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The cost of diamond varies directly as the square of its weight. Once,

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The cost of diamond varies directly as the square of its weight. Once,  [#permalink]

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New post 27 Mar 2020, 02:34
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The cost of 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 $70,000 less. What was the original price of the diamond.

A. $100,000
B. $120,000
C. $140,000
D. $200,000
E. $250,000


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The cost of diamond varies directly as the square of its weight. Once,  [#permalink]

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New post 27 Mar 2020, 03:08
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The cost, \(c\), varies directly as the weight, \(w\), of the diamond.
So \(c=w^2\) -----(1)
Diamond broke into pieces in the ratio 1:2:3:4
Hence total cost of the pieces will be: \((w/10)^2 + (2w/10)^2 + (3w/10)^2 + (4w/10)^2 = 30w^2/100 = 3w^2/10 \)----(2)
But we are given that the merchant got $70,000 less from the sale of the 4 pieces.
Hence \(w^2 - 3w^2/10 = 70,000\)
\(10w^2 - 3w^2 = 700,000\)
\(7w^2=700,000\)
\(w^2=100,000\)

From (1), we know that \(c=w^2\), hence the original price of the diamond is $100,000

The answer is A.
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The cost of diamond varies directly as the square of its weight. Once,  [#permalink]

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New post 27 Mar 2020, 03:13
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Bunuel wrote:
The cost of 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 $70,000 less. What was the original price of the diamond.

A. $100,000
B. $120,000
C. $140,000
D. $200,000
E. $250,000


Let the original weight = \(10a\)
--> Cost of original diamond = \((10a)^2 = 100*a^2\), for some value '\(a\)'

Diamond broke into four pieces with weights in the ratio 1:2:3:4
--> Weights of smaller diamonds = {\(a, 2a, 3a, 4a\)}
--> Cost of all small diamonds combined = \(a^2 + (2a)^2 + (3a)^2 + (4a)^2 = 30*a^2\)

Difference of costs = \(70,000\)
--> \(100*a^2 - 30*a^2 = 70,000\)
--> \(70*a^2 = 70,000\)
--> \(a^2 = 1000\)

Original Cost = \(100*a^2 = 100*1000\) = $\(100,000\)

Option A
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Re: The cost of diamond varies directly as the square of its weight. Once,  [#permalink]

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New post 27 Mar 2020, 04:25
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Bunuel wrote:
The cost of 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 $70,000 less. What was the original price of the diamond.

A. $100,000
B. $120,000
C. $140,000
D. $200,000
E. $250,000


Project PS Butler


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Solution:


Let 10w units be the original weight of the diamond.
    • The cost of the diamond = \(k*100w^2\) , where k is the proportionality constant
After division the cost of the diamond will be directly proportional to \(w^2\), \(4w^2\), \(9 w^2\), and \(16 w^2\)
    • The total cost = \(kw^2 + 4kw^2 + 9kw^2 + 16kw^2 = 30kw^2\)
Now, the difference in values is $ 70000.
    • \(100kw^2 – 30kw^2 = 70000\)
      o \(kw^2 = \frac{70000}{70} = 1000\)
    • The original cost of diamond = \(100*kw^2=100*1000 = $ 100000\)
Hence, the correct answer is Option A.
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The cost of diamond varies directly as the square of its weight. Once,  [#permalink]

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New post 28 Mar 2020, 14:11
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The equation for the price of the diamond is: C2 = C1 * (W2/W1)^2
Where c2: new cost; c1: previous cost; w2: new weight; w1: previous weight

Let's work backwards, from the smallest diamond and up.

We know the ratio (1:2:3:4) and thus we know the weight of the whole diamond (1+2+3+4=10)

W(diamond a) = w1
W (diamond b) = 2
W (diamond c) = 3
W (diamond d) = 4

W (original diamond) = 10

Define the cost of diamond A (w1) as c

Cost (diamond b) = c * (2/1)^2 = 4c
Cost (diamond c) = c * (3/1)^2 = 9c
Cost (diamond d) = c * (4/1)^2 = 16c
Cost (original diamond) = c * (10/1)^2 = 100c

Cost difference = 70000 = 100c - (1c + 4c + 9c + 16c) = 100c - 30c = 70c
70000 = 70c
1000 = c

Orignal diamond cost = 100c = 100000

(edit: small typo)
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Re: The cost of diamond varies directly as the square of its weight. Once,  [#permalink]

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New post 03 Jun 2020, 00:21
1:2:3:4 means weights of new pieces are x, 2x, 3x, 4x
Price = k.(weight)^2

Total weight earlier is x+2x+3x+4x = 10x

So deltaprice = k.100x^2 - k(x^2+4x^2+ 9x^2 + 16x^2)
k*70x^2 = 70000

kx^2 = 1000
Hence A
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Re: The cost of diamond varies directly as the square of its weight. Once,  [#permalink]

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New post 10 Jun 2020, 21:31
eakabuah

I totally understand this but my question is why k is not part of the equation at all. Shouldn't it be c = kw^2? Can you please explain how you eliminated k?
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Re: The cost of diamond varies directly as the square of its weight. Once,   [#permalink] 10 Jun 2020, 21:31

The cost of diamond varies directly as the square of its weight. Once,

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