# Small cubes with edge lengths of 1/4 inch will be packed into a rectangular prism (shown below). How many small cubes are needed to completely fill the rectangular prism?

## Thank you for the answer!

Jan 6, 2018

it's a volume question

#### Explanation:

First, find the volume of the small shape. $\left(\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}\right) = \frac{1}{64}$

Then, divide the volume of the bigger shape. $\left(\frac{4.1}{2} \times 5 \times \frac{3.3}{4}\right) = \frac{675}{8}$

Finally, divide the big shape volume by the small shape volume. $\left(\frac{675}{8} \div i \mathrm{de} \frac{1}{64}\right) = 5400$

Smaller shapes will fit into bigger shapes.

Jan 6, 2018

5400

#### Explanation:

Using a word incorrectly for mathematics; it is a matter of how many 'towers' of $\frac{1}{4}$inch blocks you can fit it.

Consider the height of one 'tower'

Height $\to 3 \frac{3}{4} \text{ inches } \to \frac{15}{4}$

$\textcolor{b r o w n}{\text{So for the height we have 15 blocks}}$

Consider the 5 inches depth $\to \frac{20}{4}$

$\textcolor{b r o w n}{\text{So for the depth we have 20 'towers'}}$

Consider the $4 \frac{1}{2}$ width $\to \frac{18}{4}$

$\textcolor{b r o w n}{\text{So for the width we have 18 'towers'}}$

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One approach for working it out without a calculator.
Using a calculator is much easier.

Note that 20 is the same as $10 \times 2$ Multiplying by 10 is easy and doubling is straight forward. Taking advantage of this we have:

Total count of 'towers' $\to 18 \times 20 = 18 \times 10 \times 2 = 360$

$\textcolor{w h i t e}{}$
$\textcolor{w h i t e}{}$

Total count of blocks at 15 per 'tower' $\to 15 \times 360 = 5400$

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What follows is just a way of thinking!

 color(white)("ddddd") 15color(white)("d")->color(white)("d")[10color(white)("dv")color(white)("dd")+color(white)("dddd")5color(white)("ddd"2/2)]

color(white)("ddddd")15color(white)("d")->color(white)("d")[10color(white)("dd")color(white)("dd")+ color(white)("d") (1/2xx10)]

$360 \times 15 \to 360 \left[10 \textcolor{w h i t e}{/ \text{d")color(white)("d.")+color(white)("d}} \left(\frac{1}{2} \times 10\right)\right]$

$360 \times 15 \to \left[10 \times 360\right] + \left[\frac{1}{2} \times 10 \times 360\right]$

$\textcolor{w h i t e}{\text{d}}$

$\left(10 \times 360\right) \textcolor{w h i t e}{\text{dd.}} = 3600$
1/2xx10xx360=ul(1800 larr" Add"
$\textcolor{w h i t e}{\text{dddddddddddddd}} 5400$