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Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
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WE:Education (Education)
The cost of gold varies directly as the cube of its weight. A gold pie
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21 Feb 2021, 02:25
21 Feb 2021, 02:25
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The cost of gold varies directly as the cube of its weight. A gold piece weighing 27 decigrams costs $1000. If it is broken into two pieces whose weights are in ratio 4 : 5, then what is the profit or loss percent?
A. 26% profit
B. 26% loss
C. 74% profit
D. 74% loss
E. None of the above
A. 26% profit
B. 26% loss
C. 74% profit
D. 74% loss
E. None of the above
The cost of gold varies directly as the cube of its weight. A gold pie
[#permalink]
21 Feb 2021, 04:46
21 Feb 2021, 04:46
1
Solution:
Let the cost be C, Weight be W & k be constant
C∝\(W^3\)
Thus, C= k\(W^3\)
Firstly, we need to find k. Substitue the given value to find k
C=1000$
W= 27
1000=k\((27)^3\)
You can use a calculator and save the value of k by using M+
\(\frac{1000}{19683}\)=k
k~0.050805
Now, we are asked to split the weight in the ratio 4:5 so that would be 12 decigrams and 15 decigrams
Substitute the above value of k and the weight to find cost.
C=0.50805*\((12)^3\)
Cost of 12 decigram gold~$87.7914
Cost of 15 decigram gold=050805*\((15)^3\)~171.46$
87.7914+171.46~259.26
Thus, \(\frac{259.26-1000}{1000}\)*100~-74%
IMO D
Hope this helps!
Let the cost be C, Weight be W & k be constant
C∝\(W^3\)
Thus, C= k\(W^3\)
Firstly, we need to find k. Substitue the given value to find k
C=1000$
W= 27
1000=k\((27)^3\)
You can use a calculator and save the value of k by using M+
\(\frac{1000}{19683}\)=k
k~0.050805
Now, we are asked to split the weight in the ratio 4:5 so that would be 12 decigrams and 15 decigrams
Substitute the above value of k and the weight to find cost.
C=0.50805*\((12)^3\)
Cost of 12 decigram gold~$87.7914
Cost of 15 decigram gold=050805*\((15)^3\)~171.46$
87.7914+171.46~259.26
Thus, \(\frac{259.26-1000}{1000}\)*100~-74%
IMO D
Hope this helps!
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
The cost of gold varies directly as the cube of its weight. A gold pie
[#permalink]
21 Feb 2021, 23:30
21 Feb 2021, 23:30
OA Explanation:
\(C ∝ W^3\)
\(C ∝ 27^3\)
\(C ∝ 19683\)
NOTE: We don't have to worry about the proportionality symbol
Now, the piece is divided into 4 : 5 i.e. 12 and 15 decigrams
\(\frac{4}{(4+5)}(27)\) and \(\frac{5}{(4+5)}(27)\)
So, \(C_1 ∝ 12^3\)
\(C_1 ∝ 1728\)
\(C_2 ∝ 15^3\)
\(C_2 ∝ 3375\)
\(C_1 + C+2 = 5103\)
Calculate % change = \(\frac{(19683-5103)}{(19683)}(100) = 74.07\) % Loss
As, \(C > C_1 + C_2\)
Hence, option D
\(C ∝ W^3\)
\(C ∝ 27^3\)
\(C ∝ 19683\)
NOTE: We don't have to worry about the proportionality symbol
Now, the piece is divided into 4 : 5 i.e. 12 and 15 decigrams
\(\frac{4}{(4+5)}(27)\) and \(\frac{5}{(4+5)}(27)\)
So, \(C_1 ∝ 12^3\)
\(C_1 ∝ 1728\)
\(C_2 ∝ 15^3\)
\(C_2 ∝ 3375\)
\(C_1 + C+2 = 5103\)
Calculate % change = \(\frac{(19683-5103)}{(19683)}(100) = 74.07\) % Loss
As, \(C > C_1 + C_2\)
Hence, option D
