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Chapter 3 Exercises and Summary

🎓 Class 6 Mathematics CBSE Theory Ch 3 — Number Play ⏱ ~35 min
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This MCQ module is based on: Chapter 3 Exercises and Summary

This mathematics assessment will be based on: Chapter 3 Exercises and Summary
Targeting Class 6 level in Number Theory, with Basic difficulty.

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Figure it Out — End-of-Chapter Exercises

Solve each problem. Reveal the worked-out solution when ready.

Q1. There is only one supercell (number greater than all its neighbours) in this grid. If you exchange two digits of one of the numbers, there will be 4 supercells. Figure out which digits to swap. Grid: 16,200 | 39,344 | 29,765 | 23,609 | 62,871 | 45,386 | 19,381 | 50,319 | 38,408
Exchange the digits 1 and 6 in 16,200: 16,200 → 61,200 (rearranging digits 1 and 6). New grid: 61,200, 39,344, 29,765, 23,609, 62,871, 45,386, 19,381, 50,319, 38,408. Now supercells are: 61,200, 62,871, 45,386, 50,319 → 4 supercells.
Q2. How many rounds does your year of birth take to reach the Kaprekar constant?
Example: birth year 2011. Digits 2, 0, 1, 1. Largest = 2110, Smallest = 0112 = 112 (treated as 4-digit 0112). 2110 − 0112 = 1998. 9981 − 1899 = 8082. 8820 − 0288 = 8532. 8532 − 2358 = 6174. 4 rounds. For 1980: 1980 → 9810 − 0189 = 9621 → 9621 → 9621-1269 = 8352 → 8532 − 2358 = 6174. 3 rounds. (Try your own birth year.) Most reach 6174 in ≤ 6 rounds.
Q3. We are the group of 5-digit numbers between 35,000 and 75,000 such that all our digits are odd. Who is the largest number in our group? Who among us is the closest to 50,000?
Odd digits: 1, 3, 5, 7, 9.
Largest (with repetition): 73,999 — no wait, must be <75,000 and all digits odd: 73,999? 9 is odd ✓, 7 odd ✓, 3 odd ✓. So 73,999. With no repeating digits: 73,591. (NCERT answer with repeating: 73,999. Without repeating: 73,591.)
Closest to 50,000: 50,000 itself has even digits. Next odd-digit options: 51,111 (diff 1,111) or 35,111 no — must be in range. 51,111 (diff = 1,111). Without repeating: 51,379 (diff = 1,379) or 35,117 etc. Answer: 51,111 closest with repetition; 51,379 without repetition.
Q4. Estimate the number of holidays you get in a year including weekends, festivals and vacation. Then, try to get an exact number and see how close your estimate was.
Weekends: 52 × 2 = 104. National + festival holidays: ~20. Summer vacation: ~30. Winter vacation: ~10. Total ≈ 164 days out of 365. Actual may vary per school/state.
Q5. Estimate the number of litres a mug, a bucket and an overhead tank can hold.
Mug ≈ 0.5–1 litre; Bucket ≈ 15–20 litres; Overhead tank ≈ 500–1000 litres.
Q6. Write one 5-digit number and two 3-digit numbers such that their sum is 18,670.
18,000 + 300 + 370 = 18,670. So 18,000 (5-digit), 300, 370. Try any others: 18,250 + 200 + 220 = 18,670.
Q7. Choose a number between 210 and 390. Create a number pattern similar to those shown in Section 3.9 that will sum up to this number.
Choose 250. Pattern A: rows of 25 and 50. 25 + 25 = 50, then 50 + 50 + 50 = 150, then 25 + 25 = 50. Total = 50 + 150 + 50 = 250 ✓.
Pattern B: five rows of 10 × 5 = 50 each → 50 × 5 = 250 ✓ using 25 tens.
Q8. Recall the sequence of Powers of 2 from Chapter 1, Table 1: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, ... Why is the Collatz conjecture correct for all the starting numbers in this sequence?
A power of 2, like \(2^n\), is always even. Applying Collatz: \(2^n \div 2 = 2^{n-1}\). Each step halves and stays a power of 2. Eventually we reach 2, then 1. So the conjecture trivially holds.
Q9. Check if the Collatz conjecture holds for the starting number 100.
100 → 50 → 25 → 76 → 38 → 19 → 58 → 29 → 88 → 44 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1. Yes, reaches 1 in 25 steps.
Q10. Starting with 0, players alternate adding numbers between 1 and 3. The first person to reach 22 wins. What is the winning strategy now?
Step = 1 + 3 = 4. Target ladder (mod 4 = 22 mod 4 = 2): 22, 18, 14, 10, 6, 2. First player says 2 first, then mirrors opponent's move to keep on ladder. First player wins.

Bonus Puzzle — Palindrome Riddle

Puzzle (p. 62)
"I am a 5-digit palindrome. I am an odd number. My 't' digit is double my 'u' digit. My 'h' digit is double my 't' digit. Who am I?"
Solve the palindrome riddle.
A 5-digit palindrome reads u t h t u. Conditions: u is odd (number ends in u), t = 2u, h = 2t = 4u.
Try u = 1: t = 2, h = 4 → 12421 ✓ (odd, palindrome, t = 2×1, h = 2×2 = 4).
Try u = 2 (not odd, skip). Try u = 3 (wait, u must make h = 12 which isn't a digit). So only u = 1 works.
Answer: 12,421.
Activity: Predict → Observe → Explain — Chapter Check
Predict: For 3-digit numbers, does Kaprekar's process also give a fixed constant?
  1. Pick any 3-digit number with at least two different digits (e.g. 482).
  2. Form largest − smallest arrangements of its digits.
  3. Repeat the subtraction until you see repetition.
  4. Try at least 5 different starting numbers.
Observe: Every 3-digit number with at least two distinct digits reaches 495 in at most 6 steps. Example: 482 → 842−248 = 594 → 954−459 = 495 → 954 − 459 = 495 (fixed). Explain: 495 is the 3-digit Kaprekar constant — just as 6174 is the 4-digit one.
Competency-Based Questions
Scenario: At a number carnival, visitors play three games: (i) spot supercells in a grid, (ii) apply Kaprekar's process, (iii) play the 21-game. Each is scored on strategy and accuracy.
Q1. Apply: In a row 480, 920, 350, 870, 210, 660, 120, identify the supercells.
L3 Apply
Compare each to neighbours: 480 vs 920 (end, need just >920 ✗); 920 > 480 and > 350 ✓; 350 ✗; 870 > 350 and > 210 ✓; 210 ✗; 660 > 210 and > 120 ✓; 120 end ✗. Supercells: 920, 870, 660.
Q2. Analyse: In Kaprekar's process, why must there be at least two different digits in the starting number?
L4 Analyse
If all digits are the same (e.g. 7777), largest and smallest arrangements are identical, so A − B = 0 — not a valid 4-digit number, and the process cannot proceed.
Q3. Evaluate: Manvi says "In any number game where players add 1 to N and try to reach a target T, the first player wins iff T is divisible by (1 + N + 1) = N+2." Evaluate her claim.
L5 Evaluate
Partially correct but slightly off. The ladder step is (1 + N) not (N + 2). The first player wins iff T is divisible by (N + 1) — they start on the ladder, so the first move must reach T mod (N+1). If T mod (N+1) = 0, the first player has no safe first move and the second player wins. So her formula's direction is inverted.
Q4. Create a new number-pattern puzzle: design a palindromic 5-digit number that is also divisible by 11, and explain how you found it.
L6 Create
All 5-digit palindromes abcba have alternating digit sum (a − b + c − b + a) = 2a − 2b + c. For divisibility by 11, this difference must be 0 or ±11. Try a = 1, b = 1, c = 0 → 11011 ÷ 11 = 1001 ✓. 11011 is a 5-digit palindrome divisible by 11. Another: 12321? Alt sum = 1−2+3−2+1 = 1 (not divisible by 11). So 12321 NO. 13431 → 1−3+4−3+1 = 0 → 13431 ÷ 11 = 1221 ✓.
Assertion–Reason Questions
A: The digit '7' appears 20 times from 1 to 100.
R: Each digit from 0 to 9 appears an equal number of times in the units place from 1 to 100.
(a) Both true, R explains A
(b) Both true, R doesn't explain A
(c) A true, R false
(d) A false, R true
Answer: (b) — 7 appears 10 times in units (7, 17, …, 97) and 10 times in tens (70–79). Total 20. R is true but describes only units, not both places. Does not fully explain A.
A: Kaprekar's process for any 3-digit number (with 2 distinct digits) ends at 495.
R: 495 is a fixed point: 954 − 459 = 495.
(a) Both true, R explains A
(b) Both true, R doesn't explain A
(c) A true, R false
(d) A false, R true
Answer: (a) — Both true, R explains why 495 is the terminal point.
A: Estimation is always inferior to exact calculation.
R: Estimation saves time when precise answers are not needed.
(a) Both true, R explains A
(b) Both true, R doesn't explain A
(c) A true, R false
(d) A false, R true
Answer: (d) — A is false (estimation is valuable in its own right). R correctly states its value.

Chapter 3 — Summary

Key Takeaways

  • Numbers can tell us things — about heights, counts, positions — beyond mere counting.
  • Supercells are cells larger than all their neighbours. Two supercells cannot be adjacent. Maximum supercells in a row of n cells = \(\lceil n/2 \rceil\).
  • On a number line, the scale must be chosen first; numbers are placed by their size.
  • Digit sum is the sum of a number's digits; it is the same under digit rearrangement.
  • Palindromes read the same forwards and backwards. The reverse-and-add process usually produces a palindrome in a few steps.
  • Kaprekar's constant 6174 is reached by every 4-digit number (with at least 2 distinct digits) within 7 steps. For 3-digit numbers, the constant is 495.
  • Mental math strategies: round, split and regroup numbers for faster arithmetic.
  • Collatz conjecture: even → ÷2, odd → 3n+1. Every tested number reaches 1 — but this is still unproven.
  • Estimation is a valuable skill for real-life situations where exact counts are unnecessary or impractical.
  • Number games like the 21-game have winning strategies based on a target ladder with step size (min + max of allowed moves).
  • Thinking about and formulating procedures to use numbers is a skill called computational thinking.

Frequently Asked Questions

What exercises are in Class 6 Ganita Prakash Chapter 3?

Chapter 3 exercises cover supercells identification, palindrome generation, Kaprekar's routine practice, Collatz sequence steps, mental math computations, and clock arithmetic problems. Each exercise reinforces a concept from the main chapter of NCERT Class 6 Ganita Prakash.

How do you solve supercell exercises?

For each cell in the given grid, compare its number with every neighbour. If the cell is strictly greater than all neighbours, mark it as a supercell. Check boundary cells carefully because they have fewer neighbours. Follow this method for NCERT Class 6 Chapter 3 exercises.

What is the main summary of Number Play chapter?

Number Play teaches that numbers have rich patterns beyond basic operations. Key ideas include supercells, palindromes, Kaprekar's constant 6174, the Collatz conjecture, and mental math strategies. NCERT Class 6 Ganita Prakash Chapter 3 emphasises exploration and reasoning.

How many steps to reach 6174 for 1234?

Start 1234. 4321 - 1234 = 3087. 8730 - 0378 = 8352. 8532 - 2358 = 6174. Three steps to reach Kaprekar's constant. NCERT Class 6 Chapter 3 exercises ask students to count such steps.

Why is the final part called Exercises and Summary?

The final part of each NCERT Class 6 Ganita Prakash chapter consolidates learning through practice problems and a summary of key ideas. Solving Chapter 3 exercises tests understanding of supercells, palindromes, Kaprekar, Collatz and mental math.

How do Chapter 3 exercises prepare students for higher classes?

Chapter 3 exercises build number sense, logical reasoning, and pattern recognition - skills essential for algebra, sequences, and problem-solving in later classes. The Number Play approach of NCERT Class 6 makes mathematics exploratory and engaging.

Frequently Asked Questions — Number Play

What is Chapter 3 Exercises and Summary in NCERT Class 6 Mathematics?

Chapter 3 Exercises and Summary is a key concept covered in NCERT Class 6 Mathematics, Chapter 3: Number Play. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Chapter 3 Exercises and Summary step by step?

To solve problems on Chapter 3 Exercises and Summary, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 6 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 3: Number Play?

The essential formulas of Chapter 3 (Number Play) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Chapter 3 Exercises and Summary important for the Class 6 board exam?

Chapter 3 Exercises and Summary is part of the NCERT Class 6 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Chapter 3 Exercises and Summary?

Common mistakes in Chapter 3 Exercises and Summary include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Chapter 3 Exercises and Summary?

End-of-chapter NCERT exercises for Chapter 3 Exercises and Summary cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 3, and solve at least one previous-year board paper to consolidate your understanding.

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