How many cubical boxes of side 9 cm can be accommodated in a cuboidal box of dimensions 18 cm x 27 cm x 9 cm ?

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Solution

The correct option is C 6
Volume of cubical box
$= 9 \times 9 \times 9 = 729 c u c m$
Volume of carton
$= 18 \times 27 \times 9 = 4, 374 c u c m$

$No. of boxes = \frac{V o l u m e o f c u b o i d a l b o x}{V o l u m e o f e a c h c u b i c a l b o x}$

$No. of boxes = 4374 \div$ 729

$No. of boxes = 6$

Let's see this using repeated subtraction

4374 - 729 = 3645
3645 - 729 = 2916
2916 - 729 = 2187
2187 - 729 = 1458
1458 - 729 = 729
729 - 729 = 0

After repeated subtraction of 729 from 4374 for 6 times, we will get 0.

So, 4374 $\div$ 729 = 6

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\times

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