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06 Dec 2021, 04:15
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
54% (02:13) correct
46%
(02:18)
wrong
based on 78
sessions
History
Date
Time
Result
Not Attempted Yet
A cubic meter of copper weighing 9000 kilograms is rolled into a square bar 9 metres long. An exact cube is cut off from the bar. How much will it weigh?
(A) 27.5 kg
(B) 90.6 kg
(C) 333.3 kg
(D) 900.5 kg
(E) 9000.0 kg
Are You Up For the Challenge: 700 Level Questions
(A) 27.5 kg
(B) 90.6 kg
(C) 333.3 kg
(D) 900.5 kg
(E) 9000.0 kg
Are You Up For the Challenge: 700 Level Questions
07 Dec 2021, 02:06
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Volume of cube:\(1 cm^3\)
Volume of rolled out square bar: As we don't know the the two sides, only that the bar is 9m long, let the two unknown sides = x.
Volume = \(9*x*x\) or \(9*x^2\)
As the volume doesn't change, make the two equations equal to one another: \(9*x^2 = 1\)
Solving for x: \(9*x^2 = 1\)
\(x^2 = \frac{1}{9}\)
\(x = \frac{1}{3}\)
A cube cut from the bar will have the dimensions: \(\frac{1}{3}*\frac{1}{3}*\frac{1}{3}\) with a volume of \(\frac{1}{27} cm^3\)
Which means that it's weight will be \(\frac{1}{27}th\) of the original 9000kgs.
\(\frac{9000}{27} = 333.3333\)
Answer C
Volume of rolled out square bar: As we don't know the the two sides, only that the bar is 9m long, let the two unknown sides = x.
Volume = \(9*x*x\) or \(9*x^2\)
As the volume doesn't change, make the two equations equal to one another: \(9*x^2 = 1\)
Solving for x: \(9*x^2 = 1\)
\(x^2 = \frac{1}{9}\)
\(x = \frac{1}{3}\)
A cube cut from the bar will have the dimensions: \(\frac{1}{3}*\frac{1}{3}*\frac{1}{3}\) with a volume of \(\frac{1}{27} cm^3\)
Which means that it's weight will be \(\frac{1}{27}th\) of the original 9000kgs.
\(\frac{9000}{27} = 333.3333\)
Answer C
22 Jun 2024, 03:49
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