Question

The internal dimensions of a box are 1.2 m, 80 cm and 50 cm. How many cubes each of edge 7 cm can be packed in the box with faces parallel to the sides of the box. Also, find the space left empty in the box.

Answer 1

Hint: Here as we are given that the focus of the cubes is parallel to sides of the box, first of all, calculate the number of cubes that can be placed along length, breadth, and height of the box. Then multiply them to get the total number of boxes. The space left empty in the box would be the difference in the volume of the box and all the cubes.

Complete step-by-step answer:

We are given a box whose internal dimensions are 1.2 m, 80 cm and 50 cm. We have to find the number of cubes of edge 7 cm which can be packed in the box with faces of the cube parallel to the side of the box. Also, we have to find the space left empty in the box.
Let us first consider the dimensions of the given box we have.
The value of the length of the box = 1.2 m = 1.2 x 100 cm = 120 cm
The value of the breadth of the box = 80 cm
The value of the height of the box = 50 cm
We know that the volume of cuboid = L x B x H, where L, B and H are length, breadth and height of the cuboid. Since we know that our box is in cuboidal shape. So, we get the volume of the box = L x B x H.
By substituting the values of length, breadth and height of the box, we get,
The volume of the box \[=\left( 120\times 80\times 50 \right)c{{m}^{3}}\]
\[=480000\text{ }c{{m}^{3}}\]
Now, we are given that the side of the small cube is 7 cm. We know that the volume of the cube \[={{a}^{3}}\] where ‘a’ is the side of the cube.
By substituting the value of a = 7, we get,
The volume of the small cube \[={{\left( 7 \right)}^{3}}=343\text{ }c{{m}^{3}}\]
Now, we are given that the small cubes are packed such that their faces are parallel to each side of the box.
So, we get the number of cubes that are placed lengthwise \[=\dfrac{\text{Length of the box}}{\text{Side of the cube}}\]
By substituting the value of the length of the box = 120 cm and side of the cube = 7 cm, we get,
Number of cubes that are placed lengthwise \[=\dfrac{120}{7}=17.14\]
Since we know that number of cubes would be in a whole number. So, we get, number of cubes that are placed in lengthwise = 17…..(i)
Also, we get, the number of cubes that are placed breadth wise \[=\dfrac{\text{Breadth of the box}}{\text{Side of the cube}}\]
By substituting the value of the breadth of the box = 80 cm and side of the cube = 7 cm, we get,
Number of cubes that are placed breadth wise \[=\dfrac{80}{7}\approx 11....\left( ii \right)\]
Similarly, we get the number of cubes that are placed height-wise \[=\dfrac{\text{Height of the box}}{\text{Side of the cube}}\]
By substituting the values of the height of the box = 50 cm and side of the cube = 7 cm, we get,
Number of cubes that are placed height-wise \[=\dfrac{50}{7}\approx 7....\left( iii \right)\]
So, we get the total cubes that are packed in the box = (Number of cubes that are placed in lengthwise) x (Number of cubes that are placed breadthwise) x (Number of cubes that are placed height-wise)
By substituting the values of RHS from equation (i), (ii) and (iii), we get,
Total cubes that packed in the box = 17 x 11 x 7 = 1309 cubes.
Therefore, we get a total of 1309 cubes packed inside the box.
Now, we have already calculated the volume of each of the cube as \[343\text{ }c{{m}^{3}}\].
So, we get the volume of 1309 cubes \[=1309\times 343=448987\text{ }c{{m}^{3}}\]
Also, we have already calculated the volume of the box as \[\text{480000 }c{{m}^{3}}\]
So, we get the space empty in the box = (Total Volume of the box) – (Total Volume of all the cubes)
\[\begin{align}
  & =\left( 480000 \right)-\left( 448987 \right) \\
 & =31013\text{ }c{{m}^{3}} \\
\end{align}\]
Therefore, we get the total space empty in the box as \[31013\text{ }c{{m}^{3}}\].

Note: Students must remember to convert the given length in meter to centimeters because the other values are also given in centimeters. Also, some students make this mistake of taking the number of cubes in fractions but they must note that the number of cubes would be in whole numbers and that too lengthwise, breadthwise and height wise individually.