CBSE Class 6 Mathematics Patterns in Mathematics Practice Paper Worksheet Set 02

Work through Class 6 Maths Ganita Prakash Chapter 1 Patterns in Mathematics Practice Paper 03, a full 40-mark paper of 20 questions (Sections A to E) with clear step-by-step solutions. This paper tests square, cube and triangular numbers, matchstick patterns of squares and triangles, the rule that adding odd numbers gives squares, letter and colour patterns, and growing patterns of lanterns, staircase tiles and festival lights.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Write the 9th term of the pattern formed by adding consecutive odd numbers starting from 1.
Answer:
The terms are the running totals: 1, 1 + 3 = 4, 1 + 3 + 5 = 9, 1 + 3 + 5 + 7 = 16, …
These are the square numbers: the sum of the first n odd numbers is n × n.
9th term = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 9 × 9 = 81.
In simple words: Adding the first 9 odd numbers gives 9 × 9 = 81.

Teacher's Note:
1. The 9th odd number is 2 × 9 − 1 = 17.
2. You can add step by step or use n × n.

Question 12. In a tile pattern, the colours repeat in this order:
Red, Blue, Green, Red, Blue, Green, ...
What will be the colour of the 16th tile?
Answer:
The pattern repeats every 3 tiles.
16 = 5 × 3 + 1, so after 5 full groups (15 tiles) the 16th tile is the 1st colour of a new group.
The 16th tile is Red.
In simple words: 16 leaves remainder 1 when divided by 3, so it is the first colour, red.

Teacher's Note:
1. Remainder 0 would mean the last colour, Green.
2. Write the division with remainder.

Question 13. In a pattern of letters: A, C, E, G, I, ...
Which letter will be the 7th in the pattern?
Answer:
The pattern skips one letter each time: A, (B), C, (D), E, (F), G, (H), I, …
Continuing: 6th letter = K (skip J), 7th letter = M (skip L).
So the 7th letter is M.
In simple words: Skip one letter each time: A, C, E, G, I, K, M.

Teacher's Note:
1. The letters are in alphabet positions 1, 3, 5, 7, … (odd numbers).
2. The 7th odd number is 13, and the 13th letter is M.

Question 14. Form a number pattern where each term is twice the previous term, starting from 3. Write the 4th and 5th terms.
Answer:
1st term = 3; 2nd = 3 × 2 = 6; 3rd = 6 × 2 = 12.
4th term = 12 × 2 = 24; 5th term = 24 × 2 = 48.
Pattern: 3, 6, 12, 24, 48, …
In simple words: Keep doubling from 3: 3, 6, 12, 24, 48.

Teacher's Note:
1. Write out the full pattern, as the question asks you to form it.
2. This is 3 times the powers of 2.

Section C: Short Answer Questions (3 Marks Each)

Question 15. A pattern of shapes follows:
Write the number of shapes in the 10th row and the total number of shapes till the 10th row.
Answer:
Each row has as many shapes as its row number: 1st row 1, 2nd row 2, 3rd row 3.
So the 10th row has 10 shapes.
Total till the 10th row = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 shapes.
Quick way: pair the numbers as (1 + 10) + (2 + 9) + … + (5 + 6) = 5 × 11 = 55.
In simple words: The 10th row has 10 shapes, and 1 + 2 + … + 10 = 55.

Teacher's Note:
1. 55 is the 10th triangular number.
2. Pairing from both ends makes the sum quick.

Question 16. A garden has tile designs increasing by square numbers in each row.
Row 1: 1 tile
Row 2: 4 tiles
Row 3: 9 tiles
Row 4: 16 tiles
Find number of tiles in 6th and 8th rows and total tiles till 6th row.
Answer:
Row n has n × n tiles.
6th row = 6 × 6 = 36 tiles; 8th row = 8 × 8 = 64 tiles.
5th row = 5 × 5 = 25, so total till the 6th row = 1 + 4 + 9 + 16 + 25 + 36 = 91 tiles.
In simple words: Rows hold square numbers: 36 and 64 tiles; the first six rows hold 91 in all.

Teacher's Note:
1. Do not forget the 5th row in the total.
2. Add in steps: 1, 5, 14, 30, 55, 91.

Question 17. A block pattern has the rule: Multiply the term number by itself and subtract 1.
Term 1: 0
Term 2: 3
Term 3: 8
Find the 5th and 6th terms. What is the rule?
Answer:
Rule: term = (term number × term number) − 1, that is, one less than a square number.
Check: 1 × 1 − 1 = 0, 2 × 2 − 1 = 3, 3 × 3 − 1 = 8, and 4th term = 4 × 4 − 1 = 15.
5th term = 5 × 5 − 1 = 24.
6th term = 6 × 6 − 1 = 35.
In simple words: Square the term number and take away 1: 24 and 35.

Teacher's Note:
1. The jumps 3, 5, 7, 9, 11 are odd numbers, a useful check.
2. Write the rule in words as well as with numbers.

Section D: Long Answer Question (5 Marks)

Question 18. A lantern pattern is built as follows:
The 1st lantern has 5 paper pieces
The 2nd has 9
The 3rd has 13
The 4th has 17
(a) Write the rule for the pattern.
(b) Find how many pieces are in the 10th lantern.
(c) Find the total pieces used to make 10 lanterns.
(d) If you have only 200 paper pieces, how many lanterns can you make?
Answer:
(a) Start with 5 and add 4 for each next lantern (5, 9, 13, 17, …). The nth lantern has 4 × n + 1 pieces.
(b) 5, 9, 13, 17, 21, 25, 29, 33, 37, 41. The 10th lantern has 4 × 10 + 1 = 41 pieces.
(c) 5 + 9 + 13 + 17 + 21 + 25 + 29 + 33 + 37 + 41 = 230 pieces. (Pairing: (5 + 41) × 5 = 46 × 5 = 230.)
(d) 9 lanterns use 230 − 41 = 189 pieces, which is within 200.
10 lanterns need 230 pieces, which is more than 200.
So 9 lanterns can be made (11 pieces are left over).
In simple words: Add 4 each time: the 10th lantern has 41 pieces, 10 lanterns use 230, and 200 pieces make 9 lanterns.

Teacher's Note:
1. Find the running total until it passes 200.
2. Pair the first and last terms for a quick sum.
3. An incomplete 10th lantern does not count.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. At a newly built park, the stone staircase is decorated with beautiful tiles. The first step has 2 tiles, the second has 5 tiles, the third 10 tiles, and the fourth 17 tiles arranged in a specific pattern. Raj, the park supervisor, wants to estimate how many tiles the sixth step will require and the total number of tiles needed to decorate the first six steps, so that he can order the right amount from the tile supplier.
Questions based on above information:
(a) What is the pattern rule?
(b) How many tiles on the 6th step?
(c) Total tiles for 6 steps?
(d) If Raj has only 100 tiles, how many steps can he decorate fully?
Answer:
(a) The increases are 3, 5, 7, …: add the next odd number each time. Equivalently, step n has n × n + 1 tiles (1 + 1 = 2, 4 + 1 = 5, 9 + 1 = 10, 16 + 1 = 17).
(b) 5th step = 17 + 9 = 26; 6th step = 26 + 11 = 37 tiles (6 × 6 + 1 = 37).
(c) Total = 2 + 5 + 10 + 17 + 26 + 37 = 97 tiles.
(d) 6 steps use 97 tiles, which fits within 100. The 7th step needs 7 × 7 + 1 = 50 more tiles, making 147, which is too many.
So Raj can decorate 6 steps fully (3 tiles are left over).
In simple words: Each step has a square number plus 1 tiles: the 6th step needs 37, six steps need 97, so 100 tiles cover 6 steps.

Teacher's Note:
1. Spot the odd-number jumps first.
2. For (d), keep a running total and stop before it passes 100.

Question 20. During a local festival, the event organizer arranges decorative lights in rows following a triangular pattern. The first row has 1 light, the second row has 3 lights, the third has 6 lights, and the fourth row has 10 lights, increasing with each row. The organizer needs to know how many lights will be in the ninth row and the total number of lights used across all nine rows to manage the electrical load and budget effectively.
Questions based on above information:
(a) Identify the pattern.
(b) How many lights in the 9th row?
(c) Total lights in 9 rows?
(d) If only 120 lights are available, how many rows can be lit fully?
Answer:
(a) These are the triangular numbers: add 2, then 3, then 4, and so on (each row adds the next counting number).
(b) Rows 5 to 9: 10 + 5 = 15, 15 + 6 = 21, 21 + 7 = 28, 28 + 8 = 36, 36 + 9 = 45 lights in the 9th row.
(c) Total = 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 = 165 lights.
(d) Rows 1 to 8 use 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 = 120 lights exactly. Row 9 would need 45 more.
So 8 rows can be lit fully.
In simple words: Rows hold triangular numbers: the 9th row has 45 lights, nine rows need 165, and 120 lights fill exactly 8 rows.

Teacher's Note:
1. Keep a running total: 1, 4, 10, 20, 35, 56, 84, 120, 165.
2. These running totals are called tetrahedral numbers.