We at Headache through integrated teaching approach help students to learn every topic to it’s depth bypassing all the constrains in learning

@2023. All rights reserved.

Sat Sep 19, 2020

Cubes is one of the most frequently asked puzzle themes in CAT. These puzzles expect us to contemplate the puzzle, which makes it to some degree harder than other kind of puzzles. We will examine the fundamental ideas of the block and not many things that we would utilize is greater part of the puzzles identified with cubes. Block related puzzles are solicited in a number from ways dissimilar to choice or plan puzzles, where the essential example continues as before. We will outline a portion of the sorts of puzzles and the methodology one should use to take care of such issues. We will zero in our conversation on what might befall a 3D square when we cut it with a specific number of cuts, state of the 3D shape after the cuts, the number of sub-parts are framed and so forth.

At the point when we state Cube, we allude to the three dimensional solid with equivalent faces (complete 6 countenances).

Barely any things to recollect while managing a block:

# A solid shape has 6 countenances

# A solid shape has 12 edges

# A solid shape has 8 vertices

Let us make a slice corresponding to the base of the solid shape. We get two solids. On the off chance that we make another slice corresponding to the base, we get three solids. For n slices commonly corresponding to the base (or some other face), we get n+1 solid pieces.

In the event that we make one slice corresponding to the base (say x-pivot) and another slice perpendicular to the principal cut (say y hub), the subsequent cut will separate every one of the two solids we get from the initial cut into two pieces. We get complete 4 pieces or 4 pieces of the underlying huge solid shape. Correspondingly, another slice corresponding to both the past cuts (z-pivot) will additionally separate these 4 pieces into two pieces every, in this manner making the complete pieces as 8. At the point when we examine this, we see that when we make X commonly equal cuts along x-hub, Y commonly equal cuts along y pivot and Z commonly paralel cuts along z hub we get (X+1)*(Y+1)*(Z+1) pieces of the first block. X cuts along x itself will deliver X+1 pieces and when we cut every one of those pieces with Y cuts, every one of them will create Y+1 pieces and comparable with Z cuts.

Absolute number of pieces acquired with X cuts in x pivot, Y cuts in y hub and Z cuts in z hub =

(X+1)*(Y+1)*(Z+1)

Note that, all the pieces won't really be cubes. They can be cubes or cuboids. All the pieces are cubes when X = Y = Z and all the cuts are equidistant.This is the situation when the quantity of slices are given to us. At times, we are given the entirety of X + Y + Z rather than singular qualities.

At the point when we are given all out number of cuts, the cuts can be made in any pivot. State on the off chance that we make absolute 12 cuts, every one of the 12 can be in x-hub, or 6 in x and 6 in y-hub and so forth. For a given number of cuts, we can have various mixes. We get the base number of pieces when all the cuts are made in a similar pivot. In the event that we make each of the 12 cuts along x-pivot, absolute 13 pieces are framed. All different mixes will bring about in excess of 13 pieces. Suppose, 11 cuts along x-hub and 1 cut along y. The complete number of pieces shaped is (11+1)*(1+1) = 24.

MAXIMUM NUMBER NUMBER OF PIECES WITH A GIVEN TOTAL NUMBER OF CUTS

So as to get most extreme number of pieces with given number of cuts, isolate the all out cuts similarly among the three hub. On the off chance that we partition 12 cuts similarly among the three hub, we get 4 cuts in every pivot bringing about (4+1)*(4+1)*(4+1) = 125 pieces generally speaking. In any case, imagine a scenario in which the absolute number of cuts can't be partitioned similarly among the three pivot. For a number like 14, we can't appropriate similarly among all thethree hub. In such case, to boost the complete number of pieces, we have to limit the contrast between the quantity of cuts in any two hub is least. For 14 cuts, we can isolate it as X = 5, Y = 5, Z = 4. Some other mix of numbers will bring about a lower number of pieces. In the event that we break down, we see that if there should be an occurrence of equivalent cuts, the distinction between the quantities of cuts is zero that outcomes in the greatest number of pieces.

The inquiry can likewise specify the total number of pieces framed after a specific number of cuts. We have to locate the number of cuts in various pivot dependent on the prerequisite of the inquiry. Assume, a solid shape is cut into 10 pieces. Presently 10 = (X+1)*(Y+1)*(Z+1). Presently 10 is to be spoken to as the result of 3 numbers or factors. We can have, 10 = 1*1*10 or 1*2*5.

Least numbers of cuts is acquired in the event that we factorize the number of pieces into three equivalent variables. On the off chance that three equivalent components are impractical, at that point we ought to limit the contrast between the numbers.In instance of 10 pieces, 1*2*5 gives least distinction between factors.

To amplify the total number of cuts, all the cuts are made in one hub. For example, for 10 pieces, factors are 10*1*1 that implies the x-pivot cuts are 9 and no cut is made in other two hub.

PAINTING THE OUTER SURFACE OF THE CUBE AFTER CUTTING

The most common type of cube question is painting the faces of the cube. Let us say, a cube is painted black, then it is cut into 64 pieces by 3 cuts in each directions. As we can see, there are 4 rows of 16 pieces (small cubes) each. Few conclusions from the

figure1. Number of cubes with 3 faces painted black = all the corner cubes which is equal to 8 cubes.

2. Number of cubes with exactly 2 faces painted = (n-2)*(number of edges) = (4-2)*(12) = 24.

3. Number of cubes with exactly 1 face painted red = cubes on the faces (except on the edge and the corner) = (n – 2)2*6=24

4. Number of cubes with 0 faces painted black = (n-2)3 = 8

The number of cubes that remain unaffected with the painting are those, which are not on the outer side. In other words, if we remove the outer painted shell then the inner structure is what we desire or we reduce the dimension of the cube by 2 units (remove a painted layer from all the sides or faces) to get a inner cube of side dimension = n-2 units.

If the number of cuts are not equal, let’s say if X =5, Y = 6, Z = 7, then:

1. Number of cubes with 3 faces painted black = all the corner cubes which is equal to 8 cubes.

2. Number of cubes with exactly 2 faces painted = (X-2)*4 + (Y-2)*4 + (Z-2)*4 {since 4 edges will have each of (X-2),(Y-2),(Z-2)}= 12 + 16 + 20 = 48.

3. Number of cubes with exactly 1 face painted red = cubes on the faces (except on the edge and the corner) = 2*[(X-2)(Y-2) + (Y-2)(Z-2) + (X-2)(Z-2)] = 2*(12 + 20 +15) = 94

4. Number of cubes with 0 faces painted black = the inner cuboid = (X-2)*(Y-2)*(Z-2)= 3*4*5 = 60

In the previous case, we saw the case when all the faces are painted with the same color. The question can also paint faces with a different color. Let us say the three pair of adjacent faces are painted with with Red, Blue and Green respectively. The cube is cut into 64 identical cubes with identical number of cuts along the three axis.

1. Number of cubes with all the three color faces = 2 cubes diagonally opposite to each other.

2. Number of cubes with only same color faces on two faces = three edges without corner cubes = 3*(n-2) = 6

3.Total cubes on the edges = 4*n + 8*(n-2) = 32

4. Number of cubes with only 2 color faces = total cubes on edges – same color on two faces cubes – all three color face cubes = (3) – (2) = 32 – 6 – 2 = 24

5. From the above observation, we can say cubes with only Red and Green faces or only blue and green = 24/3 = 8

6. Number of cubes with exactly one face red = cubes on the two faces of the painted red surface = 2*(n-2)2 = 8

7. Number of cubes with faces only red = (2)/3 + (6) = (n-2) + 2*(n-2)2 = 10

8. Number of cubes with exactly one face painted = 6 faces with (n-2)2 cubes painted on one face = 6*(n-2)2 = 150.

9. Number of cubes with exactly two surfaces painted with different colors = 9 edges with (n-2) cubes = 9*(n-2) = 18

Cubes can also be painted like opposite sides with similar colors or each layer with a different layer. The approach to every kind of question depends upon visualization of the figure. The 2-D diagram would assist a lot in solving but one needs to think more on the three faces hidden in the figure. Cube, being a symmetrical figure allows us to think in symmetrical terms in majority of the questions.

Another type of cube question is the numbering of the smaller cubes after cutting the larger cube. Remember, the concept of cut could also be framed as ‘125 identical cubes are placed on one another in a manner such that they form a larger cube’,the approach for these questions would remain the same as discussed. Let us say there are 125 identical unit cubes numbered from 1 to 125 combined to form a larger cube to side 5 units. The cubes are always placed from left to right in ascending order of their numbers starting from the bottommost layer and the first row. After placing the first row, the next row is placed behind the first row such that the first cube of the next row comes directly behind the first cube of the first row. After completing the bottommost layer, the next layer is placed in a similar way the first layer was placed. Find the sum of the numbers of the cubes placed at the vertices.

From the above statement, we know that each row have 5 unit cubes. Let us make the bottom most layer of the larger cube.

1 2 3 4 5

6 7 8 9 10

11 12 13 14 15

16 17 18 19 20

21 22 23 24 25

Layer 1 (bottommost)

26 27 28 29 30

31 32 33 34 35

36 37 38 39 40

41 42 43 44 45

46 47 48 49 50

Layer 2

From the above placement of cubes we see that in bottom most layer cubes numbered 1, 5, 21, 25 are the vertex cubes. The topmost layer will have the other 4 cubes. To get the numbers of those cubes, we can either draw all the five layers or from the pattern we see that the last cube is a multiple of 25 in every layer. So the last cube in layer 4 will be 100. Let us draw the fifth layer.

101 102 103 104 105

106 107 108 109 110

111 112 113 114 115

116 117 118 119 120

121 122 123 124 125

Layer 5 (topmost)

The other 4 cubes are 101, 105, 121, 125. The sum of the 8 cubes is (1+5+21+25+101+105+121+125) = 504. If we want to save time, we analyze that the sum of the opposite vertex cubes is 126 or from the first layer itself, the sum of the diagonal cubes is 25*(layer number) + 1.

**Team Headache**

@2023. All rights reserved.

Mail Us

Email: guptajitesh.jg@gmail.com

Launch your Graphy

100K+ creators trust Graphy to teach online

Headache Tutorials Best online and offline coaching IPMAT and CAT 2023
Privacy policy
Terms of use
Contact us
Refund policy