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

Practice Paper 02 for Class 6 Maths Ganita Prakash Chapter 1 Patterns in Mathematics carries 40 marks and has 20 questions spread over Sections A to E, each solved step by step below. Its questions focus on square-plus-one and triangular number sequences, double-and-add-one and double-and-subtract-one rules, a doubling pattern of road poles, sides of decagons and octagons, and repeating tile and stadium-light patterns.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Complete and explain: 800, 400, 200, 100, ___, ___.
Answer:
Rule: divide by 2 (each term is half of the one before).
Check: 800 ÷ 2 = 400, 400 ÷ 2 = 200, 200 ÷ 2 = 100.
100 ÷ 2 = 50 and 50 ÷ 2 = 25.
In simple words: Halve each number: 100, then 50, then 25.

Teacher's Note:
1. Check the rule on every given pair.
2. Reading the pattern backwards, it doubles.

Question 12. Pattern: ◆ ■ ◆◆ ■■ ◆◆◆. Predict next two terms. Explain.
Answer:
The terms take turns between diamonds and squares.
Each time a shape returns, it has one more symbol than before: ◆, ◆◆, ◆◆◆ and ■, ■■.
Next term (squares): ■■■.
Term after that (diamonds): ◆◆◆◆.
In simple words: The shapes alternate and grow by one, so ■■■ then ◆◆◆◆.

Teacher's Note:
1. Track the shape and the count separately.
2. The 6th term has 3 symbols, the 7th has 4.

Question 13. Pattern: 3, 7, 15, __, 63
(a) Find the missing number.
(b) Justify your answer.
Answer:
(a) The missing number is 31.
(b) Rule: multiply by 2 and add 1. 3 × 2 + 1 = 7, 7 × 2 + 1 = 15, 15 × 2 + 1 = 31, and 31 × 2 + 1 = 63, which matches the last term.
In simple words: Double and add 1: 15 gives 31, and 31 gives 63.

Teacher's Note:
1. Use the term after the blank to confirm.
2. The jumps 4, 8, 16, 32 also double.

Question 14. How many sides in total are present in 12 octagons? Show your working.
Answer:
An octagon has 8 sides.
Total sides = 12 × 8 = 96.
In simple words: 12 octagons with 8 sides each make 96 sides.

Teacher's Note:
1. Octa means eight, as in octopus.
2. Show the multiplication as working.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Pattern: 2, 5, 10, 17, 26, __, __
(a) Find next two terms.
(b) Write the differences between terms.
(c) Identify the logic and complete the pattern.
Answer:
(b) Differences: 5 − 2 = 3, 10 − 5 = 5, 17 − 10 = 7, 26 − 17 = 9. They are the odd numbers 3, 5, 7, 9, increasing by 2.
(a) Next differences are 11 and 13: 26 + 11 = 37 and 37 + 13 = 50.
(c) Logic: add consecutive odd numbers (3, 5, 7, …). Each term is also one more than a square number: 1 × 1 + 1 = 2, 2 × 2 + 1 = 5, …, 7 × 7 + 1 = 50.
Completed pattern: 2, 5, 10, 17, 26, 37, 50.
In simple words: Add 3, 5, 7, 9, 11, 13 in turn; the terms are squares plus 1.

Teacher's Note:
1. Work out part (b) first; it gives part (a).
2. Square-plus-one is a quick check.

Question 16. Pattern: 1, 3, 6, 10, 15, ...
(a) What are the 6th and 7th numbers?
(b) Find the sum of the 5th and 6th terms.
(c) What is the difference between 4th and 2nd term?
Answer:
These are the triangular numbers; the jumps are 2, 3, 4, 5, …
(a) 6th = 15 + 6 = 21; 7th = 21 + 7 = 28.
(b) 15 + 21 = 36 (a square number, 6 × 6).
(c) 4th term − 2nd term = 10 − 3 = 7.
In simple words: Add the next counting number each time: 21, 28; then 15 + 21 = 36 and 10 − 3 = 7.

Teacher's Note:
1. Two consecutive triangular numbers always add to a square.
2. Read 'the 2nd term' as 3, not 2.

Question 17. Pattern: 2, 3, 5, 9, 17, ...
(a) Find the next two terms.
(b) What is the 8th term?
(c) Find the sum of the 4th and 5th terms.
Answer:
Rule: multiply by 2 and subtract 1. Check: 2 × 2 − 1 = 3, 3 × 2 − 1 = 5, 5 × 2 − 1 = 9, 9 × 2 − 1 = 17.
(a) 17 × 2 − 1 = 33 and 33 × 2 − 1 = 65.
(b) 6th = 33, 7th = 65, 8th = 65 × 2 − 1 = 129.
(c) 4th + 5th = 9 + 17 = 26.
In simple words: Double and subtract 1: 33, 65, then 129 is the 8th term; 9 + 17 = 26.

Teacher's Note:
1. The jumps 1, 2, 4, 8, 16 double each time.
2. List the terms with their positions to avoid counting slips.

Section D: Long Answer Question (5 Marks)

Teacher's Note:
1. Check the rule on both given pairs.
2. Add carefully: 3 + 6 + 12 = 21, 21 + 24 = 45, 45 + 48 = 93.

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

Question 19. At a large banquet hall construction site, a tiling company was hired to design a stylish floor. The architect chose a repeating tile pattern to reduce cost and avoid mismatched designs. The company used a sequence of 3 black tiles followed by 7 white tiles, continuing this till 500 tiles were laid. The workers needed to estimate how many tiles of each color were required to avoid shortages during the installation.
Questions based on above information:
(a) How many full tile pattern sets can be formed using 500 tiles?
(b) How many black tiles are used in total?
(c) How many white tiles are used in total?
(d) What will be the color of the 456th tile?
Answer:
(a) One set = 3 black + 7 white = 10 tiles. Number of sets = 500 ÷ 10 = 50.
(b) Black tiles = 50 × 3 = 150.
(c) White tiles = 50 × 7 = 350. (Check: 150 + 350 = 500.)
(d) 456 = 45 × 10 + 6. After 45 full sets (450 tiles), the 456th tile is the 6th tile of the next set.
In a set, tiles 1–3 are black and 4–10 are white, so the 456th tile is white.
In simple words: 500 tiles make 50 sets: 150 black and 350 white; the 456th is the 6th in a set, which is white.

Teacher's Note:
1. Sets of 10 make division easy.
2. The remainder gives the place inside the set.

Question 20. For a night football match, a stadium lighting company was tasked with installing decorative lights across the upper deck. The lights were programmed in a repeating pattern: 6 red lights followed by 10 blue lights. This pattern was followed continuously for a total of 256 lights. The operations team had to plan the wiring and light order carefully to avoid last-minute confusion and ensure a synchronized lighting show.
Questions based on above information:
(a) How many complete lighting pattern cycles can be made using 256 lights?
(b) How many red lights are used in total?
(c) How many blue lights are used in total?
(d) What will be the colour of the 255th light in the sequence?
Answer:
(a) One cycle = 6 red + 10 blue = 16 lights. Cycles = 256 ÷ 16 = 16.
(b) Red lights = 16 × 6 = 96.
(c) Blue lights = 16 × 10 = 160. (Check: 96 + 160 = 256.)
(d) 255 = 15 × 16 + 15. After 15 full cycles (240 lights), the 255th light is the 15th light of the next cycle.
Lights 1–6 of a cycle are red and 7–16 are blue, so the 255th light is blue.
In simple words: 256 lights make 16 cycles: 96 red and 160 blue; the 255th is the 15th in a cycle, which is blue.

Teacher's Note:
1. 16 × 16 = 256, a square number.
2. The 255th light is the second-last light, so it must be blue.