Lesson Plan

How to solve a 3x3 Rubik's Cube

2026-07-05T07:08:27.163+05:30

How to solve a 3 x 3 Rubik's Cube

A Rubik’s Cube is a popular 3D combination puzzle invented in 1974 by Ernő Rubik. It is made of small colored squares arranged on a cube. The most common version is the 3 × 3 × 3 cube, which has six colored sides.

How it works

Each side has 9 squares of the same color when solved.
The rows and columns can be twisted and turned.
The goal is to mix the colors and then return every side to a single color.

Main Features

Improves memory and concentration
Develops problem-solving skills
Enhances logical thinking and patienc

Guidelines to solve the Rubik's Cube

Step 1: Put middle white at the base so that middle yellow is at the top

Step 2: Arrange all the white around the yellow middle as shown in the figure above.

Step 3: Match blue edge of white with the blue middle on the side face and then rotate this face down and match with the middle white at the bottom. Similarly repeat the same with other colors and then complete the "+" of white color at the base.

Step 4: Now keep the "+" white at the base and bring all the white at the base to complete one face with white color. For this do the following step.

Step 5: Bring white "+" at the base and bring the corner with white face as shown above and match the other blue face of the corner with middle blue of side face and apply the following algorithm.

Upper ← Right ↑ Upper → Right ↓

Similarly repeat the same algorithm after putting all white corner at the top so that all whites are arranged at the bottom as shown below.

Upper → Right ↑ Upper ← Right ↓

Step 6 : Now we have to arrange all the edges at the proper places (match with the middle one). Put the green edge at the top and match the color with the middle green. Top red face should shift away the side of the cube whose middle red matched with the top red

Here we use the following algorithm.

Upper → Right ↑ Upper ← Right ↓

Step 6 : Now again one white reached at the top. Place it at the proper place by using the algorithm of step 5

Step 7: Now repeat the same algorithm with all the edges till the edges are properly arranged and the second layer completely solved.

Making + on the top yellow face

Front → Top ←Right ↑ Top →Right ↓Front←

Front ← Top →Left ↑ Top ←Left ↓Front→

After making + at the top with yellow colour, now we have to correct the middle edges of the third row

R↑ U← R'↓U←R↑U2← ←R'↓

R↑U2←←R'↓U'→R↑U'→R'↓

These algorithms make the top edges at the proper place

Now we have to correct the top cormers

I we start from the correct corner then use the following algorithm

U←R↑U'→L↑U←R'↓U'→L'↓

Start this algorithm from U if left corner is correct otherwise we can start from R

When all the corners are correct then invert the cube (white up and yellow down) and keep the unsolved corner at the right side the use the following algorithm till the corner is solved.

R↑U←R'↑U'→

When corner at the bottom is solved then rotate the bottom and bring the unsolved corner at the right bottom again use the algorithm untill the corner is solved

RUR'U'

Use the same algorithm with all the corner till the cube is solved.

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Mathematics Lab Activity-17 Class IX | Statistics

2024-12-30T17:53:00.001+05:30

Lab Activity-17 Class iX

Mathematics Lab Activities on Statistics for class IX students, with complete observation tables, strictly according to the CBSE syllabus.

Chapter - 11 statistics

Activity - 17

Objective:

To draw histograms for classes of equal widths and varying widths.

Material Required:

Graph paper, geometry box, sketch pens, scissors, adhesive, cardboard.

Procedure:

1. Collect data from day to day life such as weights of students in a class and

make a frequency distribution table.

Case I : For classes of equal widths.

C-I	10-20	20-30	30-40	40-50	50-60
Frequency	50	10	30	20	50

Case II : For classes of varying widths

C-I	10-20	20-40	40-60
Frequency	50	40	70

2. Take a graph paper ( 20 cm × 20 cm) and paste it on a cardboard.

3. Draw two perpendicular axes X'OX and YOY' on the graph paper.

4. Mark classes on x-axis and frequencies on y-axis at equal distances as shown in Fig. 1.

Figure 1

5. Draw the histogram for case-1 as shown in figure 2

Figure 2

6. For the case-2 width of class interval are not equal. So first of all we adjust the frequency by using the formula.

7. Draw the histogram for the case 2 as shown in the figure 2

Figure 3

Observations

For Case 1: We take class interval along the x-axis and frequencies along the y-axis. We find the continuous graph which is called Histogram.

For Case 2: Here Class intervals are continuous but class widths are not same. So first of all we need to adjust the frequency by using the formula.

So adjusted frequency table is become as below

C-I	Frequency	Adjusted Frequency
10-20	50
20-40	40
40-60	70

Result :

In this activity we learn to made a Histogram for equal class widths and for for varying class widths.

Applications

Histograms are used in presenting large data in a concise form pictorially.

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Mathematics Lab Activity-16 Class IX | Surface Area & Volume

2024-12-29T07:52:00.001+05:30

Lab Activity-16 Class iX

Mathematics Lab Activities on Surface area and volume for class IX students, with complete observation tables, strictly according to the CBSE syllabus.

Chapter - 11 surface area & volume

Activity - 16

Objective:

To obtain the formula for the surface area of a sphere.

Material Required:

A ball, cardboard/wooden strips, thick sheet of paper, ruler, cutter, string, measuring tape, adhesive.

Procedure:

1. Take a spherical ball and find its diameter by placing it between two vertical boards (or wooden strips) [see Fig. 1]. Denote the diameter as d.

Figure 1

2. Mark the topmost part of ball and fix a pin [see Fig. 2].

Figure-2

3. Taking support of pin, wrap the ball (spirally) with string completely, so that on the ball no space is left uncovered [see Fig. 2].
4. Mark the starting and finishing points on the string, measure the length between these two marks and denote it by l. Slowly, unwind the string from the surface of ball.
5. On the thick sheet of paper, draw 4 circles of radius ‘r’ (radius equal to the radius of ball).
6. Start filling the circles [see Fig. 3] one by one with string that you have wound around the ball.

Figure 3

7. Let the length of string which covers a circle (radius r) be denoted by a.

8. The string which had completely covered the surface area of ball has been used completely to fill the region of four circles (all of the same radius as of ball or sphere).

9. This suggests:

Length of string needed to cover sphere of radius r = 4 × length of string needed to cover one circle i.e., l = 4a

or, surface area of sphere = 4 × area of a circle of radius r

So, surface area of a sphere = 4πr².

Observations

Diameter d of the spherical ball =14 cm.

Radius r =7 cm.

Length l of string used to cover ball = 480 cm.

Length a of string used to cover one circle =120 cm.

So length of the string (l) = 4 × 120 cm = 480 cm

Surface area of a sphere of radius r = 4 × Area of a circle of radius r = 4πr².

Result :

Surface area of sphere = 4πr².

Applications

This result is useful in finding the cost of painting, repairing, constructing spherical and hemispherical objects.

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