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18 Jun 2019, 00:46
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D
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The value of a precious stone is directly proportional to the cube of its weight. If a big stone broke into three parts in the ratio 1:4:5, what was the percentage drop in the value of the stone?
A. 10%
B. 40%
C. 71%
D. 81%
E. 93%
A. 10%
B. 40%
C. 71%
D. 81%
E. 93%
18 Jun 2019, 01:48
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Bunuel
Previous weight = 1x+4x+5x=10x; value= 1000x^3
Now value = x^3+64X^3+125x^3=190x^3
Deop in value is 810x^3 as against 1000x^3 previously so IMO D
18 Jun 2019, 06:11
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This is a very good question on proportionality concepts. It also tests your knowledge of interpretation of ratios and percentage change concepts.
Let V be the value of the precious stone whose weight is w. Then,
V \(\alpha\) \(w^3\).
Introducing a constant of proportionality, this becomes,
V = k * \(w^3\).
The big stone has been broken into three parts in the ratio of 1:4:5. The actual weights of these three parts could be x, 4x and 5x. This means, the weight of the big stone should have been 10x (because it’s the sum of these 3 pieces). Therefore,
Value of big stone, \(V_b\) = k \((10x)^3\) = 1000k\(x^3\)
Value of 1st piece, \(V_1\) = k\((x)^3\)
Value of 2nd piece, \(V_2\) = k\((4x)^3\) = 64k\(x^3\)
Value of 3rd piece, \(V_3\) = k\((5x)^3\) = 125k\(x^3\)
The drop in the value of the stone = 1000k\(x^3\) – (k\(x^3\) + 64k\(x^3\) + 125k\(x^3\))
= 1000k\(x^3\) – 190k\(x^3\)
.
= 810k\(x^3\).
Therefore, percentage value in the drop of the stone = 810k\(x^3\) / 1000 k\(x^3\) = 81%.
The correct answer option is D.
A point to note is that you do not have to calculate the values of k or x to solve this question. So, when you draw the first equation, do not set out with the mindset that you solve for the variables because that’s not what is required to answer this question.
Hope this helps!
Let V be the value of the precious stone whose weight is w. Then,
V \(\alpha\) \(w^3\).
Introducing a constant of proportionality, this becomes,
V = k * \(w^3\).
The big stone has been broken into three parts in the ratio of 1:4:5. The actual weights of these three parts could be x, 4x and 5x. This means, the weight of the big stone should have been 10x (because it’s the sum of these 3 pieces). Therefore,
Value of big stone, \(V_b\) = k \((10x)^3\) = 1000k\(x^3\)
Value of 1st piece, \(V_1\) = k\((x)^3\)
Value of 2nd piece, \(V_2\) = k\((4x)^3\) = 64k\(x^3\)
Value of 3rd piece, \(V_3\) = k\((5x)^3\) = 125k\(x^3\)
The drop in the value of the stone = 1000k\(x^3\) – (k\(x^3\) + 64k\(x^3\) + 125k\(x^3\))
= 1000k\(x^3\) – 190k\(x^3\)
.
= 810k\(x^3\).
Therefore, percentage value in the drop of the stone = 810k\(x^3\) / 1000 k\(x^3\) = 81%.
The correct answer option is D.
A point to note is that you do not have to calculate the values of k or x to solve this question. So, when you draw the first equation, do not set out with the mindset that you solve for the variables because that’s not what is required to answer this question.
Hope this helps!
