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24 Mar 2014, 16:20
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E
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:
57% (02:23) correct
43%
(02:20)
wrong
based on 354
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For a certain cylinder, the diameter equals the height. If every length in this cylinder is decreased by 60%, then to the nearest integer, by what percent does the volume decrease?
(A) 22%
(B) 40%
(C) 60%
(D) 84%
(E) 94 %
(A) 22%
(B) 40%
(C) 60%
(D) 84%
(E) 94 %
This is how I am doing it but getting the incorrect answer. Can someone please help?
Case 1: Assume diameter = 10 then radius = 5 and height will be 10
Volume = pi*r^2*h = 250pi ---------------------------------------------------------(1)
Decrease height by 60 %
New height = 4
New volume = pi*25*4 = 100pi-----------------------------------------------------(2)
Volume decrease in percent = Old - new / old *100
250 pi - 100 pi/250pi * 100 = 60 % which is not correct.
Case 1: Assume diameter = 10 then radius = 5 and height will be 10
Volume = pi*r^2*h = 250pi ---------------------------------------------------------(1)
Decrease height by 60 %
New height = 4
New volume = pi*25*4 = 100pi-----------------------------------------------------(2)
Volume decrease in percent = Old - new / old *100
250 pi - 100 pi/250pi * 100 = 60 % which is not correct.
24 Mar 2014, 19:47
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Answer = E = 94%
Area of Cube = \(\pi r^2 h\)
Given that 2r = h
So the Area of cube = \(2 \pi r^3\) ................. (1)
All dimension are decreased by 60%, so new dimension is \(\frac{40r}{100}= \frac{2r}{5}\)
Now the area of cube = \(2 \pi (\frac{2r}{5})^3\)
= \(2 \pi r^3 . (\frac{2}{5})^3\) ......................... (2)
Reduction in the area = (1) - (2)
=\(2 \pi r^3 (1 - \frac{8}{125})\)
= \(2 \pi r^3 (\frac{117}{125})\) ................ (3)
Comparing equation (1) & (3) ; finding the percentage reduction
= \(\frac{\frac{117}{125} . 2 \pi r^3 . 100}{2 \pi . r^3}\)
\(= 93.6 =\approx{94}\)
Answer = E
Area of Cube = \(\pi r^2 h\)
Given that 2r = h
So the Area of cube = \(2 \pi r^3\) ................. (1)
All dimension are decreased by 60%, so new dimension is \(\frac{40r}{100}= \frac{2r}{5}\)
Now the area of cube = \(2 \pi (\frac{2r}{5})^3\)
= \(2 \pi r^3 . (\frac{2}{5})^3\) ......................... (2)
Reduction in the area = (1) - (2)
=\(2 \pi r^3 (1 - \frac{8}{125})\)
= \(2 \pi r^3 (\frac{117}{125})\) ................ (3)
Comparing equation (1) & (3) ; finding the percentage reduction
= \(\frac{\frac{117}{125} . 2 \pi r^3 . 100}{2 \pi . r^3}\)
\(= 93.6 =\approx{94}\)
Answer = E
Updated on: 08 Aug 2014, 03:28
Originally posted by PareshGmat on 24 Mar 2014, 20:00.
Last edited by PareshGmat on 08 Aug 2014, 03:28, edited 1 time in total.
Last edited by PareshGmat on 08 Aug 2014, 03:28, edited 1 time in total.
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enigma123
In this case, the equation (2) is incorrect
Problem says "If every length in this cylinder is decreased by 60%"; means height & radius both are decreased by 60%
So the new volume would be \(2 . \pi . (\frac{40}{100})^3 . 5^3\)
New volume = \(16 \pi\) ............... (2)
Reduction = \(250 \pi - 16 \pi = 234 \pi\)
Percentage reduction = \(234 * \frac{100}{250} = 93.6\)
\(\approx{94}\)
