Cubes

A cube is a 3-dimensional diagram with all sides equal.  If we divide it into the size (¹⁄_n)^th   part of its side, we get n³ smaller cubes. 
Shown below is a cube which is painted on all the sides and the cut into (¹⁄₄)^th of its original side. 

Some observations: A cube has 6 faces, 12 edges and 8 corners. We can see that the cubes which got all the three sides painting lies at the corners. So the number of cubes which got painted all the three sides is equal to 8. Cubes with 2 sides painting lie on the edges (see the diagram). But the cubes which are on the left and right side of the edge matches with the corners. So we have to substract these two cubes from the number of cubes lying on the edge to get the number of cubes with 2 sides painting. Cubes with 1 side painting lies on the surfaces. Since, the top row, bottom row, left column, and right column matches with the edges, We must exclude these cubes while calculating the single side painted cubes.
The following rules may be helpful: If a cube is divided into the size $\left( {\displaystyle\frac{{\rm{1}}}{{\rm{n}}}} \right)^{{\rm{th}}} $ of its original side after get painted all the sides, Then

Total number of cubes = ${\rm{n}}^3 $
Cubes with 3 sides painting = 8
Cubes with 2 sides painting = $12 \times (n - 2)$
Cubes with 1 sied painting = ${\rm{6 \times (n - 2)}}^{\rm{2}} $
Cubes with no painting = ${\rm{(n - 2)}}^{\rm{3}} $

Solved Examples (Level - 1)

1. A cube whose two adjacent faces are colored is cut into 64 identical small cubes. How many of these small cubes are not coloured at all?

Assume the top face of the cube and its right side are colored green and orange respectively.

Now If we remove the colored faces, we left with a cuboid, whose front face is indicated with dots. 

So on the front face there are 9 cubes, and behind it lies 4 stacks.  So total 9 x 4 = 36

2. A cube, painted yellow on all-faces is cut into 27 small cubes of equal size. How many small cubes got no painting?

Assume we have taken out the front 9 cubes.  Then the cube looks like the one below.

Now the cube which is in the middle has not got any painting.  The cubes on the Top row, bottom row, left column and right column all got painting on atleast one face.
Alternative method:
Use formula: ${\rm{(n - 2)}}^{\rm{3}} $  Here n = 3 So ${\rm{(3 - 2)}}^3 $ = 1

3.  All surfaces of a cube are colored. If a number of smaller cubes are taken out from it, each side 1/4 the size of the original cube's side, Find the number of cubes with only one side painted.

The original (colored) cube is divided into 64 smaller cubes as shown in the figure.  The four central cubes on each face of the larger cube, have only one side painted.  Since, there are six faces, therefore total number of such cubes = 4 x 6 = 24.

Alternative Method: 
Use formula : ${\rm{6 \times (n - 2)}}^{\rm{2}} $ = ${\rm{6}} \times {\rm{(4 - 2)}}^2 $ = 24

Level - 2

4. Directions: One hundred and twenty-five cubes of the same size are arranged in the form of a cube on a table. Then a column of five cubes is removed from each of the four corners. All the exposed faces of the rest of the solid (except the face touching the table) are coloured red. Now, answer these questions based on the above statement:
(1) How many small cubes are there in the solid after the removal of the columns?
(2) How many cubes do not have any coloured face?
(3) How many cubes have only one red face each?
(4) How many cubes have two coloured faces each?
(5) How many cubes have more than 3 coloured faces each?
The following figure shows the arrangement of 125 cubes to form a single cube followed by the removal of 4 columns of five cubes each.

When the corner columns of the original cube are removed , and the resulting block is coloured on all the exposed faces (except the base) then we get the right hand side diagram.  We labelled the various columns from a to u as shown in the figure
(1): Since out of 125 total number of cubes, we removed 4 columns of 5 cubes each, the remaining number of cubes = 125 - (4 x 5) = 125 - 20 = 105.
(2): Cubes with no painting lie in the middle.  So cubes which are blow the cubes named as s, t, u, p, q, r, m, n, o got no painting.    Since there are 4 rown below the top layer, total cubes with no painting are (9 x 4) = 36.
(3): There are 9 cubes namaed as m, n, o, p, q, r, s, t and u in layer 1, and 4 cubes (in columns b, e, h and k) in each of the layers 2, 3, 4 and 5 got  one red face. Thus, there are 9 + (4 x 4) = 25 cuebs.
(4)  the columns (a, c, d, f, g, i, j, l) each got 4 cubes in the layers 2, 3, 4, 5.  Also in the layer 1, h, k, b, e cubes got 2 faces coloured.  so total cubes are 32 + 4 = 36
(5): There is no cube in the block having more than three coloured faces. There are 8 cubes (in the columns a, c, d, f, g, i, j and l) in layer 1 which have 3 coloured faces. Thus, there are 8 such cubes.
Thus, there are 8 such cubes.

5.  Directions: A cube of side 10 cm is coloured red with a 2 cm wide green strip along all the sides on all the faces. The cube is cut into 125 smaller cubes of equal size. Answer the following questions based on this statement:
(1) How many cubes have three green faces each?
(2) How many cubes have one face red and an adjacent face green?
(3) How many cubes have at least one face coloured?
(4) How many cubes have at least two green faces each?
Clearly, upon colouring the cube as stated and then cutting it into 125 smaller cubes of equal size we get a stack of cubes as shown in the following figure.

The figure can be analysed by assuming the stack to be composed of 5 horizontal layers.
(1): All the corner cubes are painted green.  So there are 8 cubes with 3 sides painted green.
(2): There is no cube having one face red and an adjacent face green as all the green painted cubes got paint on atleast 2 faces.
(3): Let us calculate the number of cubes with no painting.  By formula,  $\left( {{\rm{n - 2}}} \right)^{\rm{3}} $ = $\left( {{\rm{5 - 2}}} \right)^{\rm{3}} $ =  27
Therefore, there are 125 - 27 = 98 cubes having at least one face coloured.
(4): From the total cubes, Let us substract the cubes with red painting, cubes with no painting. 
125 - (9 x 6) - 27 = 44

Non-Verbal Reasoning (Analogy)

Analogy means relationship.  Let us have a look at an example:

Teacher : Pen  : : Soldier : _________   

What should come in the blank? If teacher's main tool is pen, Soldier's main tool is a gun.  

Similarly, we have to identify the relationship between in figures A and B so that to identify the option which got relationship with figure C. 

Just look at few examples and you can easily understand these problems:

1. 

The square in PF(A) rotated 90$^0$ clockwise along with dot.  So option 5 is correct. 

2. 

 Pentagon in PF(A) became small and circumscribed with Square in PF(B).  So If a square has to become small and to be circumscribed with triangle.  So option 1 is correct.  option 5 is rules out as the square rotated 45$^0$ instead of 90$^0$.

3. 

 
Here the hexagon becae pentagon and the dots came out of the diagram, and a new darkened dot appeared in the middle.  So PF(C) should become triangle and two dots must come out and a darkened dot must appear in the iddle.  So correct Option 5

4. 

  Here Bottom square became big, and the figure above it, came into it and pentagon appeared in the triangle.  So in PF(C) pentagon must become big, and square must be inside it and a hexagon should appear in it.  so correct option 2

5. 

  Simple one.  Two circles became a single square, and the square became two squares. So two triangles must become single triangle and circle must become two circles. So answer option 4.

6. 

 Another simple one.  the directions of the arrows changed their positions.  So answer option 4.

7. 

 In PF(1), top half darkened rectangle turned 90$^0$ clockwise, middle half darkened rectangle turned anti-clockwise 90$^0$ and bottom half darkened rectangle turned clockwise by 90$^0$.  So turn the rectangles in PF(C), clockwise, anti-clockwise, and clockwise. So correct Option 1

8. 

 In PF(A), the square has three dots each at the middle of its sides.  In PF(B), square became pentagon, and number of dots got increased by one and one dot occupied the vertex.  So PF(C) must become hexagon and there must be 5 dots and one dot should occupy the vertex. Correct option 4

9. 

 The entire diagnol rotated by about 180$^0$ clock wise and the open circle became darkened.  So the square must be darkened and the entire diagnol should rotate by 180$^0$.  So correct option 3

10. 

 Simple observation.  Pentagon at the top became bigger and square came inside of it.  So hexagon in PF(C) should become big and circle should enter into it.  So correct option 3

Questions for Practice

11. 

 Answer: Option 2

12. 

 Answer: Option 3

13. 

 Answer: Option 5

14. 

 Answer: Option 4

15. 

 Answer: Option 4

16. 

 Answer: Option 4

17. 

 Answer: Option 4

18. 

 Answer: Option 3

19. 

 Answer: Option 1

20. 

 Answer: Option 5

21. 

 Answer: Option 2

22. 

 Answer: Option 5

23.

 Answer: Option 1

24.

Answer: Option 3

Non - verbal reasoning (Series)

Non-Verbal reasoning appears in Bank exams, Infosys, MAT exams constantly. There are 5 Problem Figures (PF) will be given with 5 Answer Figures (AF).  We need to determine the next figure in the series.  There are certain rules which make solving these problems easy.  So study the rules and solved examples.

How to answer these questions: 

Step 1: 

For all the series problems the following rules apply.  If problem figures A and E are equal our answer is problem figure B. Similarly, the other rules as follows. 

1. PF(A) = PF(E) then answer is PF(B)

2. PF(D) = PF(E) then answer is PF(C)

3. PF(A) = PF(C) = PF(E) then answer is PF(B)

4. PF(A) = PF(D), PF(B) = PF(E) then answer is PF(C)

5. PF(D) = inverse of PF(A) and PF(E) = inverse of PF(2) then answer is inverse of PF(C)

Step 2: 

In general, the items in the box takes different positions in the subsequent figures.  They may rotate certain degrees either clock wise or anti-clockwise.  Look at the following diagram. In some problems new items add to the existing figures and some existing figures vanish.  

Solved Examples

1. 

In this problem If PF(A) = PF(E) then answer is PF(B).  In the answer options AF(4) is same as PF(B) so option 4

2. 

 Here PF(C) and PF(E) are equal.  So Answer figure should be PF(B).  So correct option is c.

3. 

The arrow is changing its positions clock wise 90^o, 45^o, 135^o, 45^o, ....next should be 180^o. So option 3.

4. 

 Simple one.  A new arrow and a new line are adding alternatively.  In PF(E) a new line has added.  So in the next figure a new arrow must be added.  And total lines should be 6.  Option 5

5. 

 Small hand is moving anticlock wise 90^o, 45^o, 90^o, 45^o,... and Big hand is moving clock wise 135^o constantly.  So in the next figure, small hand must move 90^o anti clockwise, and big hand must move 135^o.  So option 4

 6.

Here the symbol is changing positions anti clockwise by  45^o and every time a new symbol is adding. The "C"s in the middle are rotating clock wise by 90^o. So the next figure must be option 4

7. 

 This is a simple analogy.  There is a relationship between 1 and 2, 3 and 4.  the small figures in the first diagram are getting bigger and vice-versa.  So Option 3

8. 

 All the three symbols in the dice are rotating clockwise.  So option 3

Alternative method:

We know that if  PF(A) = PF(D), PF(B) = PF(E) then answer is PF(C).  So option 3

9. 

 A new symbols is appearing in the middle of the previous figure and the previous figure is getting bigger. So option 4 is the right option.  3 and 5 options are ruled out as the figures in the middle are appeared already.

10. 

 A dot and line are adding constantly to the figures in left and right sides alternatively.  So option 3

11.

There appears to be no pattern on immediate look, but his problem can be solved by simple observation.  Have a look at the diagram below..

The positions of two symbols are not changing in 2 consecutive figures.  So option 5

12. 

 the arrow and small line inside the small square are rotating constantly anti clockwise and clockwise respectively by 90^o, 45^o, 90^o, 45^o,... and 45^o, 90^o, 45^o, 90^o .  So next figure would be option 3.

13.

 The line is rotating anti clock wise by 90^o, 180^o, 270^o, 360^o  so next figure should be 90^o from figure E and a new symbol must appear.  So option 1 is the correct.

14. 

 Symbols X is rotating clockwise by 45^o, 90^o, 45^o, 90^o.  So our options will be either 1 or 3 as in the next figure symbol X must move 45^o.  A new symbols is being added to X each time one at front and next time at back.  So option 3 is right one.

15.

 Simple.  Observe PF(A) and PF(E) are equal.  So next figure will be PF(B).  So option 5

16. 

 the symbols are changing constantly in clockwise direction and a new symbol is being added.

The red rounded circle is a place whenever a symbol appear in that position must not appear in the next.  And remaining positions are moving clockwise by 90^o.  A new symbol must come at the place shown by green arrow.  So our option will be 1.  Option 2 is ruled out as + symbol appeared earlier.

17.

 Circle is moving diagonally and triangle is moving clockwise by 90^o. So option 1 is correct one.


18. 

Here you can easily observe that the lines are rotating 90^o clockwise. also in PF(B) and PF(D), half line has added at the right most side and in figures PF(C) and PF(E) a new line has added. So in our answer half line has to be added and lines should rotate 90^o.  So answer option 2.

19. 

 Simple one.  Figures A and B changed their symbols opposite them.  C and D also did So.  So option 1

20. 

Symbols in  A, B are same except Symbols at bottom.  A new symbol is coming there.  Similarly in C, D.  So option 3.  Option 2 is ruled out as C appeared earlier.