CBSE Class 6 Mathematics Patterns in Mathematics Worksheet Set 01

Long Answer Questions

Teacher's Note:
1. Any three sequences from Table 1 are acceptable.
2. Check each 'next three' by applying the rule.

Question 2. Many number sequences become easier to understand when they are drawn as pictures of dots or cubes.
(i) Describe, with dots, how the first four triangular numbers are drawn.
(ii) Describe, with dots, how the first four square numbers are drawn.
(iii) The number 36 can be arranged perfectly as a triangle of dots and also as a square of dots. Explain what this tells us about the way the same number can play different roles.
Answer:
(i) Start with 1 dot. Below it place a row of 2 dots (3 in all), then a row of 3 dots (6 in all), then a row of 4 dots (10 in all). Each picture is a triangle, and the next one adds a longer row at the bottom. So the numbers are 1, 3, 6, 10.
(ii) Start with 1 dot. Then 2 rows of 2 dots (4), 3 rows of 3 dots (9), and 4 rows of 4 dots (16). Each picture is a square grid with as many rows as columns. So the numbers are 1, 4, 9, 16.
(iii) 36 dots make a triangle of 8 rows, since 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36, and a square of 6 rows of 6, since 6 × 6 = 36.
So 36 is the 8th triangular number and also the 6th square number. A number is not tied to one picture; it can play different roles, and such double roles often hint at a hidden link between two sequences.
In simple words: Dot pictures show the shape behind a number, and one number can have more than one shape.

Teacher's Note:
1. Small sketches make the answer clearer.
2. Learn 36 as the key example of a double role.

Question 3. Table 3 of the chapter shows four shape sequences: Regular Polygons, Complete Graphs, Stacked Squares and the Koch Snowflake.
(i) State the rule for getting the next shape in the Regular Polygons sequence, and name the regular polygons that have 5, 6, 7 and 8 sides. Also explain the word ‘regular’.
(ii) State the rule for getting the next complete graph after K₄, and draw or describe K₅.
(iii) State the rule for getting the next shape of the Koch Snowflake, and explain why it becomes very hard to draw after a few steps.
Answer:
(i) Rule: each next shape has one side more than the one before, with all sides equal. ‘Regular’ means all sides are equal in length and all corners look exactly alike.
5 sides: pentagon; 6 sides: hexagon; 7 sides: heptagon; 8 sides: octagon.
(ii) Rule: add one more dot and join it to every dot already there. So K₅ has 5 dots with every pair joined. It looks like a pentagon with all its 5 diagonals, and has 10 lines.

(iii) Rule: replace every straight segment by a speed bump of four smaller segments.
After a few steps the drawing becomes very hard, because the number of segments becomes 4 times larger each step while each segment becomes only one-third as long. Soon the pieces are too tiny to draw with a pencil.
In simple words: Polygons gain a side, graphs gain a dot, and the snowflake's bumps soon get too tiny to draw.

Teacher's Note:
1. Draw K₅ neatly as a pentagon with a star inside.
2. Mention both 'more segments' and 'shorter segments' for (iii).

Teacher's Note:
1. Each new counting number is one new row.
2. Tables make running totals easy to read.

Teacher's Note:
1. Compare terms at the same position.
2. Multiplying grows faster than adding.

Question 6. The dots of a square grid have been separated into L-shaped groups as shown below.
(i) Write the addition sum shown by the picture, and give its total.
(ii) Explain why a picture like this can be drawn for a square grid of any size.
(iii) Using this idea, write the sum of the first 10 odd numbers and the sum of the first 50 odd numbers.
Answer:
(i) The grid has 7 dots in each row and each column, cut into 7 L-shaped bands, one inside the other. The bands hold 1, 3, 5, 7, 9, 11 and 13 dots.
So the picture shows 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49.
(ii) Suppose a square grid has k − 1 rows and k − 1 columns. To make the next square we add a new row of k dots and a new column of k dots, but the corner dot is shared. So the new L band holds k + k − 1 dots, which is always odd.
Starting from one dot, the bands hold 1, 3, 5, 7, … dots for a square of any size. That is why odd numbers always add up to a square number.
(iii) First 10 odd numbers fill a 10 by 10 grid: 10 × 10 = 100. First 50 odd numbers fill a 50 by 50 grid: 50 × 50 = 2500.
In simple words: Each L band is the next odd number, and the bands always build a square.

Teacher's Note:
1. Count the dots in each band from the corner dot.
2. The shared corner dot makes the band odd.

Teacher's Note:
1. Stacked squares and stacked triangles give the same row of numbers.
2. Build the complete graphs one dot at a time.

Question 8. The hexagonal numbers are 1, 7, 19, 37, 61, …
(i) Show how multiplying each triangular number by 6 and adding 1 produces the hexagonal numbers after 1, and write the sixth hexagonal number.
(ii) Find the totals 1, 1 + 7, 1 + 7 + 19 and 1 + 7 + 19 + 37. Which number sequence do these totals form?
(iii) Explain how thinking about a cube made of small cubes helps you believe the result of part (ii).
Answer:
(i) Triangular numbers 1, 3, 6, 10, 15: 6 × 1 + 1 = 7, 6 × 3 + 1 = 19, 6 × 6 + 1 = 37, 6 × 10 + 1 = 61, 6 × 15 + 1 = 91.
These are the hexagonal numbers after 1. A hexagonal arrangement is one centre dot with six equal triangular groups of dots around it. So the sixth hexagonal number is 91.
(ii) 1 = 1, 1 + 7 = 8, 1 + 7 + 19 = 27, 1 + 7 + 19 + 37 = 64. These are the cube numbers.
(iii) Build a cube from small cubes. A cube of side 2 is made from a cube of side 1 by fitting a corner shell over three faces; this shell needs 8 − 1 = 7 cubes. The next shell needs 27 − 8 = 19, and the one after it 64 − 27 = 37.
Looking straight down the long diagonal of the big cube, each shell appears as a hexagonal ring. So each shell holds a hexagonal number of cubes, and piling the shells gives the cubes 1, 8, 27, 64, …
In simple words: Each shell around a cube holds a hexagonal number of small cubes, so hexagonal numbers add up to cubes.

Teacher's Note:
1. Six triangles around a centre dot make a hexagon.
2. Differences of cubes are hexagonal numbers.

Question 9. The first four shapes of the Stacked Triangles sequence are shown below.
(i) Count the little triangles in each shape and write the number sequence you get.
(ii) Using the number of little triangles in each row, write each of these counts as a sum of odd numbers, and state the rule you notice.
(iii) How many little triangles will the 10th shape of this sequence have? Explain why the Stacked Triangles and the Stacked Squares give the same number sequence even though the shapes look quite different.
Answer:
(i) Shape 1 has 1 little triangle, Shape 2 has 4, Shape 3 has 9 and Shape 4 has 16. The sequence 1, 4, 9, 16, … is the square numbers.
(ii) Counting row by row from the top:
Shape 1: 1 = 1
Shape 2: 1 + 3 = 4
Shape 3: 1 + 3 + 5 = 9
Shape 4: 1 + 3 + 5 + 7 = 16
Rule: each new row holds 2 more little triangles than the row above, so every count is a sum of odd numbers, and the sum of the first n odd numbers is n × n.
(iii) The 10th shape has rows of 1, 3, 5, …, 19 triangles, the first 10 odd numbers, so it has 10 × 10 = 100 little triangles.
Both sequences start at 1 and grow by the same odd number each step: going from a square of side n − 1 to side n needs 2n − 1 extra little squares, and the new bottom row of the nth stacked triangle holds exactly 2n − 1 little triangles. Same start and same additions give the same totals.
In simple words: Rows of 1, 3, 5, 7, … triangles add up to squares, just like stacked squares.

Teacher's Note:
1. Count the upward and downward triangles in each row.
2. Same start, same steps, same sequence.

Question 10. To get the next shape of the Koch Snowflake, every line segment is replaced by a ‘speed bump’ made of four smaller segments.
(i) Write the number of line segments in the first five shapes of the sequence and state the rule connecting one shape to the next.
(ii) Find the number of line segments in the 7th shape, showing your steps.
(iii) Compare how fast this sequence grows with the triangular numbers by writing the 5th number of each sequence, and comment on what you notice.
Answer:
(i) The shapes have 3, 12, 48, 192 and 768 segments.
Rule: every segment is replaced by four smaller ones, so the next count is 4 times the previous count. The sequence is 3 times the powers of 4.
(ii) 5th shape: 768; 6th shape: 768 × 4 = 3072; 7th shape: 3072 × 4 = 12288.
So the 7th shape has 12,288 line segments (3 × 4 × 4 × 4 × 4 × 4 × 4).
(iii) The 5th Koch number is 768, but the 5th triangular number is only 15.
The Koch sequence grows far faster: triangular numbers grow by adding the next counting number, while the Koch counts are multiplied by 4 each time, and multiplying makes a sequence shoot up much more quickly than adding.
In simple words: Multiplying by 4 each time makes the Koch numbers grow much faster than triangular numbers.

Teacher's Note:
1. Show each multiplication step.
2. Compare the same position in both sequences.

Case Study Based Questions

Question 11. Before Diwali, Meera draws a rangoli at her doorstep. She first marks dots with rice flour and then joins them with colours. She places 1 dot in the top row, 2 dots in the next row, 3 dots in the row below it, and so on, each row having one dot more than the row above. When the dots are placed in this way, the whole design has the look of a neat triangle. Her first design had 4 rows. Her younger brother liked it so much that he asked her to make bigger designs on the next two days.
(i) How many dots are there in Meera’s design with 4 rows?
(ii) How many dots will the design have if she makes 5 rows?
(iii) Which number sequence do these totals belong to?
(iv) On the third day Meera used exactly 28 dots. How many rows did that design have?
Answer:
(i) 1 + 2 + 3 + 4 = 10 dots.
(ii) 10 + 5 = 15 dots.
(iii) 1, 3, 6, 10, 15, … are the triangular numbers, since the dots form a triangle.
(iv) Triangular numbers: 1, 3, 6, 10, 15, 21, 28. 28 is the 7th, so the design had 7 rows.
In simple words: Rows of 1, 2, 3, … dots make triangular numbers; 28 dots need 7 rows.

Teacher's Note:
1. Each new row adds one more dot than the last.
2. List triangular numbers to work backwards.

Question 12. The courtyard of a school is being repaired. The workers lay square patches of tiles. The first patch is made of a single tile. The second patch is a square of 2 tiles by 2 tiles. The third patch is a square of 3 tiles by 3 tiles, and the patches keep growing in the same way. All tiles are of the same size, and no tile is cut. The headmaster notices that each new patch looks exactly like one of the shape sequences printed in the Class 6 mathematics book.
(i) How many tiles are needed for the third patch?
(ii) How many tiles are needed for the fourth patch?
(iii) Which shape sequence of the chapter do these patches match, and which number sequence do the tile counts form?
(iv) How many extra tiles are needed to change a patch of 6 tiles by 6 tiles into a patch of 7 tiles by 7 tiles? What kind of number is this?
Answer:
(i) 3 × 3 = 9 tiles.
(ii) 4 × 4 = 16 tiles.
(iii) They match the Stacked Squares, and the counts 1, 4, 9, 16, 25, … are the square numbers.
(iv) 7 × 7 − 6 × 6 = 49 − 36 = 13 extra tiles. 13 is an odd number, which fits the rule that odd numbers add up to squares.
In simple words: Square patches need square numbers of tiles, and each step up adds an odd number.

Teacher's Note:
1. The extra tiles form an L shape.
2. 13 = 2 × 7 − 1.

Question 13. In the craft period, Ravi takes a large sheet of paper and folds it exactly in half. He then folds the folded paper in half again, and again, each time pressing the fold flat. Before any fold there is 1 layer of paper. After the first fold there are 2 layers, after the second fold 4 layers, and after the third fold 8 layers. Ravi writes these numbers in his notebook and sees a familiar sequence from Table 1 of his textbook. His friend then adds up all the layer counts he has written so far.
(i) How many layers of paper are there after the fourth fold?
(ii) How many layers are there after the fifth fold?
(iii) Name the number sequence formed by 1, 2, 4, 8, 16, …
(iv) Find 1 + 2 + 4 + 8 + 16, then add 1 to your total. What do you notice about the answer?
Answer:
(i) Layers double each fold: 1, 2, 4, 8, 16. After the fourth fold there are 16 layers.
(ii) 16 × 2 = 32 layers.
(iii) The Powers of 2.
(iv) 1 + 2 + 4 + 8 + 16 = 31, and 31 + 1 = 32, the next power of 2. So adding powers of 2 always gives one less than the next power of 2.
In simple words: Each fold doubles the layers, and the sum of the counts is one short of the next count.

Teacher's Note:
1. Folding in half is doubling.
2. Link (iv) to the result 'one less than the next power of 2'.

Question 14. A town park has six flower beds. Each bed is in the shape of a regular polygon, that is, all its sides are of equal length and all its corners look alike. The first bed has 3 equal sides, the second has 4, the third has 5, and so on up to the sixth bed. A small wooden pole is fixed at every corner of every bed, and a chain is tied along every side. The gardener is proud that the beds, placed in a row, show a neat shape sequence from the mathematics textbook.
(i) What is the name of the bed that has 5 equal sides?
(ii) How many poles are fixed around the bed that has 8 equal sides?
(iii) Which number sequence is formed by the numbers of sides of the six beds?
(iv) The gardener wants to add one more bed with 10 equal sides. What is this bed called, and how many poles will it need?
Answer:
(i) A regular pentagon.
(ii) A regular octagon has 8 corners, so 8 poles.
(iii) 3, 4, 5, 6, 7, 8: the counting numbers starting from 3.
(iv) A regular decagon. A closed figure has as many corners as sides, so it needs 10 poles.
In simple words: Poles go at corners, and a polygon has as many corners as sides.

Teacher's Note:
1. One pole per corner, one chain per side.
2. Revise polygon names up to 10.

Question 15. For the Independence Day assembly, chairs are placed on the stage in rows. The first row has 1 chair, the second row has 3 chairs, the third row has 5 chairs, and every row after that has 2 chairs more than the row just before it. The chairs are placed so that the whole arrangement keeps the shape of a square as the rows are added. The teacher in charge keeps a record of the running total of chairs after each new row is placed.
(i) How many chairs are there in the 6th row?
(ii) How many chairs are there altogether in the first 6 rows?
(iii) Which number sequence is formed by the running totals?
(iv) The school hall has exactly 196 chairs placed in this very way. How many rows are there?
Answer:
(i) Rows: 1, 3, 5, 7, 9, 11, so the 6th row has 11 chairs.
(ii) 1 + 3 + 5 + 7 + 9 + 11 = 36 chairs.
(iii) 1, 4, 9, 16, 25, 36, … are the square numbers, since the sum of the first n odd numbers is n × n.
(iv) n × n = 196 and 14 × 14 = 196, so there are 14 rows.
In simple words: Odd rows of chairs add up to squares; 196 = 14 × 14 means 14 rows.

Teacher's Note:
1. The nth row has 2n − 1 chairs.
2. Use the square rule backwards for (iv).

Question 16. The Mathematics Club of a school has 5 members. At the first meeting of the year, the teacher asks every member to shake hands with every other member exactly once. To help the children count, she marks 5 dots on the board, one for each member, and joins every dot to every other dot with a straight line. She tells the class that such a picture is called a complete graph, and that the number of lines in such pictures follows one of the sequences in Table 1.
(i) How many handshakes does one member make?
(ii) How many handshakes take place in all?
(iii) Which complete graph does the teacher’s picture show, and which number sequence do the line counts of complete graphs form?
(iv) Next week, 7 members attend the meeting. How many handshakes will take place then?
Answer:
(i) Each member shakes hands with the other 4 members.
(ii) Each line is one handshake: 1 + 2 + 3 + 4 = 10 handshakes.
(iii) The picture is K₅; the line counts 1, 3, 6, 10, 15, … form the triangular numbers.

(iv) 1 + 2 + 3 + 4 + 5 + 6 = 21 handshakes.
In simple words: Every handshake is one line of the complete graph, so the counts are triangular numbers.

Teacher's Note:
1. Do not count a handshake twice.
2. Check (iv): 7 × 6 ÷ 2 = 21.

Question 17. On Diwali night, Latika arranges clay diyas on her terrace in a diamond shape. She puts 1 diya in the top row, 2 in the next row, 3 in the next and 4 in the middle row. After the middle row the rows become smaller again: 3 diyas, then 2 diyas, then 1 diya. So the number of diyas in the rows, from top to bottom, is 1, 2, 3, 4, 3, 2, 1. Her brother says the total is a number he has seen in the chapter on patterns.
(i) How many diyas are there in Latika’s diamond altogether?
(ii) Which number sequence do such totals belong to?
(iii) Next year Latika makes a bigger diamond whose longest row has 8 diyas. How many diyas does she need?
(iv) A friend makes a diamond of the same kind using 121 diyas in all. How many diyas are there in its longest row?
Answer:
(i) 1 + 2 + 3 + 4 + 3 + 2 + 1 = 16 diyas.
(ii) The square numbers 1, 4, 9, 16, 25, …; adding counting numbers up and then down always gives a square.
(iii) 8 × 8 = 64 diyas.
(iv) 121 = 11 × 11, so the longest row has 11 diyas.
In simple words: An up-and-down diamond holds the square of its longest row.

Teacher's Note:
1. The middle row decides the square.
2. A diamond of dots is a tilted square.

Question 18. A craftsman makes table mats out of flat six-sided tiles. He starts with one tile in the middle. Around this tile he fits a ring of 6 tiles, so that the mat still looks like a six-sided shape. Around that he fits a ring of 12 tiles, and then a ring of 18 tiles, each new ring needing 6 tiles more than the ring before it. He writes down the total number of tiles in the mat after each ring is completed, and finds a sequence that is not printed in Table 1 of his son’s textbook.
(i) How many tiles are in the mat after the first ring is fitted?
(ii) How many tiles are in the mat after the second ring is fitted?
(iii) Name the sequence 1, 7, 19, 37, … and show how 19 can be obtained from a triangular number.
(iv) How many tiles are in the mat after the fourth ring is fitted?
Answer:
(i) 1 + 6 = 7 tiles.
(ii) 7 + 12 = 19 tiles.
(iii) These are the hexagonal numbers. The 2nd triangular number is 3, and 6 × 3 + 1 = 19.
(iv) Rings hold 6, 12, 18 and 24 tiles: 1 + 6 + 12 + 18 + 24 = 61 tiles.
In simple words: Each ring adds 6 more tiles than the last, giving the hexagonal numbers 1, 7, 19, 37, 61.

Teacher's Note:
1. Rings grow by 6 each time.
2. Check: 6 × 10 + 1 = 61.

Teacher's Note:
1. Check the table: 48 × 4 = 192.
2. Mention the shrinking edge length.

Question 20. A flight of steps leads up to the school library. Arjun climbs it in his own way: at each move he either steps on to the very next step, or he skips one step and lands on the step after it. His teacher asks the class to count the different ways of climbing a flight. For a flight of 1 step there is only 1 way. For 2 steps there are 2 ways. For 3 steps there are 3 ways, and for 4 steps there are 5 ways. The class writes down 1, 2, 3, 5, … and the teacher says that this sequence is named after the ancient Indian scholar Virahānka.
(i) In how many ways can Arjun climb a flight of 5 steps?
(ii) In how many ways can he climb a flight of 6 steps?
(iii) State the rule that gives each number of this sequence from the numbers before it.
(iv) Which is greater: the 9th number of this sequence, or the 9th triangular number? Show both numbers.
Answer:
(i) 3 + 5 = 8 ways.
(ii) 5 + 8 = 13 ways.
(iii) The first two numbers are 1 and 2; after that each number is the sum of the two numbers just before it. (To reach a step, Arjun comes either from the step just below or from the one below that, so the two earlier counts are added.)
(iv) Virahānka: 1, 2, 3, 5, 8, 13, 21, 34, 55, so the 9th is 55. Triangular: 1, 3, 6, 10, 15, 21, 28, 36, 45, so the 9th is 45.
Since 55 > 45, the 9th Virahānka number is greater.
In simple words: Add the last two counts to get the next one; the 9th Virahānka number, 55, beats 45.

Teacher's Note:
1. The last move is 1 step or 2 steps.
2. Write both sequences side by side.