8th Class Mathematics Cube and Dice Cube and Dice

Cube and Dice

Category : 8th Class

Introduction

In this chapter students will learn the features of cube and Dice and also learn how to solve problems with the help of Cube and Dice.

 

CUBES:

A cube is a three dimensional figure, having 8 corners, 6 surfaces and 12 edges. If a cube is painted on all of its surfaces with any colour and further divided into various smaller cubes, we get the following results.

                    

(a) Smaller cubes with three surfaces painted will be present on the comers of the big cube.

(b) Smaller cubes with two surface painted will be present on the edges of the big cube.

(c) Smaller cubes with one surface painted will be present on the outer surfaces of the big cube.

(d) Smaller cubes with no surface painted will be present inside the big cube.

If a cube is painted on all of its surfaces with a color and then divided into smaller cubes of equal size, then after separation, number of smaller cubes so obtained will be calculated as under:

Number of smaller cubes with three surfaces painted = 8

Number of smaller cubes with two surfaces painted \[=(n-2)\times 12\]

Number of smaller cubes with one surface painted \[={{(n-2)}^{2}}\times 6\]

Number of smaller cubes with no surfaces painted \[={{(n-2)}^{3}}\]

Where n = No. of divisions on the surfaces of the bigger cube

\[=\frac{length\text{ }of\text{ }edge\text{ }of\text{ }big\text{ }cube}{length\text{ }of\text{ }edge\text{ }of\text{ }one\text{ }smaller\text{ }cube}\]

 

TYPE-I:  

A cube is painted on all of its surfaces with a

single colour and then divided into various smaller

cubes of equal size.

DIRECTIONS: A cube of side 4 cm. is painted black on all of its surface and then divided into various smaller cubes of side 1 cm. each. The smaller cubes so obtained are separated.

Total cubes obtained \[=\frac{4\times 4\times 4}{1\times 1\times 1}=64\]

Here,  \[\text{n=}\frac{\text{side of big cube}}{\text{side of small cube}}\text{=}\frac{\text{4}}{\text{1}}\text{=4}\]

  1. Number of smaller cubes with three surfaces painted = 8
  2. Number of smaller cubes with two surfaces painted

                \[=(n-2)\times 12=(4-2)\times 12=24\]

  1. Number of smaller cubes with one surface painted

                \[={{(n-2)}^{2}}\times 6={{(4-2)}^{2}}\times 6=24\]

  1. Number of smaller cubes with no surface painted

                \[={{(n-2)}^{3}}={{(4-2)}^{3}}={{(2)}^{3}}=8\]

 

TYPE-II:

A cube is painted on its surfaces with different colors and then divided into various smaller cubes of equal size.

 

DIRECTIONS:  A cube of side 4 cm. is painted black on pair of opposite surfaces, blue on another pair of opposite surfaces and red on the remaining pair of opposite surfaces, The cube is divided into smaller cubes of equal side of 1 cm each.

 

  1. Number of smaller cubes with three surfaces painted = 8

(These smaller cubes will have all three surfaces painted with different colours blue, black and red)

  1. Number of smaller cubes with two surfaces painted = 24.

And out of this

(a) Number of cubes with two surfaces painted with black and blue colour = 8

(b) Number of cubes with two surfaces painted with blue and red colour = 8

(c) Number of cubes with two surfaces painted with black and red colour = 8

  1. Number of smaller cubes with one surface painted = 24. And out of this

(a) Number of cubes with one surface painted with black colour = 8

(b) Number of cubes with one surface painted with blue colour = 8

(c) Number of cubes with one surface painted with red colour =8

 

TYPE-III:

A cube is painted on its surfaces in such a way that one pair of opposite surfaces is left unpainted.

 

DIRECTIONS: A cube of side 4cm. is painted red on one pair of opposite surface, green on another pair of opposite surfaces and one pair of opposite surfaces is left unpainted. Now the cube is divided into 64 smaller cubes of side 1 cm. each

  1. Number of smaller cubes with three surfaces painted = 0

(Because each smaller cube at the comer is having a surface which is not painted)

  1. Number of smaller cubes with two surfaces painted = Number of cubes present at the comers + Numbers of cubes present at the 4 edges \[=8+(n-2)\times 4=8+8=16\]

                             

 

  1. Number of smaller cubes with one surface painted.

= Number of cubes present at 8 edges + number of cubes present at the four surfaces

\[=(n-2)\times 8+{{(n-2)}^{2}}\times 4=2\times 8+4\times 4=32\]

  1. Number of smaller cubes with no side painted.

= Number of cubes on the two unpainted surfaces + number of cubes present inside the cube

\[={{(n-2)}^{2}}\times 2+{{(n-2)}^{3}}=2\times 4+8=16\]

 

TYPE-IV:

A cube is painted on its surfaces in such a way that one pair of adjacent surfaces is left unpainted.

 

DIRECTIONS: A cube of side 4 cm. is painted red on one pair of adjacent surface, green on the other pair of adjacent surfaces and two adjacent surfaces are left unpainted. Now the cube is divided into 64 smaller cubes of side 1 cm. each.

 

  1. Number of smaller cubes with three surfaces painted = Number of smaller cubes at two comers = 2
  2. Number of smaller cubes with two surface painted = Number of smaller cubes at four corners + Numbers of smaller cubes at 5 edges

                \[=4+(n-2)\times 5=4+2\times 6=4+10=14\]

  1. Number of smaller cubes with one surface painted = Number of smaller cubes at four surfaces + Number of cubes present at 6 edges + Number of smaller cubes at two comers

                \[={{(n-2)}^{2}}\times 4+(n-2)\times 6+2\]

                \[=4\times 4+2\times 6+2=16+12=28+2=30\]

  1. Number of smaller cubes with no surfaces painted = Number of smaller cubes from inside the big cube + Number of cubes at two surfaces + Number of cubes at one edge

                \[={{(n-2)}^{3}}+{{(n-2)}^{3}}\times 2+(n-2)\]

                \[={{(2)}^{3}}+{{(2)}^{2}}+2=8+8+2=18\]

 

EXAMPLE 1:

Count the number of cubes in the given figure.

(a) 6                                       (b) 8

(c) 10                                     (d) 12

Sol. (c) Clearly, there is 1 column containing 3 cubes, 2 columns containing 2 cubes each and 3 columns containing 1 cube each.

 

EXAMPLE 2:

Select from the alternatives the box that can be formed by folding the sheet shown in figure (X):

                               

                (a)                         (b)

                (c)                         (d)

 

Sol. (d) When the sheet in fig.(X) is folded to from a box (cube) then

The number 2 will lie opposite the number 4; the number 1 will lie opposite the number 6 and the number 5 will lie opposite the number 3. Fig. (a) has the numbers 1 and 6 on adjacent face, fig. (b) has number 3 and 5 on adjacent faces and the fig. (c) has the numbers 2 and 4 on the adjacent face. So, these three alternatives are not possible .Since, the numbers 1, 3 and 4 can appear on adjacent face, so fig.(d) is possible.

Hence, only the box shown in fig. (d) can be formed by folding fig. (X).

 

EXAMPLE 3:

Count the number of cubes in the given figure.

(a) 68                                     (b) 69

(c) 70                                     (d) 71

Sol. (b) In the figure, there are 11 columns containing 4 cubes each, 7 columns containing 3 cubes each and 2 columns containing 2 cubes each .

\[\therefore \]  Total number of cubes

\[=(11\times 4)+(7\times 3)+(2\times 2)=44+21+4=69.\]

 

DICE

 

EXAMPLE4:

A dice has been thrown four times and produces the following results.    

          

Which number will appear opposite to the number 3?

(a) 4                                       (b) 5

(c) 6                                       (d) 1

Sol. (a) From the figure (i), (ii) and (iv) we find that numbers 6, 1, 5 and 2 appear on the adjacent surfaces to the number 3. Therefore, number 4 will be opposite to number 3.

Hence option (a) is the answer.

 

CATEGORY 2:

 

EXAMPLE 5:

The figures given below show the two different positions of a dice. Which number will appear opposite to number 2.

 

                         

(a) 3                                       (b) 4

(c) 5                                       (d) 6

Sol. (c) The above question, where only two positions of a dice are given, can easily be solved with the following method.

Step I:   The dice, when unfolded, will appear as shown in the figure given on the right side.

Step II: Write the common number to both the dice in the middle block. Since common number is 4, hence number 4 will appear in the central block.

Step III: Consider the figure (i) and write the first number in the anti-clockwise direction of number 4, (common number) in block I and second number in block II. Therefore, numbers 3 and 2 being the first and second number to 4 in anticlockwise directions respectively, will appear in block I and II respectively.

Step IV: Consider figure (ii) and write first and second number in the anticlockwise direction to number 4, (common number) in block (III) and (IV). Hence numbers 6 and 5 will appear in the blocks III and IV respectively.

Step V: Write remaining number in the remaining block.

Therefore, number 1 will come in the remaining block. Now, from the unfolded figure we find that number opposite to 6 is 3, number opposite to 2 is 5 and number opposite to 4 is 1. Therefore, option (c) is our answer.

 

CATEGORY 3:

EXAMPLE6:

From the following figures of dice, find which number will come in place of '?'

                                   

(a) 4                                       (b) 5

(c) 2                                       (d) 3

Sol. (d) If the above dice is unfolded, it will look like as the figure (a) given below.

.

               

 

Now the number in place of '?' can be obtained by making a slight change in the figure as given here. Now comparing figure (b) with figure (iii) as above, we get that number in place of is 3.

 

CATEGORY 4:

 

EXAMPLE 7:

Which of the following dice is identical to the unfolded figure as shown here?

(a)                    (b)

(c)                    (d)

 Sol.         (a) From the unfolded figure of dice, we find that number opposite to 2 is 4, for 5 it is 3 and for 1 it is 6. From this result we can definitely say that figure (b), (c) and (d) cannot be the answer figure as numbers lying on the opposite pair of surfaces are present on the adjacent surfaces.

Hence figure (a) is our answer.

 

Miscellaneous Solved Examples

 

EXAMPLE 1:

A dice is thrown three times and its three different positions are given below. Find the number on the face opposite the face showing 3.

  

(a) 1                                       (b) 4

(c) 5                                       (d) 6

Sol.         (d) The number 3 occurs most often in the given three figures. From these three figures it is clear that 1, 2, 5 and 4 lie adjacent to 3. Clearly, 6 lies opposite the face showing 3.

 

EXAMPLE 2:

Select from the alternatives the box that can be formed by folding the sheet shown in figure (X):

                                            

                                             

(a) A and C only            (b) B and D only

(c) C and D only            (d) A and D only

Sol.         (a) When the sheet shown in fig.(X) is folded to from a box (cuboid), then the two rectangular-shaded faces lie opposite to each other, the two rectangular white faces lie opposite to each other and the two square shaped faces (one shaded and one white) lie opposite to each other. Clearly, the cuboids shown in fig. (b) and (d) cannot be formed as in each of the two cuboids the two shaded rectangular faces appear adjacent to each other. So, only the cuboids in figures (a) and (c) can be formed.

 

EXAMPLE 3:

Amongst the following figures, find the correct one, if it is known that the total number of dots on opposite faces of cube shown is always 7.

(a)                 (b)

(c)                  (d)

Sol.         (a) Since the total number of dots on opposite faces is always 7, therefore 1 dot must lie opposite 6 dots, 2 dots must lie opposite 5 dots and 3 dots must lie opposite 4 dots. In the figures (b) and (d), 2 dots appear adjacent to 5 dots, and in fig: (c), 3 dots appear adjacent to 4 dots. Hence, figures are incorrect. Therefore, only fig. (a) is correct.

 

DIRECTIONS (Example 4-8): A cube is painted red on three adjacent sides and then cut into 8 small cubes.  Each blank face is then painted green. Then each such cube is again cut into eight cubes. Total numbers of cubes after second dissection \[=8\times 8=64.\]

After the first cut and painting green, we have

 

Coloured faces

No. of cubes

I

3 faces red and 3 green

1

II

2 faces red and 4 green

3

III

1 face red and 5 green

3

IV

No face red and 6 green

1

 

Total

8

 

After the second cut, the following types of cubes will be there:

 

 

 

Type

 

 

Colour on Faces

I

II

III

IV

A

:

3 red, none green

1

-

-

-

B

:

2 red, 1 green

3

6

-

-

C

:

1 red, 2 green

3

12

12

-

D

:

3 green only

1

6

12

8

 

EXAMPLE 4:       

How many cubes have three red faces each?

                (a) 1                                       (b) 2

                (c) 4                                       (d) 8

Sol.        (a)

 

EXAMPLE 5:

How many cubes have two green faces each?

                (a) 4                                       (b) 5

                (c) 6                                       (d) 9

Sol.        (d)

 

EXAMPLE 6:

How many cubes have three green faces each?

                  (a) 8                                          (b) 12

                  (c) 24                                     (d) 27

Sol.         (d)

 

EXAMPLE 7:

How many cubes have at least two red faces?

                (a) 4                                       (b) 6

                (c) 8                                       (d) 10

Sol.         (d)

 

EXAMPLE 8:

How many cubes have at least two blank faces?

                 (a) 24                                     (b) 48

                  (c) 42                                     (d) 64

Sol.         (d)

 

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