Understanding the Diamond Cost Problem

This question involves a concept of direct variation. We are told that the cost of a diamond is directly proportional to the square of its weight. This means if the weight increases, the cost increases much faster.

Let \(C\) be the cost of the diamond and \(W\) be its weight. The relationship can be written as:

\(\text{Cost} \propto (\text{Weight})^2\)

Or, using a constant of proportionality, let's say \(k\) (to avoid confusion with the weight ratio multiplier):

\(C = kW^2\)

where \(k\) is a constant value.

Calculating the Original Diamond Weight and Value

The original diamond broke into four pieces with weights in the ratio \(1:2:3:4\). Let the common ratio multiplier be \(x\). So, the weights of the four pieces are \(1x, 2x, 3x, 4x\).

The total weight of the original diamond is the sum of the weights of these pieces:

Original Weight \(= 1x + 2x + 3x + 4x = (1+2+3+4)x = 10x\)

Let's denote the original weight as \(W_{original} = 10x\).

Using the cost relationship \(C = kW^2\), the price of the original diamond was:

Original Price \(= k \times (W_{original})^2 = k \times (10x)^2 = k \times 100x^2 = 100kx^2\)

Calculating the Value of Broken Diamond Pieces

When the diamond breaks, its total value changes because the cost is based on the weight of *each piece* squared, not the total weight squared. The value of the four pieces are calculated separately based on their individual weights:

• Value of Piece 1 (weight \(1x\)) \(= k \times (1x)^2 = kx^2\)
• Value of Piece 2 (weight \(2x\)) \(= k \times (2x)^2 = k \times 4x^2 = 4kx^2\)
• Value of Piece 3 (weight \(3x\)) \(= k \times (3x)^2 = k \times 9x^2 = 9kx^2\)
• Value of Piece 4 (weight \(4x\)) \(= k \times (4x)^2 = k \times 16x^2 = 16kx^2\)

The total value of the broken diamond (sum of the values of the pieces) is:

Total Value of Pieces \(= kx^2 + 4kx^2 + 9kx^2 + 16kx^2 = (1 + 4 + 9 + 16)kx^2 = 30kx^2\)

Determining the Loss in Diamond Value

The loss in total value is the difference between the original price of the single diamond and the sum of the values of the broken pieces.

Loss \(= \text{Original Price} - \text{Total Value of Pieces}\)

Loss \(= 100kx^2 - 30kx^2 = 70kx^2\)

We are given that the loss in total value was Rs. 70000.

So, \(70kx^2 = 70000\)

Dividing both sides by 70, we get:

\(kx^2 = \frac{70000}{70} = 1000\)

Finding the Original Price of the Diamond

The original price of the diamond was \(100kx^2\). We have just found that \(kx^2 = 1000\).

Original Price \(= 100 \times (kx^2) = 100 \times 1000 = 100000\)

Therefore, the price of the original diamond was Rs. 100000.

Summary of Calculation Steps

Revision Table: Diamond Cost and Weight

Additional Information: Properties of Ratios and Proportion

In problems involving ratios, it is common to represent the actual quantities using a common multiplier (like \(x\) or \(k\) in this problem). If quantities are in the ratio \(a:b:c\), they can be represented as \(ax, bx, cx\) for some constant \(x\). The sum of these quantities would be \((a+b+c)x\). This method helps convert ratio relationships into algebraic equations that can be solved.

Understanding how different quantities vary (directly, inversely, directly as a square, etc.) is crucial for solving problems involving proportionality. The constant of variation (\(k\)) links the two quantities and can be determined if a pair of values for the quantities is known.