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Figure it Out — End-of

🎓 Class 7 Mathematics CBSE Theory Ch 6 — Number Play ⏱ ~16 min
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This MCQ module is based on: Figure it Out — End-of

This mathematics assessment will be based on: Figure it Out — End-of
Targeting Class 7 level in Number Theory, with Basic difficulty.

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Figure it Out — End-of-Chapter Exercises

These exercises cover everything we learnt in Chapter 6: parity, magic squares, Virahānka–Fibonacci numbers and cryptarithms. Work them out carefully, then reveal the answers to check your reasoning.

1. A light bulb is ON. Deepa toggles its switch 77 times. Will the bulb be on or off? Why?
Each toggle flips the state. Starting from ON, after an even number of toggles the bulb is ON; after an odd number it is OFF. 77 is odd ⇒ OFF.
2. Liswini has a large old encyclopaedia. When she opened it, several loose pages fell out of it. She counted 50 sheets in total, each printed on both sides. Can the sum of the page numbers of the loose sheets be 6000? Why or why not?
On each printed sheet, the two page numbers are consecutive, one odd and one even. So the two numbers on one sheet sum to an odd number. The sum of 50 odd numbers is even (even count of odds). 6000 is even, so the parity allows it. But more is required — on a loose sheet the two page numbers are of the form \((2k-1)+2k = 4k-1\). Summing 50 such expressions gives a number of the form \(4K - 50\), i.e., \(4K - 50 \equiv 2 \pmod 4\). Check: \(6000 \equiv 0\pmod 4\). So 6000 is impossible as the total of page numbers on 50 loose sheets.
3. Here is a 3 × 3 magic square. For each row and column, the parity of the sum is the same. Fill the circle with 'o' for odd, 'e' for even, and 'o+e' or 'e+o' as needed.
For the 1–9 magic square, each row contains either three odd numbers OR (two even + one odd). The magic sum 15 is odd, so each row must have either 3 odd numbers or (1 odd + 2 even). Using the standard square (2, 7, 6 | 9, 5, 1 | 4, 3, 8) parity-pattern is (e, o, e | o, o, o | e, o, e): rows are (oeo → sum odd ✓), (ooo → odd ✓), (eoe → odd ✓). Columns match similarly.
4. Make a 3 × 3 magic square with 0 as the magic sum. All 9 numbers need not be positive.
Take the 1–9 magic square and subtract 5 (the centre) from every entry. Each row/column/diagonal sum becomes \(15 - 15 = 0\). Result: 
 -3  +2  +1
 +4   0  -4
 -1  -2  +3
All distinct integers (including 0 and negatives), every line sums to 0. ✓
5. Fill in the following blanks with 'odd' or 'even':
(a) Sum of an odd number of even numbers is ______.
(b) Sum of an even number of odd numbers is ______.
(c) Sum of an even number of even numbers is ______.
(d) Sum of an odd number of odd numbers is ______.
(a) even   (b) even   (c) even   (d) odd.
6. What is the parity of the sum of the numbers from 1 to 100?
\(1 + 2 + \ldots + 100 = \frac{100 \times 101}{2} = 5050\) — even. (Alternatively, among 1 – 100 there are 50 odd numbers; 50 odds sum to an even number. Adding the 50 even numbers keeps it even.)
7. Two consecutive numbers in the Virahānka sequence are 987 and 1597. What are the next 2 numbers in the sequence?
Next = \(987 + 1597 = 2584\). After that = \(1597 + 2584 = 4181\).
8. Angaan wants to climb an 8-step staircase. His playful rule is that he can take either 1 step or 2 steps at a time. For example, one of his paths is 1, 2, 1, 2, 1, 1, in how many different ways can he reach the top?
Same as Virahānka: \(V_8 = 34\). So 34 different ways.
9. What is the parity of the 20th term of the Virahānka sequence?
The parity pattern is period 3: O, E, O, O, E, O, O, E, .... Evens occur at positions 2, 5, 8, 11, 14, 17, 20. So \(V_{20}\) is even.
10. Identify the statements that are true.
(a) The expression \(4n - 1\) always gives odd numbers.
(b) All even numbers can be expressed as \(6j - 4\).
(c) Both expressions \(2p + 1\) and \(2q - 1\) describe all odd numbers.
(d) The expression \(2f + 3\) gives both even and odd numbers.
(a) True — \(4n\) is even, minus 1 is odd.
(b) False — \(6j - 4\) produces only multiples of 2 with residue 2 mod 6 (i.e., 2, 8, 14, ...). Misses 4, 6, 10, 12, ...
(c) True — both describe all odd integers.
(d) False — \(2f + 3\) is always odd (2f is even, +3 is odd).
11. Solve this cryptarithm: \( \overline{UT} + \overline{TA} = \overline{TAT}\).
\( \overline{UT} + \overline{TA} = \overline{TAT}\). The sum of two 2-digit numbers gives a 3-digit number, so the leading carry produces T = 1 (the hundreds digit). Substituting T = 1: \(\overline{U1} + \overline{1A} = \overline{1A1}\) i.e., \(10U + 1 + 10 + A = 100 + 10A + 1\) ⇒ \(10U + A + 11 = 100 + 10A + 1\) ⇒ \(10U - 9A = 90\) ⇒ \(U = 9, A = 0\).
Answer: U = 9, T = 1, A = 0. Check: 91 + 10 = 101. ✓

SUMMARY

  • In the first activity, we saw how to represent information about how a sequence of numbers (e.g., heights) is arranged without knowing the actual values.
  • We learnt the idea of parity — numbers that can be arranged in pairs (even numbers) and numbers that cannot (odd numbers).
  • We learnt to decide the parity of sums and products using quick parity rules: even + even = even, odd + odd = even, odd + even = odd; and products are odd only when all factors are odd.
  • Exploring sums in grids, we could determine whether filling a grid is impossible by looking only at the row and column parities. We used parity to construct magic squares.
  • We saw how Virahānka numbers were first uncovered in history through ancient Indian studies of rhythm. The Virahānka–Fibonacci sequence is 1, 2, 3, 5, 8, 13, 21, 34, 55, ...
  • We became maths-detectives through cryptarithms, where digits are replaced by letters.

Key Terms

Parity
Whether a whole number is odd or even.
Magic Square
A square grid of distinct numbers in which every row, column and main diagonal has the same "magic sum".
Magic Sum
The common total of every row, column and diagonal in a magic square.
Chautisa Yantra
The famous 4 × 4 magic square from the Parshvanatha Jain temple at Khajuraho — magic sum 34.
Virahānka–Fibonacci Numbers
The sequence 1, 2, 3, 5, 8, 13, 21, 34, 55, ..., where each term is the sum of the previous two.
Cryptarithm (Alphametic)
A puzzle in which digits have been replaced by letters; each letter stands for one digit.
Activity: Staircase Paths
L4 Analyse
Materials: Paper, pencil (you can mime climbing too!).
Predict: How many different ways can you climb a 10-step staircase taking 1 or 2 steps at a time? Guess first, then compute.
  1. Write Virahānka numbers \(V_1 = 1, V_2 = 2, V_3 = 3, ..., V_{10}\).
  2. \(V_{10}\) is the number of paths for a 10-step staircase.
  3. Verify by writing explicitly the paths for \(n = 4\) (should be 5) and \(n = 5\) (should be 8).

\(V_{10} = 89\). So there are 89 different ways to climb 10 steps.

1 2 3 4 5 6 Climb with steps of size 1 or 2 — Virahānka counts the paths

Competency-Based Questions

Scenario: In a school maths club, students solve an end-of-chapter review: decoding cryptarithms, analysing stair-climb possibilities for a 12-step temple staircase (like the Chautisa Yantra plinth), and reasoning about parities of large sequences.
Q1. A bulb is OFF. A group of students toggle the switch 2024 times. Is it ON or OFF at the end? Justify using parity.
L3 Apply
An even number of toggles returns the switch to its original state. 2024 is even, so the bulb remains OFF.
Q2. How many ways can a student climb a 12-step staircase using 1-step and 2-step moves? Show the recurrence calculation.
L4 Analyse
\(V_n\): 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233. So 233 different ways.
Q3. Evaluate the statement: "Since 37 is odd, the product \(37 \times 48 \times 99\) is odd."
L5 Evaluate
Incorrect. A product is odd only when every factor is odd. 48 is even, so \(37 \times 48 \times 99\) is even. One even factor makes the whole product even.
Q4. Design your own cryptarithm using addition where at least one carry occurs, and every letter maps to a unique digit. Provide the puzzle and a complete solution.
L6 Create
Sample puzzle: \(\overline{AB} + \overline{BA} = \overline{CBC}\) with distinct digits. Solution: A = 9, B = 2, C = 1 — check: 92 + 29 = 121 ✓. Unique because \(11(A+B) = 100C + 10B + C = 101C + 10B\) ⇒ \(11A + B = 101C\). With C = 1: \(11A + B = 101\) ⇒ A = 9, B = 2. Valid, distinct digits.
Assertion–Reason Questions
Assertion (A): A bulb that is toggled 101 times from an OFF starting state ends up ON.
Reason (R): An odd number of toggles flips the bulb's state.
(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). From OFF + odd toggles = ON. R explains A.
Assertion (A): The sum \(1 + 2 + 3 + \ldots + 100\) is even.
Reason (R): It contains 50 odd numbers and an even count of odd numbers gives an even sum.
(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). Sum = 5050 (even). The sum of 50 odd numbers is even; adding 50 evens keeps it even. R directly explains A.
Assertion (A): Angaan has exactly 34 ways to climb an 8-step staircase using 1-step and 2-step moves.
Reason (R): The number of ways to climb \(n\) steps using 1-steps and 2-steps equals the number of binary sequences of length \(n\).
(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: (c). A is true (\(V_8 = 34\)). R is false — the count equals the Virahānka number \(V_n\), not \(2^n\) binary sequences.

Frequently Asked Questions — Number Play

What is Part 4 — Exercises & Summary | Class 7 Maths Ch 6 Number Play | MyAiSchool in NCERT Class 7 Mathematics?

Part 4 — Exercises & Summary | Class 7 Maths Ch 6 Number Play | MyAiSchool is a key concept covered in NCERT Class 7 Mathematics, Chapter 6: 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 Part 4 — Exercises & Summary | Class 7 Maths Ch 6 Number Play | MyAiSchool step by step?

To solve problems on Part 4 — Exercises & Summary | Class 7 Maths Ch 6 Number Play | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 7 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 6: Number Play?

The essential formulas of Chapter 6 (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 Part 4 — Exercises & Summary | Class 7 Maths Ch 6 Number Play | MyAiSchool important for the Class 7 board exam?

Part 4 — Exercises & Summary | Class 7 Maths Ch 6 Number Play | MyAiSchool is part of the NCERT Class 7 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 Part 4 — Exercises & Summary | Class 7 Maths Ch 6 Number Play | MyAiSchool?

Common mistakes in Part 4 — Exercises & Summary | Class 7 Maths Ch 6 Number Play | MyAiSchool 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 Part 4 — Exercises & Summary | Class 7 Maths Ch 6 Number Play | MyAiSchool?

End-of-chapter NCERT exercises for Part 4 — Exercises & Summary | Class 7 Maths Ch 6 Number Play | MyAiSchool 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 6, and solve at least one previous-year board paper to consolidate your understanding.

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