The price of a precious stone is directly proportional to the square of its weight. Sita has a precious stone weighing 18 units. If she breaks it into four pieces with each piece having distinct integer weight, then the difference between the highest and lowest possible values of the total price of the four pieces will be 288000. Then, the price of the original precious stone is
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Official Correct Answer: B. The price of a precious stone is directly proportional to the square of its weight. Sita has a precious stone weighing 18 units. If she breaks it into four pieces with each piece having distinct integer weight, then the difference between the highest and lowest possible values of the total price of the four pieces will be $288,000. Then, the price of the original precious stone is... and there are four options given.
Okay, so first, let's break down the problem. The price is directly proportional to the square of the weight. That means if the weight is doubled, the price becomes four times. So, mathematically, if P is the price and W is the weight, then P = k * W², where k is the constant of proportionality.
Sita has a stone weighing 18 units. She breaks it into four pieces, each with distinct integer weights. So, the total weight of the four pieces must add up to 18. Each piece has a different integer weight, so we're looking for four distinct integers that sum to 18.
The key part is that the difference between the highest and lowest possible total prices is $288,000. So, we need to find the maximum possible total price and the minimum possible total price when breaking the stone into four distinct integer weights, and their difference is 288,000.
Our goal is to find the original price of the stone, which is when it's 18 units. So, once we find k, we can compute the original price as P = k * (18)².
Let me outline the steps:
1. Express the total price as the sum of the prices of each piece. Since each piece's price is proportional to the square of its weight, the total price will be k*(w1² + w2² + w3² + w4²), where w1, w2, w3, w4 are the weights of the four pieces.
2. The difference between the maximum and minimum total prices is 288,000. So, (Max Total Price) - (Min Total Price) = 288,000.
3. To find Max Total Price, we need to maximize the sum of squares of the weights. Conversely, to find Min Total Price, we need to minimize the sum of squares.
4. Since the weights are distinct integers adding up to 18, we need to find all possible combinations of four distinct integers that sum to 18, then find which combination gives the maximum sum of squares and which gives the minimum.
But wait, that might take a lot of time. Maybe there's a smarter way.
for a given sum, the sum of squares is maximized when the numbers are as spread out as possible and minimized when they are as close together as possible.
So, to maximize the sum of squares, we should have one piece as heavy as possible and the others as light as possible, given the constraints of distinct integers.
Similarly, to minimize the sum of squares, the weights should be as close to each other as possible.
Let me try to find the maximum sum of squares first.
Maximum Sum of Squares:
We need four distinct integers adding up to 18. To maximize the sum of squares, we should have one piece as heavy as possible. Let's denote the weights as a, b, c, d, where a < b < c < d.
To make d as large as possible, the other three should be as small as possible. The smallest distinct positive integers are 1, 2, 3. So, let's try that.
1 + 2 + 3 + d = 18 => d = 18 - 6 = 12.
So, the weights would be 1, 2, 3, 12. The sum of squares is 1² + 2² + 3² + 12² = 1 + 4 + 9 + 144 = 158.
Is this the maximum? Let me check if there's a larger possible d.
If I try 1, 2, 4, then d = 18 - 7 = 11. Sum of squares would be 1 + 4 + 16 + 121 = 142, which is less than 158. So, 1,2,3,12 gives a larger sum.
Similarly, 1,2,5 would give d=10, sum of squares 1+4+25+100=130, which is even less.
So, 1,2,3,12 seems to give the maximum sum of squares.
Now, let's check if there are other combinations with d=12 but different smaller numbers. For example, 1,3,4,10: sum is 18, sum of squares is 1 + 9 + 16 + 100 = 126, which is less than 158.
Alternatively, 2,3,4,9: sum is 18, sum of squares is 4 + 9 + 16 + 81 = 110, which is also less.
So, 1,2,3,12 gives the maximum sum of squares.
Now, for the minimum sum of squares.
To minimize the sum of squares, the weights should be as close to each other as possible. Since 18 divided by 4 is 4.5, the weights should be around 4 or 5.
We need four distinct integers. Let's try 3,4,5,6. Their sum is 3+4+5+6=18. Sum of squares is 9 + 16 + 25 + 36 = 86.
Is this the minimum? Let me check another combination.
What about 2,4,5,7: sum is 18, sum of squares is 4 + 16 + 25 + 49 = 94, which is higher than 86.
Another combination: 3,4,5,6 is better.
What about 3,4,5,6: sum is 18, sum of squares 86.
Alternatively, 2,3,5,8: sum is 18, sum of squares 4 + 9 + 25 + 64 = 102, which is higher.
Another one: 1,4,6,7: sum is 18, sum of squares 1 + 16 + 36 + 49 = 102, also higher.
So, 3,4,5,6 gives the minimum sum of squares.
Therefore, the maximum sum of squares is 158, and the minimum is 86.
The difference between the