CBSE Class 6 Mathematics Number Play Practice Paper Assignment Set 02 with Answers

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Access comprehensive school assignments for Chapter 03 Number Play using the CBSE Class 6 Mathematics Number Play Practice Paper Assignment Set 02 with Answers. Designed to align with the 2026-27 CBSE academic guidelines, these practice sets help Class 6 Mathematics students reinforce core concepts and improve their problem-solving accuracy.

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This is Class 6 Maths Ganita Prakash Chapter 3 Number Play Practice Paper 07, a 40-mark test of 20 questions in Sections A to E, given here with step-by-step answers. It checks taller-neighbour counts, supercells in number grids, palindromic clock times, digit sums and divisibility by 9, the “first to 60” game, Kaprekar’s routine to 6174 and the Collatz sequence of 19.

Section A: Multiple Choice Questions (1 Mark Each)

Question 1: In a row of 6 children, the sequence of “taller neighbour counts” is 0, 1, 2, 0, 2, 0. Which is correct?
(a) Possible
(b) Impossible because two 0’s cannot be separated by a 2
(c) Impossible because ends cannot be both 0
(d) Possible only if heights repeat
Show Answer & Explanation

Answer: (a) Possible

Explanation:
1. Each child says how many of the children standing next to them are taller.
2. Try heights 6, 4, 2, 5, 1, 3 (in any unit) for children C1 to C6. 0 C1 1 C2 2 C3 0 C4 2 C5 0 C6 Red number = how many neighbours are taller
3. C1: neighbour 4 is shorter → 0. C2: 6 is taller, 2 is not → 1. C3: 4 and 5 are taller → 2.
4. C4: 2 and 1 are shorter → 0. C5: 5 and 3 are taller → 2. C6: neighbour 1 is shorter → 0.
5. This gives 0, 1, 2, 0, 2, 0 with all heights different, so the arrangement is possible.

Teacher's Note:
1. One example that works is enough to prove “possible”.
2. Children at the two ends have only one neighbour each.

Question 2: The centre of a 3×3 grid has neighbours 5320 (up), 5200 (down), 5100 (left), and 5300 (right). Which value makes it a supercell?
 5320 
5100???5300
 5200 
(a) 5201
(b) 5310
(c) 5299
(d) 5400
Show Answer & Explanation

Answer: (d) 5400

Explanation:
1. A supercell must be greater than every neighbour: 5320, 5200, 5100 and 5300.
2. The biggest neighbour is 5320, so the centre must be more than 5320.
3. 5201, 5310 and 5299 are all less than 5320. Only 5400 is greater than all four.

Teacher's Note:
1. Compare with the largest neighbour first.
2. If it beats the largest, it beats them all.

Question 3: Which of these is not a palindromic time in HH:MM 12-hour format?
(a) 03:30
(b) 04:40
(c) 11:22
(d) 05:50
Show Answer & Explanation

Answer: (c) 11:22

Explanation:
1. Write each time as digits and read it backwards.
2. 03:30 → 0330 → 0330; 04:40 → 0440 → 0440; 05:50 → 0550 → 0550. These are palindromes.
3. 11:22 → 1122 backwards is 2211, which is different. So 11:22 is not a palindromic time.

Teacher's Note:
1. Ignore the colon and read the four digits.
2. 11:11 would be a palindrome, 11:22 is not.

Question 4: What is the digit sum of the smallest 5-digit number divisible by 9?
(a) 3
(b) 9
(c) 18
(d) 27
Show Answer & Explanation

Answer: (b) 9

Explanation:
1. A number is divisible by 9 when its digit sum is a multiple of 9.
2. The smallest 5-digit number is 10000 (digit sum 1). To reach a digit sum of 9 with the smallest number, change the last digit to 8: 10008.
3. Check: 10008 ÷ 9 = 1112. Its digit sum is 1 + 0 + 0 + 0 + 8 = 9.

Teacher's Note:
1. Use the divisibility rule of 9.
2. Keep the left digits as small as possible.

Question 5: How many times does the digit '0' appear in the numbers from 1 to 50?
(a) 5
(b) 6
(c) 4
(d) 1
Show Answer & Explanation

Answer: (a) 5

Explanation:
1. The one-digit numbers 1 to 9 have no 0.
2. A 0 appears only in the units place of 10, 20, 30, 40 and 50.
3. So the digit 0 appears 5 times.

Teacher's Note:
1. 0 cannot be the first digit of a number.
2. List the multiples of 10 up to 50.

Question 6: The largest 4-digit palindrome divisible by 9 is:
(a) 9889
(b) 9779
(c) 9669
(d) 9999
Show Answer & Explanation

Answer: (d) 9999

Explanation:
1. 9999 is the largest 4-digit number and it reads the same both ways, so it is a palindrome.
2. Its digit sum is 9 + 9 + 9 + 9 = 36, which is a multiple of 9, so 9999 is divisible by 9 (9999 ÷ 9 = 1111).
3. No 4-digit number is bigger, so the answer is 9999.

Teacher's Note:
1. Test the largest candidate first.
2. Digit sums: 9889 → 34, 9779 → 32, 9669 → 30, none a multiple of 9.

Question 7: Which digit appears most often in the hundreds place from 100 to 999?
(a) 1
(b) 2
(c) 9
(d) All appear equally often
Show Answer & Explanation

Answer: (d) All appear equally often

Explanation:
1. Numbers 100 to 199 have hundreds digit 1: that is 100 numbers.
2. In the same way 200 to 299 have digit 2, and so on up to 900 to 999 with digit 9.
3. Each of the digits 1 to 9 appears 100 times, so they all appear equally often.

Teacher's Note:
1. Group the numbers in hundreds.
2. 0 never appears in the hundreds place of a 3-digit number.

Question 8: In the game “first to 60 wins”, moves allowed are 1 to 6, starting at 0. To win, your first move should be:
(a) 4
(b) 5
(c) 6
(d) 3
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. In each round, whatever your friend adds (1 to 6), you can add enough to make the round total 7.
2. So you want to say 60, and before that 53, 46, 39, 32, 25, 18, 11, 4 (each 7 less).
3. 60 ÷ 7 leaves remainder 4, so the first move should be 4.

Teacher's Note:
1. Largest move + 1 = 7 is the key number.
2. Work backwards from the target in steps of 7.

Question 9:
Assertion (A): Every 4-digit number with all digits equal does not reach 6174 by Kaprekar’s process.
Reason (R): When all digits are equal, subtracting the largest and smallest numbers gives 0, so the process stops immediately.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For 1111, the largest and smallest numbers are both 1111, and 1111 − 1111 = 0.
2. From 0 we only keep getting 0, so 6174 is never reached. A is true.
3. R is true and it is exactly why A happens. So both are true and R explains A.

Teacher's Note:
1. Kaprekar’s process needs at least two different digits.
2. Numbers like 2222 or 7777 give 0 at once.

Question 10:
Assertion (A): In the Collatz sequence, if the starting number is a power of 2, it takes a certain number of steps to reach 1 by repeatedly dividing by 2.
Reason (R): Each step divides the number by 2, making it smaller until it reaches 1.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A power of 2 is 2 × 2 × … × 2, for example 8 = 2 × 2 × 2.
2. Every number in its chain (8, 4, 2) is even, so the Collatz rule always says “divide by 2”.
3. 8 → 4 → 2 → 1 takes exactly 3 steps. Both statements are true and R explains A, so the answer is (a).

Teacher's Note:
1. For a power of 2 the “3 × n + 1” rule is never needed.
2. 16 takes 4 steps, 32 takes 5 steps.

Section B: Short Answer Questions (2 Marks Each)

Question 11. Find the smallest and largest 4-digit number whose digit sum is 9.
Answer:
Smallest: the thousands digit must be at least 1, so take 1. The other digits must add to 8; put the 8 in the units place and 0s in between: 1008. Check: 1 + 0 + 0 + 8 = 9.
Largest: make the thousands digit as big as possible, 9. Then the rest must be 0: 9000. Check: 9 + 0 + 0 + 0 = 9.
Smallest = 1008, largest = 9000.
In simple words: Small numbers keep big digits at the right end; large numbers keep them at the left end.

Teacher's Note:
1. The first digit of a 4-digit number cannot be 0.
2. Always check the digit sum at the end.

 

Question 12. Identify all supercells in the 3×3 table:
560055505700
558057205590
550056505540

Answer:
A supercell is greater than every cell directly above, below, left and right of it.
5600: neighbours 5550, 5580 → greater than both, supercell.
5700: neighbours 5550, 5590 → greater than both, supercell.
5720: neighbours 5550, 5580, 5590, 5650 → greater than all, supercell.
The other cells each have a bigger neighbour (for example 5650 is next to 5720, and 5590 is next to 5700).
Supercells: 5600, 5700 and 5720.
In simple words: Check each cell against only the cells that share a side with it.

Teacher's Note:
1. Corner cells have 2 neighbours, edge cells 3, the centre 4.
2. Diagonal cells are not neighbours.

 

Question 13. Write the smallest 5-digit number whose digit sum is 23.
Answer:
Keep the left digits as small as possible and push the big digits to the right.
The first digit must be at least 1; take 1. Then the remaining four digits must add to 22.
The last two digits can be at most 9 + 9 = 18, so the hundreds digit must be 22 − 18 = 4, and the thousands digit can be 0.
Number = 10499. Check: 1 + 0 + 4 + 9 + 9 = 23.
Smallest number = 10499.
In simple words: Fill 9s from the right and keep the left side small.

Teacher's Note:
1. Filling from the right with 9s gives the smallest number.
2. Do not forget the first digit cannot be 0.

 

Question 14. List all 3-digit palindromes made from digits 4, 6, and 9 (repetition allowed).
Answer:
A 3-digit palindrome has the form aba: the first and last digits are the same.
The first digit a can be 4, 6 or 9 (3 ways) and the middle digit b can be 4, 6 or 9 (3 ways), so there are 3 × 3 = 9 palindromes.
They are 444, 464, 494, 646, 666, 696, 949, 969, 999.
In simple words: Choose the outer digit and the middle digit; that fixes the palindrome.

Teacher's Note:
1. List them in order so none is missed.
2. Count check: 3 choices × 3 choices = 9.

 

Section C: Short Answer Questions (3 Marks Each)

Question 15. Perform the Kaprekar process for 9013 until 6174 is reached. State steps.
Answer:
In each step arrange the digits in decreasing order (largest number) and increasing order (smallest number, keeping a 0 in front if needed), then subtract.
Step 1: 9013 → largest 9310, smallest 0139 → 9310 − 0139 = 9171
Step 2: 9171 → largest 9711, smallest 1179 → 9711 − 1179 = 8532
Step 3: 8532 → largest 8532, smallest 2358 → 8532 − 2358 = 6174
Chain: 9013 → 9171 → 8532 → 6174. 6174 is reached in 3 steps.
In simple words: Largest minus smallest, again and again, until 6174 appears.

Teacher's Note:
1. Write 0139 with its 0 so the digits are not lost.
2. Once 6174 appears it repeats: 7641 − 1467 = 6174.

 

Question 16. Start with the number 59. Reverse its digits and add it to the original number. If the result is not a palindrome, repeat the process with the new number. Find the palindrome number you get after this process.
Answer:
Step 1: 59 + 95 = 154. 154 reversed is 451, so it is not a palindrome.
Step 2: 154 + 451 = 605. 605 reversed is 506, so it is not a palindrome.
Step 3: 605 + 506 = 1111. 1111 reads the same both ways.
The palindrome reached is 1111 (after 3 steps).
In simple words: Keep reversing and adding until the answer reads the same both ways.

Teacher's Note:
1. Check for a palindrome after every addition.
2. Add carefully; one carry mistake changes everything.

 

Question 17. A 4-digit number is formed using the digits 1, 2, 3, and 4 without repetition. What is the difference between the largest possible even number and the smallest possible odd number that can be formed?
Answer:
Largest even number: it must end in 2 or 4. Ending in 2, the rest in decreasing order gives 4312; ending in 4 gives 3214. The larger is 4312.
Smallest odd number: it must end in 1 or 3. Ending in 3, the rest in increasing order gives 1243; ending in 1 gives 2341. The smaller is 1243.
Difference = 4312 − 1243 = 3069.
In simple words: Fix the last digit first, then arrange the other digits.

Teacher's Note:
1. Even or odd depends only on the units digit.
2. Try every possible last digit and compare.

 

Section D: Long Answer Question (5 Marks)

Question 18. Collatz sequence starting from 19 until reaching 1. Count total steps and state the maximum number reached in the sequence.
Answer:
Rule: if the number is even, halve it; if it is odd, multiply by 3 and add 1.
19 (odd) → 3 × 19 + 1 = 58
58 → 29 → (odd) 3 × 29 + 1 = 88 → 44 → 22 → 11
11 (odd) → 3 × 11 + 1 = 34 → 17 → (odd) 3 × 17 + 1 = 52 → 26 → 13
13 (odd) → 3 × 13 + 1 = 40 → 20 → 10 → 5 → (odd) 3 × 5 + 1 = 16 → 8 → 4 → 2 → 1
Full sequence: 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.
There are 21 numbers, so the number of steps (arrows) = 21 − 1 = 20 steps.
The largest number in the sequence is 88.
In simple words: Halve the evens, triple-and-add-one the odds, and count the arrows to 1.

Teacher's Note:
1. Steps are counted as arrows, one less than the numbers listed.
2. Collatz chains can climb high before they fall.

 

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

Question 19. Supercell Upgrade: In a strategy game, players manage a 4×4 grid of cells, each represented by a five-digit number. The current grid contains some supercells with special properties. The challenge is to analyze the existing grid, identify all current supercells, and then improve the setup by changing exactly one digit in two different cells to create at least two additional supercells. This task tests your observation and problem-solving skills to optimize the grid layout effectively.
61000608006150061200
60600618006070060500
60300604006200060200
60000601005990059800
Tasks:
(a) Identify all current supercells.
(b) Change exactly one digit in two cells to create at least 2 more supercells.
Answer:
(a) Compare each cell with the cells directly above, below, left and right.
61000 > 60800, 60600. 61500 > 60800, 61200, 60700. 61800 > 60800, 60400, 60600, 60700. 62000 > 60700, 59900, 60400, 60200.
Every other cell has a larger neighbour (for example 61200 is next to 61500, 60500 is below 61200).
Current supercells: 61000, 61500, 61800 and 62000 (4 supercells).
(b) One possible answer: change the first digit 6 to 7 in two cells.
60500 → 70500: now greater than 61200, 60700 and 60200, so it becomes a supercell.
60300 → 70300: now greater than 60600, 60400 and 60000, so it becomes a supercell.
The old four supercells are not next to these cells, so they stay supercells. The grid now has 6 supercells:
61000608006150061200
60600618006070070500
70300604006200060200
60000601005990059800

In simple words: A cell is super when it beats every side neighbour; raising one digit can make it super.

Teacher's Note:
1. Changing the leading digit makes the biggest jump.
2. Other correct pairs of changes are also accepted.

 

Question 20. Kaprekar Contest: Three friends—A, B, and C—decided to challenge themselves with a fascinating number puzzle called Kaprekar’s routine. Each picked a different four-digit number: A chose 8640, B selected 7032, and C took 5421. Their goal was to apply the Kaprekar steps repeatedly and see how many iterations it takes for each number to reach the mysterious constant 6174. They also wanted to find out who among them would reach 6174 first in this exciting mathematical race.
(a) How many steps for each?
(b) Who reaches 6174 first?
Answer:
A (8640): 8640 − 0468 = 8172; 8721 − 1278 = 7443; 7443 − 3447 = 3996; 9963 − 3699 = 6264; 6642 − 2466 = 4176; 7641 − 1467 = 6174.
B (7032): 7320 − 0237 = 7083; 8730 − 0378 = 8352; 8532 − 2358 = 6174.
C (5421): 5421 − 1245 = 4176; 7641 − 1467 = 6174.
(a) A takes 6 steps, B takes 3 steps and C takes 2 steps.
(b) C needs the fewest steps, so C (5421) reaches 6174 first.
In simple words: Count the subtractions each friend needs; the fewest wins.

Teacher's Note:
1. Keep the leading 0 in numbers like 0468.
2. Each subtraction counts as one step.

 

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Download Practice Assignments: Class 6 Mathematics Chapter 03 Number Play

Class 6 Mathematics Chapter 03 Number Play Printable Assignments

Access structured practice assignments for Chapter 03 Number Play designed in alignment with the latest CBSE curriculum for Class 6 Mathematics. These printable sets cover objective and descriptive problem types to support thorough revision.

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Steps to Complete Chapter 03 Number Play Assignments Successfully

  1. Initial Reading: Begin by reading the NCERT book for Class 6 Mathematics to build a baseline understanding.
  2. Independent Testing: Attempt assignment questions unassisted, then verify work using provided answer keys.
  3. Supplementary Aids: Leverage revision notes and worksheets whenever you encounter difficult topics.

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Yes. These assignments are designed as per the latest CBSE syllabus for 2026. We have included huge variety of question formats such as MCQs, Case-study based questions and important diagram-based problems found in Chapter 03 Number Play.

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