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# A carpenter constructed a rectangular sandbox with a

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A carpenter constructed a rectangular sandbox with a [#permalink]

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13 Aug 2012, 07:02
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A carpenter constructed a rectangular sandbox with a capacity of 10 cubic feet. If the carpenter were to make a similar sandbox twice as long, twice as wide, and twice as high as the first sandbox, what would be the capacity, in cubic feet, of the second sandbox?

(A) 20
(B) 40
(C) 60
(D) 80
(E) 100

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Re: A carpenter constructed a rectangular sandbox with a [#permalink]

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13 Aug 2012, 07:02
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SOLUTION

A carpenter constructed a rectangular sandbox with a capacity of 10 cubic feet. If the carpenter were to make a similar sandbox twice as long, twice as wide, and twice as high as the first sandbox, what would be the capacity, in cubic feet, of the second sandbox?

(A) 20
(B) 40
(C) 60
(D) 80
(E) 100

Since given that $$volume=L*W*H=10$$, then $$new \ volume=(2L)*(2W)*(2H)=(2*2*2)*(L*W*H)=8*10=80$$.

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Re: A carpenter constructed a rectangular sandbox with a [#permalink]

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13 Aug 2012, 08:16
Let x,y & z be L,W & H
so xyz = 10 cu ft
As per question if we double the lengths on all dimensions
we get (2x)(2y)(2z)=8 xyz= 8*10= 80
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Re: A carpenter constructed a rectangular sandbox with a [#permalink]

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13 Aug 2012, 08:21
volume =l*b*h =10 cubic feet
Now we have twice as long, twice as wide, and twice as high as the first sandbox,
L=2*l
B=2*b
H=2*h
final volume =L*B*H =2*l*2*b*2*h=8*(l*b*h) =8*10=80 cubic feet

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Re: A carpenter constructed a rectangular sandbox with a [#permalink]

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13 Aug 2012, 09:16
V=LWH

Original Volume = 10

In order to keep things simple. I made Height = 5, Length = 2, and Width = 1

The second statement says double everything.

Height = 10, Length = 4, and Width = 2

V = 80
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Re: A carpenter constructed a rectangular sandbox with a [#permalink]

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17 Aug 2012, 01:02
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SOLUTION

A carpenter constructed a rectangular sandbox with a capacity of 10 cubic feet. If the carpenter were to make a similar sandbox twice as long, twice as wide, and twice as high as the first sandbox, what would be the capacity, in cubic feet, of the second sandbox?

(A) 20
(B) 40
(C) 60
(D) 80
(E) 100

Since given that $$volume=L*W*H=10$$, then $$new \ volume=(2L)*(2W)*(2H)=(2*2*2)*(L*W*H)=8*10=80$$.

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Re: A carpenter constructed a rectangular sandbox with a [#permalink]

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12 Apr 2013, 09:52
A quick note on doubling. When you double a length you have 2*L1. When you double all lengths of a rectangle you have (2*L1)(2*L2) = A. An increase of 2^2 or 4. When you double all lengths of a rectangular prism you have (2*L1)(2*L2)(2*L3) = V. An increase of 2^3 or 8.

This leads to the basic relationship:

Line: 2*original size
Rectangle: 4*original size
Rectangular Prism: 8*original size

You can do the math out or memorize this relationship to speed things up.
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Re: A carpenter constructed a rectangular sandbox with a [#permalink]

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01 Dec 2014, 16:17
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Re: A carpenter constructed a rectangular sandbox with a [#permalink]

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02 Dec 2014, 02:30

Original volume = 10 cubic feet

New volume = 10 * 2 * 2 * 2 = 80
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Re: A carpenter constructed a rectangular sandbox with a [#permalink]

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24 May 2016, 10:43
Bunuel wrote:
A carpenter constructed a rectangular sandbox with a capacity of 10 cubic feet. If the carpenter were to make a similar sandbox twice as long, twice as wide, and twice as high as the first sandbox, what would be the capacity, in cubic feet, of the second sandbox?

(A) 20
(B) 40
(C) 60
(D) 80
(E) 100

We are given a rectangular sandbox with a given capacity, which is the volume of the sandbox.

Therefore, we know that the volume of the sandbox is: (L)(W)(H) = 10 cubic feet

We then are told that the carpenter doubles the length, the width, and the height. We can represent this doubling as (2L)(2W)(2H). Thus

(2L)(2W)(2H) = (2)(2)(2)(L)(W)(H) = (2)(2)(2)(10) = 80 cubic feet

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Re: A carpenter constructed a rectangular sandbox with a   [#permalink] 24 May 2016, 10:43
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