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The cost of a diamond varies directly as the square of its weight. Once, this diamond broke into four pieces with weights in the ratio 1 : 2 : 3 : 4. When the pieces were sold, the merchant got Rs. 70,000 less. Find the original price of the diamond.
- a)Rs. 1.4 lakh
- b)Rs. 2 lakh
- c)Rs. 1 lakh
- d)Rs. 2.1 lakh
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
The cost of a diamond varies directly as the square of its weight. Onc...
Cost ∝ Weight2
Let original weight = 10x
Then, original cost = 100x2k (where k is proportionality constant)
After breakage weight = 1x + 2x + 3x + 4x
After breakage cost = kx2(1 + 4 + 9 + 16) = 30x2k
Now, (100 - 30)x2k = 70,000
Or, 70x2k = 70,000
Or, x2k = 1,000
Hence, original cost = 100x2k
= 100 × 1,000 = Rs. 1,00,000
Let original weight = 10x
Then, original cost = 100x2k (where k is proportionality constant)
After breakage weight = 1x + 2x + 3x + 4x
After breakage cost = kx2(1 + 4 + 9 + 16) = 30x2k
Now, (100 - 30)x2k = 70,000
Or, 70x2k = 70,000
Or, x2k = 1,000
Hence, original cost = 100x2k
= 100 × 1,000 = Rs. 1,00,000
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Community Answer
The cost of a diamond varies directly as the square of its weight. Onc...
To solve this problem, we need to use the concept of direct variation and the given ratio of weights of the diamond pieces. Let's break down the solution step-by-step:
1. Let's assume the original weight of the diamond is 'w' and its original cost is 'c'. According to the given information, the cost of the diamond varies directly as the square of its weight. So we can write the equation as:
c = kw^2, where k is the constant of variation.
2. We are given the weights of the broken diamond pieces in the ratio 1:2:3:4. Let's assume the weights of the pieces are x, 2x, 3x, and 4x respectively.
3. Since the pieces are broken from the same diamond, the sum of their weights should be equal to the original weight:
x + 2x + 3x + 4x = w
10x = w ----(equation 1)
4. Now, let's find the cost of each piece. Since the cost varies directly as the square of the weight, we can write the equations for each piece:
c1 = kx^2
c2 = k(2x)^2 = 4kx^2
c3 = k(3x)^2 = 9kx^2
c4 = k(4x)^2 = 16kx^2
5. According to the given information, when the pieces were sold, the merchant got Rs. 70,000 less. So the total cost of the pieces is reduced by Rs. 70,000:
(c1 + c2 + c3 + c4) - 70,000 = c
6. Substituting the values of c1, c2, c3, c4 from step 4 and the value of c from step 1, we get:
(kx^2 + 4kx^2 + 9kx^2 + 16kx^2) - 70,000 = kw^2
(30kx^2) - 70,000 = kw^2 ----(equation 2)
7. Now, substitute the value of w from equation 1 into equation 2:
(30k(10x)) - 70,000 = k(10x)^2
300kx - 70,000 = 100kx^2
8. Simplify the equation:
100kx^2 - 300kx + 70,000 = 0
9. Divide the equation by 100k:
x^2 - 3x + 700 = 0
10. Use the quadratic formula to solve for x:
x = (-(-3) ± √((-3)^2 - 4(1)(700))) / (2(1))
x = (3 ± √(9 - 2800)) / 2
x = (3 ± √(-2791)) / 2
11. Since we cannot have a negative value under the square root, the equation has no real solutions. This means that the given ratio of weights is not possible.
12. Therefore, there is no valid solution to this problem and we cannot find the
1. Let's assume the original weight of the diamond is 'w' and its original cost is 'c'. According to the given information, the cost of the diamond varies directly as the square of its weight. So we can write the equation as:
c = kw^2, where k is the constant of variation.
2. We are given the weights of the broken diamond pieces in the ratio 1:2:3:4. Let's assume the weights of the pieces are x, 2x, 3x, and 4x respectively.
3. Since the pieces are broken from the same diamond, the sum of their weights should be equal to the original weight:
x + 2x + 3x + 4x = w
10x = w ----(equation 1)
4. Now, let's find the cost of each piece. Since the cost varies directly as the square of the weight, we can write the equations for each piece:
c1 = kx^2
c2 = k(2x)^2 = 4kx^2
c3 = k(3x)^2 = 9kx^2
c4 = k(4x)^2 = 16kx^2
5. According to the given information, when the pieces were sold, the merchant got Rs. 70,000 less. So the total cost of the pieces is reduced by Rs. 70,000:
(c1 + c2 + c3 + c4) - 70,000 = c
6. Substituting the values of c1, c2, c3, c4 from step 4 and the value of c from step 1, we get:
(kx^2 + 4kx^2 + 9kx^2 + 16kx^2) - 70,000 = kw^2
(30kx^2) - 70,000 = kw^2 ----(equation 2)
7. Now, substitute the value of w from equation 1 into equation 2:
(30k(10x)) - 70,000 = k(10x)^2
300kx - 70,000 = 100kx^2
8. Simplify the equation:
100kx^2 - 300kx + 70,000 = 0
9. Divide the equation by 100k:
x^2 - 3x + 700 = 0
10. Use the quadratic formula to solve for x:
x = (-(-3) ± √((-3)^2 - 4(1)(700))) / (2(1))
x = (3 ± √(9 - 2800)) / 2
x = (3 ± √(-2791)) / 2
11. Since we cannot have a negative value under the square root, the equation has no real solutions. This means that the given ratio of weights is not possible.
12. Therefore, there is no valid solution to this problem and we cannot find the
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The cost of a diamond varies directly as the square of its weight. Once, this diamond broke into four pieces with weights in the ratio 1 : 2 : 3 : 4. When the pieces were sold, the merchant got Rs. 70,000 less. Find the original price of the diamond.a)Rs. 1.4 lakhb)Rs. 2 lakhc)Rs. 1 lakhd)Rs. 2.1 lakhCorrect answer is option 'C'. Can you explain this answer?
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