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?

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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?

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Answer 1 · 2018-01-06T16:53:02

it's a volume question

Explanation:

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

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

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

Smaller shapes will fit into bigger shapes.

Answer 2 · 2018-01-06T17:32:25

5400

Explanation:

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

Consider the height of one 'tower'

Height $\to 3 \frac{3}{4} inches \to \frac{15}{4}$ $->3 3/4" inches "-> 15/4$

$So for the height we have 15 blocks$ $color(brown)("So for the height we have 15 blocks")$

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

$So for the depth we have 20 'towers'$ $color(brown)("So for the depth we have 20 'towers'")$

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

$So for the width we have 18 'towers'$ $color(brown)("So for the width we have 18 'towers'")$

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

Note that 20 is the same as $10 \times 2$ $10xx2$ Multiplying by 10 is easy and doubling is straight forward. Taking advantage of this we have:

Total count of 'towers' $\to 18 \times 20 = 18 \times 10 \times 2 = 360$ $->18xx20 = 18xx10xx2 = 360$

$color(white)()$
$color(white)()$

Total count of blocks at 15 per 'tower' $\to 15 \times 360 = 5400$ $->15xx360 = 5400$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
What follows is just a way of thinking!

$ddddd 15 d \to d [10 dv dd + dddd 5 ddd \frac{2}{2}]$ $color(white)("ddddd") 15color(white)("d")->color(white)("d")[10color(white)("dv")color(white)("dd")+color(white)("dddd")5color(white)("ddd"2/2)]$

$ddddd 15 d \to d [10 dd dd + d (\frac{1}{2} \times 10)]$ $color(white)("ddddd")15color(white)("d")->color(white)("d")[10color(white)("dd")color(white)("dd")+ color(white)("d") (1/2xx10)]$

$360 \times 15 \to 360 [10 / d d. + d (\frac{1}{2} \times 10)]$ $360xx15->360[10color(white)(/"d")color(white)("d.")+color(white)("d")(1/2xx10)]$

$360 \times 15 \to [10 \times 360] + [\frac{1}{2} \times 10 \times 360]$ $360xx15 ->[10xx360]+[1/2xx10xx360]$

$d$ $color(white)("d")$

$(10 \times 360) dd. = 3600$ $(10xx360)color(white)("dd.")=3600$
$1/2xx10xx360=ul(1800 larr" Add"$
$color(white)("dddddddddddddd")5400$

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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!

2 Answers

Explanation:

Explanation:

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