CBSE Class 6 Mathematics Number Play Practice Paper Assignment Set 03 with Answers

Here is Practice Paper 09 of Class 6 Maths Ganita Prakash Chapter 3 Number Play: 20 questions for 40 marks in Sections A to E, each followed by a clear step-by-step solution. The questions cover multiples of 13, forming the largest even and smallest odd numbers from given digits, counting 4-digit numbers with a digit sum of 3, the reverse-and-add route to palindromes, a magic square with sum 15, the Collatz sequence from 20 and the Kaprekar steps for 9210.

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 '8' appears in the numbers from 1 to 90?
Answer:
Units place: 8, 18, 28, 38, 48, 58, 68, 78, 88 → 9 times.
Tens place: 80, 81, 82, …, 89 → 10 times.
88 has two 8s, one counted in each list. Total = 9 + 10 = 19 times.
In simple words: Add the 8s in the units place to the 8s in the tens place.

Teacher's Note:
1. 90 has no 8.
2. Write the list in order to avoid missing one.

Question 12. Find the smallest and largest 3-digit numbers whose digits are all different and odd.
Answer:
The odd digits are 1, 3, 5, 7, 9.
Smallest: take the three smallest, 1, 3, 5, in increasing order: 135.
Largest: take the three largest, 9, 7, 5, in decreasing order: 975.
Smallest = 135, largest = 975.
In simple words: Pick the smallest or largest odd digits and arrange them in order.

Teacher's Note:
1. 0 is even, so it cannot be used here.
2. Digits must not repeat.

Question 13. Find the palindrome that is reached by repeating the reverse-and-add method, starting with the number 48.
Answer:
Step 1: 48 + 84 = 132. 132 reversed is 231, so it is not a palindrome yet.
Step 2: 132 + 231 = 363. 363 reads the same both ways.
The palindrome reached is 363 (after two steps).
In simple words: Reverse and add until the answer reads the same both ways.

Teacher's Note:
1. Check for a palindrome after each addition.
2. One step gives only 132.

Question 14. How many 4-digit numbers have a digit sum of 3?
Answer:
The thousands digit is 1, 2 or 3 (it cannot be 0).
Thousands digit 3: 3000 (1 number).
Thousands digit 2: the rest add to 1 → 2100, 2010, 2001 (3 numbers).
Thousands digit 1: the rest add to 2 → 1200, 1020, 1002, 1110, 1101, 1011 (6 numbers).
Total = 1 + 3 + 6 = 10 numbers.
In simple words: Fix the first digit, then list the ways to share what is left.

Teacher's Note:
1. Organise the list by the first digit.
2. Counting in groups prevents repeats and gaps.

Section C: Short Answer Questions (3 Marks Each)

Question 15. List all 3-digit palindromes using digits 2, 4, and 7.
Answer:
A 3-digit palindrome is aba; a and b can each be 2, 4 or 7 (repetition allowed), giving 3 × 3 = 9 numbers.
a = 2: 222, 242, 272
a = 4: 424, 444, 474
a = 7: 727, 747, 777
The palindromes are 222, 242, 272, 424, 444, 474, 727, 747, 777.
In simple words: Choose the outer digit and the middle digit in every possible way.

Teacher's Note:
1. The first and last digits must match.
2. 3 choices × 3 choices = 9 numbers.

Question 16. Using digits 3, 5, 6, 9, find largest and smallest 4-digit numbers, their sum, and difference.
Answer:
Largest: digits in decreasing order → 9653.
Smallest: digits in increasing order → 3569.
Sum: 9653 + 3569 = 13222.
Difference: 9653 − 3569 = 6084.
In simple words: Arrange the digits both ways, then add and subtract.

Teacher's Note:
1. Check the difference: 6084 + 3569 = 9653.
2. Line up the place values when adding.

Teacher's Note:
1. 5 is always the centre of a 1 to 9 magic square.
2. Check every line once the square is full.

Section D: Long Answer Question (5 Marks)

Question 18. Explain the Collatz sequence with a detailed example. Show the complete sequence starting from 20 until it reaches 1.
Answer:
Collatz rules: start with any counting number. If it is even, divide it by 2; if it is odd, multiply it by 3 and add 1. Keep going with each new number.
Lothar Collatz guessed that every starting number finally reaches 1. Nobody has proved it yet, but it has been checked for very large numbers.
Example: 20
20 (even) → 20 ÷ 2 = 10
10 (even) → 10 ÷ 2 = 5
5 (odd) → 3 × 5 + 1 = 16
16 (even) → 8 → 4 → 2 → 1
Sequence: 20, 10, 5, 16, 8, 4, 2, 1 (7 steps; largest number 20).
In simple words: Halve the evens, do 3 × n + 1 for the odds, and stop at 1.

Teacher's Note:
1. Write “even” or “odd” before choosing the rule.
2. The conjecture is still unproved.

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

Question 19. Case Study: The Palindrome Puzzle
Dr. Anya's lab was a maze of computers, all guarded by a single code. Her assistant, Leo, had to enter the password to access the research. He remembered the clues Dr. Anya had given him. "It's a five-digit palindrome," he muttered to himself, "The sum of its digits is 25, the hundreds digit is 7, and the thousands digit is 8." Leo's fingers hovered over the keypad, ready to unlock the secret.
(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: first digit = last digit, second digit = fourth digit, and c is the middle (hundreds) digit.
(b) Hundreds digit c = 7 and thousands digit b = 8, so the number has the form a 8 7 8 a.
Digit sum: a + 8 + 7 + 8 + a = 25, so 2a + 23 = 25, which gives 2a = 2 and a = 1.
Password = 18781. Check: 1 + 8 + 7 + 8 + 1 = 25 and it reads the same both ways.
In simple words: Put the known digits into abcba and use the digit sum to find the last one.

Teacher's Note:
1. The hundreds digit is the middle digit of a 5-digit number.
2. Without the thousands-digit clue, a + b = 9 would allow nine passwords.

Question 20. Case Study: Kaprekar's sequence
A student had always found math problems a bit tedious, but this one was different. His teacher had given him a challenging puzzle: start with the number 9210 and find its Kaprekar's sequence. "Kaprekar's sequence?" he thought. "It sounds like a secret code." As he began arranging the digits and performing the subtractions, he felt like a detective, getting closer to a hidden secret with each step.
(a) What is Kaprekar's constant for 4-digit numbers? [1 mark]
(b) Show the steps for the number 9210 to reach the constant. [3 marks]
Answer:
(a) Kaprekar’s constant for 4-digit numbers is 6174.
(b)
Step 1: 9210 → largest 9210, smallest 0129 → 9210 − 0129 = 9081
Step 2: 9081 → largest 9810, smallest 0189 → 9810 − 0189 = 9621
Step 3: 9621 → largest 9621, smallest 1269 → 9621 − 1269 = 8352
Step 4: 8352 → largest 8532, smallest 2358 → 8532 − 2358 = 6174
Sequence: 9210 → 9081 → 9621 → 8352 → 6174, reached in 4 steps.
In simple words: Largest minus smallest, four times, takes 9210 to 6174.

Teacher's Note:
1. Write 0129 with the 0 in front.
2. 6174 repeats itself, so stop there.