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diamond broke into four

by sanju09 » Fri May 07, 2010 2:23 am
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 $70,000 less. What is the original price of the diamond?
(A) $100000
(B) $140000
(C) $160000
(D) $200000
(E) $210000
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by liferocks » Fri May 07, 2010 2:27 am
let price of diamond as kx^2 where k is a contant

total price for 4 pieces

kx^2[1+4+9+16]=30kx^2

price of original diamond=100kx^2

difference 70kx^2=$70000 or kx^2=1000

original price of the diamond=100*1000=100000

Ans option A
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by clock60 » Sat May 08, 2010 1:42 pm
sanju09 wrote: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 $70,000 less. What is the original price of the diamond?
(A) $100000
(B) $140000
(C) $160000
(D) $200000
(E) $210000
the same answer but a little bit more detailed
we are given that cost=(weight)^2-directly as the square of the weight
and total piece was broken into four pieces
first let us find the total weight of initial piece
1k+2k+3k+4k=10k where k is +ve integer

so cost of intial piece=(10k)^2

then, let us find the cost of other pieces
p1. with weight k
cost=(k)^2

p2, with weight 2k
cost=(2k)^2=4k^2

p3. with weight 3k
cost=(3k)^2=9k^2

p4, with weight 4k
cost=(4k)^2=16k^2

total cost of pieces=k^2+4k^2+9k^2+16k^2=30k^2

and we are given that the value of broken pieces is 70000 less than initial total one

30k^2=100k^2-70000
70k^2=70000
k^2=1000

to find value of initial diamond simply insert k^2=1000 in expression fo its value
cost=100k^2=100*1000=100000
so A
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by shashank.ism » Sat May 08, 2010 6:55 pm
sanju09 wrote: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 $70,000 less. What is the original price of the diamond?
(A) $100000
(B) $140000
(C) $160000
(D) $200000
(E) $210000
Let the price of the diamond be mw^2 where w is the weight.
Let W be the weight of original diamone
so original price = mW^2
also weights of pieces = W/10, 2W/10, 3W/10,4W/10
so mW^2 - [m(W/10)^2+m(2W/10)^2+m(3W/10)^2+m(4W/10)^2] =70000
mW^2 [ (100-1-4-9-16)/100 ] = 70000
--> mW^2 = 70000x100/70[spoiler] = 100000 $ Ans A[/spoiler]
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by harshavardhanc » Sun May 09, 2010 4:24 am
sanju09 wrote: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 $70,000 less. What is the original price of the diamond?
(A) $100000
(B) $140000
(C) $160000
(D) $200000
(E) $210000
If the original weight is 1 unit, the new weights will be 1/10 , 2/10, 3/10, and 4/10 of the original value.

hence, if 1^2 = C (price at which it was sold) ---- (1)

(1/10)^ + (2/10)^2 + (3/10)^2 + (4/10)^2 = C-70 (let 10000 be invisible ;) ) --- (2)

(2)/(1) will yield

(1^2 + 2^2 + 3^2 + 4^2)/10^2 = (C-70)/C

30/100 = 1 - 70/C

70/C = 7/10

C = 100 (with the invisible 10,000)

or the price was $100,000 . option A
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by Stuart@KaplanGMAT » Sun May 09, 2010 12:53 pm
sanju09 wrote: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 $70,000 less. What is the original price of the diamond?
(A) $100000
(B) $140000
(C) $160000
(D) $200000
(E) $210000
Same answer with a lot less algrebra!

We can see from the ratio that there are a total of 10 parts. So, the original value is 10^2 = 100

The new values are 1^2 + 2^2 + 3^+ 4^2 = 30

Therefore, 70000 represents a 70% loss of value.

70000 = 7/10(x)

100000 = x
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by sanju09 » Sun May 09, 2010 9:59 pm
Stuart Kovinsky wrote:
sanju09 wrote: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 $70,000 less. What is the original price of the diamond?
(A) $100000
(B) $140000
(C) $160000
(D) $200000
(E) $210000
Same answer with a lot less algrebra!

We can see from the ratio that there are a total of 10 parts. So, the original value is 10^2 = 100

The new values are 1^2 + 2^2 + 3^+ 4^2 = 30

Therefore, 70000 represents a 70% loss of value.

70000 = 7/10(x)

100000 = x
well...that's the kewlest one
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by shashank.ism » Sun May 09, 2010 10:26 pm
Stuart Kovinsky wrote:
sanju09 wrote: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 $70,000 less. What is the original price of the diamond?
(A) $100000
(B) $140000
(C) $160000
(D) $200000
(E) $210000
Same answer with a lot less algrebra!

We can see from the ratio that there are a total of 10 parts. So, the original value is 10^2 = 100

The new values are 1^2 + 2^2 + 3^+ 4^2 = 30

Therefore, 70000 represents a 70% loss of value.

70000 = 7/10(x)

100000 = x
thats really a nice approach. Its same thing except you omitted the variables before writing it...it would really save time..thanks..
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