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A metallic cubical box was hammered and moulded into a rectangular box

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A metallic cubical box was hammered and moulded into a rectangular box  [#permalink]

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New post 14 Apr 2019, 02:47
1
1
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

57% (02:18) correct 43% (01:56) wrong based on 82 sessions

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A metallic cubical box was hammered and moulded into a rectangular box such that the only change was that the breadth got tripled. What was the percentage change in total surface area?

A. 33.33%
B. 66.67%
C. 100%
D. 133.33%
E. 200%

Source: Experts Global

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A metallic cubical box was hammered and moulded into a rectangular box  [#permalink]

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New post Updated on: 16 Apr 2019, 01:36
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Although the scenario is too hypothetical to think of, wherein just one side of the new cuboid has tripled after hammering and everything else remains unchanged, yet considering the hypothesis to be real the answer goes as below.

Tripling one side of the cuboid the surface area becomes 14x of the original surface area 6x.

Percent Change in Volume=\(\frac{8x}{6x}*100\) = 133.33%

Hence, the answer is D
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Originally posted by saukrit on 14 Apr 2019, 03:06.
Last edited by saukrit on 16 Apr 2019, 01:36, edited 1 time in total.
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A metallic cubical box was hammered and moulded into a rectangular box  [#permalink]

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New post 15 Apr 2019, 02:10
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themindful wrote:
Although the scenario is too hypothetical to think of, wherein just one side of the new cuboid has tripled after hammering and everything else remains unchanged, yet considering the hypothesis to be real the answer goes as below.

Tripling one side of the cuboid the volume become 3x of the original volume x.

Change in Volume=\(\frac{(3x-x)}{x}*100\) = 200%

Hence, the answer is E


I think the answer is D. They have asked change in surface area and not volume

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Re: A metallic cubical box was hammered and moulded into a rectangular box  [#permalink]

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New post 15 Apr 2019, 12:28
Hi the question stem asks % change in surface area.

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Re: A metallic cubical box was hammered and moulded into a rectangular box  [#permalink]

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New post 15 Apr 2019, 12:30
Will the answer be option D

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Re: A metallic cubical box was hammered and moulded into a rectangular box  [#permalink]

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New post 16 Apr 2019, 01:39
Sanjeetgujrall wrote:
themindful wrote:
Although the scenario is too hypothetical to think of, wherein just one side of the new cuboid has tripled after hammering and everything else remains unchanged, yet considering the hypothesis to be real the answer goes as below.

Tripling one side of the cuboid the volume become 3x of the original volume x.

Change in Volume=\(\frac{(3x-x)}{x}*100\) = 200%

Hence, the answer is E


I think the answer is D. They have asked change in surface area and not volume

Posted from my mobile device

Thanks for highlighting. Updated my response. My bad :)
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A metallic cubical box was hammered and moulded into a rectangular box  [#permalink]

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New post 19 Jul 2019, 04:18
Interestingly, some students wrote to us saying that if the width is changed, the volume also changed and that is not possible by hammering (as volume of the solid must remain constant). The mistake in that approach is that one is confusing volume of the "box" (not a solid, as such) with the volume of the "metal" (or solid). Imagine, the box is empty inside and practically, the volume of the box is the volume of the empty space and not the volume of the metal enclosing the empty space :-)
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Re: A metallic cubical box was hammered and moulded into a rectangular box  [#permalink]

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New post 21 Jul 2019, 09:41
Tripling one side of the cuboid the surface area becomes 14x of the original surface area 6x.
How??? Can you explain???

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Re: A metallic cubical box was hammered and moulded into a rectangular box  [#permalink]

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New post 21 Jul 2019, 22:44
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surface area = 2(Lw+wh+Lh)
original S.A. = 6a^2
new = 2(a*3a + 3a*a + a*a) = 14a^2

%change = (14a^2 - 6a^2)/ 6a^2 *100% = (8/6)*100 = 133.33% (D)
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A metallic cubical box was hammered and moulded into a rectangular box  [#permalink]

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New post 21 Jul 2019, 22:48
Saurabhminocha wrote:
surface area = 2(Lw+wh+Lh)
original S.A. = 6a^2
new = 2(a*3a + 3a*a + a*a) = 14a^2

%change = (14a^2 - 6a^2)/ 6a^2 *100% = (8/6)*100 = 133.33% (D)

thanks a lot...I get it now.

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A metallic cubical box was hammered and moulded into a rectangular box   [#permalink] 21 Jul 2019, 22:48
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