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

Practice Paper 08 for Class 6 Maths Ganita Prakash Chapter 3 Number Play has 20 questions worth 40 marks, arranged in Sections A to E, and every question below comes with a worked answer. This paper focuses on counting how often a digit appears, smallest and largest numbers with given digits or digit sums, 4-digit palindromes and a palindrome password puzzle, a 3×3 magic square, Kaprekar’s constants 495 and 6174, and the Collatz sequences of 10 and 13.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. What is the number of times the digit '2' appears in the numbers from 1 to 30?
Answer:
Units place: 2, 12, 22 → 3 times.
Tens place: 20, 21, 22, …, 29 → 10 times.
22 has two 2s, one in each list. Total = 3 + 10 = 13 times.
In simple words: Count the 2s in the units place and the tens place separately and add.

Teacher's Note:
1. 30 itself has no 2.
2. Do not forget the second 2 in 22.

Question 12. Find the smallest and largest 3-digit numbers whose digits are all different.
Answer:
Smallest: the first digit cannot be 0, so it is 1. Then use the smallest unused digits, 0 and 2: 102.
Largest: use the three largest digits 9, 8, 7 in decreasing order: 987.
Smallest = 102, largest = 987.
In simple words: Smallest digits at the front for the smallest number, largest digits at the front for the largest.

Teacher's Note:
1. 100 and 101 repeat digits, so they are not allowed.
2. 0 can be used, but not in the first place.

Question 13. Write down the complete Collatz sequence for the starting number 10.
Answer:
10 is even → 10 ÷ 2 = 5.
5 is odd → 3 × 5 + 1 = 16.
16 → 8 → 4 → 2 → 1 (each even, so halve).
Sequence: 10, 5, 16, 8, 4, 2, 1.
In simple words: Halve when even, multiply by 3 and add 1 when odd, until you get 1.

Teacher's Note:
1. Stop as soon as you reach 1.
2. This chain takes 6 steps.

Question 14. How many 2-digit numbers have a digit sum of 1?
Answer:
The tens digit of a 2-digit number is at least 1.
If the digit sum is 1, the tens digit must be 1 and the units digit must be 0, giving 10.
So there is only 1 such number, namely 10.
In simple words: Only 10 has digits adding to 1.

Teacher's Note:
1. 01 is not a 2-digit number.
2. Think about the smallest possible tens digit.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Find the smallest and largest 4-digit numbers whose digit sum is 3.
Answer:
Smallest: first digit 1 (it cannot be 0); the remaining 2 goes to the units place: 1002. Check: 1 + 0 + 0 + 2 = 3.
Largest: put the whole sum in the thousands place: 3000. Check: 3 + 0 + 0 + 0 = 3.
Smallest = 1002, largest = 3000.
In simple words: Push the digit sum to the right for the smallest number and to the left for the largest.

Teacher's Note:
1. The first digit must be at least 1.
2. Check the digit sum of each answer.

Question 16. Explain why the number 693 will reach the Kaprekar's constant of 495. Show the first two steps.
Answer:
The Kaprekar steps reach 495 for every 3-digit number whose digits are not all the same. 693 has three different digits (6, 9, 3), so it will reach 495.
Step 1: largest 963, smallest 369 → 963 − 369 = 594.
Step 2: largest 954, smallest 459 → 954 − 459 = 495.
So 693 reaches the constant 495 in two steps.
In simple words: Different digits always lead to 495; here it takes two subtractions.

Teacher's Note:
1. Rearrange the digits of the new number in every step.
2. 495 gives 954 − 459 = 495 again.

Teacher's Note:
1. Start with a line that has only one blank.
2. Check both diagonals at the end.

Section D: Long Answer Question (5 Marks)

Question 18. Explain the Kaprekar's constant for 4-digit numbers with a detailed example. Choose a starting number of your choice and show the complete sequence of steps.
Answer:
Take any 4-digit number whose digits are not all the same. Form the largest number (digits in decreasing order) and the smallest number (digits in increasing order, keeping any 0 in front) and subtract. Repeat with the answer.
D. R. Kaprekar found that this always ends at 6174, in at most 7 steps. It is called Kaprekar’s constant, because 7641 − 1467 = 6174 brings us back to the same number.
Example: start with 3456.
Step 1: 3456 → largest 6543, smallest 3456 → 6543 − 3456 = 3087
Step 2: 3087 → largest 8730, smallest 0378 → 8730 − 0378 = 8352
Step 3: 8352 → largest 8532, smallest 2358 → 8532 − 2358 = 6174
Chain: 3456 → 3087 → 8352 → 6174, so 6174 is reached in 3 steps.
One more step gives 7641 − 1467 = 6174 again, so the process stops changing.
In simple words: Largest minus smallest, repeated, always lands on 6174 for 4-digit numbers.

Teacher's Note:
1. Any starting number with at least two different digits is fine.
2. Keep zeros when writing 0378.

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

Question 19. Case Study: The Palindrome Puzzle
An ancient diary lay on a dusty shelf, its pages brittle with age. A young history enthusiast, Alex, stumbled upon a riddle inside: "To open the secret compartment, enter a five-digit password." The only clues were cryptic—it was a palindrome, the hundreds digit was 7, the sum of all its digits was 21, and it was the smallest number fitting these clues. Alex knew this was the key to an untold story.
(a) What is the structure of a 5-digit palindrome? [1 mark]
(b) Find the password. Show your step-by-step reasoning. [3 marks]
Answer:
(a) A 5-digit palindrome has the form abcba: the 1st and 5th digits are equal, the 2nd and 4th digits are equal, and c is the middle (hundreds) digit.
(b) The hundreds digit is the middle digit, so c = 7.
Digit sum: a + b + 7 + b + a = 21, so 2a + 2b = 14, which gives a + b = 7.
Possible (a, b): (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (7, 0). Note a cannot be 0.
The smallest number needs the smallest first digit, a = 1, so b = 6.
Password = 16761. Check: 1 + 6 + 7 + 6 + 1 = 21 and it reads the same both ways.
In simple words: Fix the middle digit as 7, share the rest of the sum equally between the matching pairs, then pick the smallest.

Teacher's Note:
1. In abcba the hundreds digit is the middle one.
2. Without the “smallest” clue there would be seven answers.

Question 20. Case Study: The Collatz Sequence
The teacher had just introduced the fascinating concept of the Collatz sequence, and it immediately captured young Sarah's imagination. Unlike other math problems, this one felt like a journey with no map. Her challenge was to take the number 13 and follow the mysterious rules: if the number was even, she had to halve it; if it was odd, she had to multiply it by three and add one. Sarah eagerly grabbed her pencil, ready to see where the sequence would lead her.
(a) What are the rules for the Collatz sequence? [1 mark]
(b) Show the steps for the number 13 to reach 1. [3 marks]
Answer:
(a) If the number is even, divide it by 2. If the number is odd, multiply it by 3 and add 1. Repeat until you reach 1.
(b) 13 (odd) → 3 × 13 + 1 = 40
40 (even) → 20 → 10 → 5
5 (odd) → 3 × 5 + 1 = 16
16 (even) → 8 → 4 → 2 → 1
Sequence: 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 (9 steps).
In simple words: Apply the even and odd rules one number at a time until 1 appears.

Teacher's Note:
1. Say “even” or “odd” before each step.
2. The largest number reached here is 40.