × (2k)2 + n × (2k)2 = 12nk2

Because there is a ‘further’ loss of Rs. 4500, we have 14nk2 – 12nk2 i.e. 2nk2 = 4500 … eqn(1)

And we need to find the value of 36kn2.

Directly multiplying by (1) by 18, we get the answer as 4500 × 18 = Rs. 81,000.

In the above explanation, if you notice … nk2 is just acting as the constant that changes the weight (not the actual, just the reduced weight given in the ratio) to value; and is present in all the terms. Thus, when you solve such a question second time or in exam, the nk2 can be completely ignored and the answer can be found as …

Weights: 1, 2, 3 …… 6 …… 2, 2, 2

Values: 1+4+9=14 …… 36 …… 4+4+4=12

2 parts ≡ Rs. 4500

36 parts ≡ ?

And use direct proportionality, 4500 × 36/2.