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A cashier retailer used to pack content in wooden cubical boxes painte

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A cashier retailer used to pack content in wooden cubical boxes painte  [#permalink]

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New post Updated on: 11 Feb 2020, 23:39
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A cashier retailer used to pack content in wooden cubical boxes painted on all the six outer sides. The retailer decided to triple the volume of cashews per box by tripling the width of the boxes. What was the percentage increase in the cost of paint required per box ?

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

Originally posted by BABITA100810 on 11 Feb 2020, 23:37.
Last edited by Bunuel on 11 Feb 2020, 23:39, edited 1 time in total.
Renamed the topic and edited the question.
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Re: A cashier retailer used to pack content in wooden cubical boxes painte  [#permalink]

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New post 12 Feb 2020, 00:20
the initial box had sides of (x,x,x) and volume of \(x^3\) and area of \(6x^2\)
the later box had sides of (x,x,3x) and volume of \(3x^3\) and area of \(14x^2\)

as the painting cost is proportional with the outer area,
the outer area percent increase = \(\frac{14x^2-6x^2}{6x^2} *100 = 133.33\) %
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Re: A cashier retailer used to pack content in wooden cubical boxes painte  [#permalink]

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New post 12 Feb 2020, 02:08
the width is increases from a to 3a.
so the entire surface area is changed from 6a^2 to 14a^2 (2 sides of a^2 surface area, and 4 sides of 3a^2 Surface area)
change is 8a^2
% change will be 133.33%
Option D
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Re: A cashier retailer used to pack content in wooden cubical boxes painte  [#permalink]

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New post 13 Feb 2020, 08:41
BABITA100810 wrote:
A cashier retailer used to pack content in wooden cubical boxes painted on all the six outer sides. The retailer decided to triple the volume of cashews per box by tripling the width of the boxes. What was the percentage increase in the cost of paint required per box ?

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


Let the initial dimension of the wooden box, (l, b, h) = (1, 1, 1)
—> Initial surface area = 6*1^2 = 6

Now, the width of the box is tripled.
—> New dimensions = (1, 3, 1)
—> New surface area = 2(1*3 + 3*1 + 1*1) = 14

—> % increase in cost = % increase in surface area = (14 - 6)/6*100 = 400/3 = 133.33%

Option D

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Re: A cashier retailer used to pack content in wooden cubical boxes painte   [#permalink] 13 Feb 2020, 08:41
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