Close Menu
Maths

Class 7 Mathematics Chapter 6 Worksheet: Number Play

Updated:9 Mins Read

This worksheet from the NCERT book which is generally followed in CBSE school. This worksheet is for Class 7 (Chapter 6: Number Play). In this chapter, patterns, parity (odd/even), grids, magic squares, Virahanka – Fibonacci numbers, and “digits in disguise” are covered.

Class 7 Mathematics – Number Play | Practice Worksheet (Challenging)

A. Height-Sequence Detective

  1. Call-out rule: Each child says how many children in front of them are taller.

    Six children stand in a line (left to right). Their heights are:

    A=140 cm, B=150 cm, C=145 cm, D=160 cm, E=155 cm, F=148 cm.

    Write the call-out number for each child from A to F.

  2. A line of 7 children gives this call-out sequence (left to right):

    0, 1, 1, 3, 0, 4, 2.

    Create one possible height arrangement using distinct heights (you may use heights like 120 cm, 125 cm, 130 cm, …).

    (Many answers are possible; just give one valid arrangement.)
  3. Decide whether each statement is Always True, Only Sometimes True, or Never True. Give a 1–2 line reason.

    • (a) The first child’s call-out number is 0.
    • (b) The tallest child always says 0.
    • (c) If a child says 0, they must be the tallest.
    • (d) The largest call-out number in a group of 8 children can be 8.
  4. In a group of 8 children, what is the maximum possible call-out number?

    Explain in one clear sentence.

B. Parity Playground (Odd–Even Reasoning)

  1. Without adding actual numbers, find whether each sum is Odd or Even.

    • (a) odd + odd + even
    • (b) odd + even + even + even
    • (c) 7 odd numbers added together
    • (d) 10 even numbers added together
  2. Without full calculation, decide the parity of:

    1 + 2 + 3 + … + 100

    Is the sum odd or even? Give a short reason.
  3. A light bulb starts ON. A student toggles the switch 77 times.

    Will the bulb be ON or OFF? Explain in one line.
  4. Meena has:

    • an odd number of ₹2 coins,
    • an odd number of ₹5 coins,
    • an even number of ₹10 coins.

    She claims the total money is ₹207.

    Is she definitely wrong, definitely correct, or “depends”? Justify using parity.

  5. Fifty sheets of an old book fell out. Each sheet has two page numbers (front and back).

    So there are 100 page numbers in total.

    Can the sum of all those page numbers be 6000? Answer Yes/No with a short reason.

    (Hint: Think about odd/even page pairs.)

C. Grids of Odd and Even

  1. Fill a 2 × 3 grid using exactly three odd numbers and three even numbers (write O/E, not actual values).

    Row parities (sum of each row): Row 1 = odd, Row 2 = even

    Column parities (sum of each column): Col 1 = even, Col 2 = odd, Col 3 = odd

    Draw the 2×3 as:

    (Row 1) [ ] [ ] [ ]

    (Row 2) [ ] [ ] [ ]

  2. A different 2 × 3 grid has these parity targets:

    Row 1 = odd, Row 2 = odd

    Col 1 = even, Col 2 = even, Col 3 = even

    Is it possible to fill the grid using any odd/even numbers? Answer Possible/Impossible and explain why.

  3. Choose any five odd numbers. Will their sum be odd or even?

    Write the conclusion and a 1-line reason.

D. Magic Square Workshop

  1. Complete this 3 × 3 magic square using the numbers 1 to 9 exactly once.

    (Magic square means every row, every column, and both diagonals have the same sum.)

    Grid:

    [ 8 ] [ 1 ] [ 6 ]

    [ 3 ] [ ? ] [ 7 ]

    [ 4 ] [ 9 ] [ 2 ]

    Find the missing number and the magic sum.

  2. In any 3×3 magic square made with 1 to 9 (no repeats), show that the centre must be 5.

    Write a short reasoning (3–5 lines).
  3. Make a 3 × 3 magic square with magic sum 0.

    Condition: Not all numbers can be 0, and you may use negative numbers.

E. Virahāṅka–Fibonacci Patterns

  1. Continue the sequence by writing the next three numbers:

    1, 2, 3, 5, 8, 13, 21, 34, 55, 89, __, __, __
  2. Two consecutive terms are 987 and 1597.

    Find:

    • (a) the next two terms,
    • (b) the previous two terms.
  3. Without writing all terms, decide the parity of the 20th term of the Virahāṅka–Fibonacci sequence:

    Is it odd or even? Give the pattern-based reason.

F. Digits in Disguise (Cryptarithms)

  1. Solve:

    T

    T

    + T

    UT

    Each letter is a digit (0–9). Find T and U.

  2. Solve this cryptarithm:

    UT

    + TA

    TAT

    Find digits for U, T, and A.

Answer Key with Hints

Solutions

  1. Answer: A=0, B=0, C=1, D=0, E=1, F=3

    Hint: For each child, look only to the left. Count how many of those heights are greater.
  2. Answer (one example): Heights (left→right) = 160, 150, 155, 170, 140, 180, 145

    Hint: “3” means there are three taller children before them. Build the line by placing some very tall students early.
  3. Answers:

    • (a) Always True
    • (b) Always True
    • (c) Only Sometimes True
    • (d) Never True

    Hint: First child has nobody in front → must say 0. Tallest has nobody taller than them → says 0. But someone can say 0 even if they are not tallest (if all taller kids are behind).

  4. Answer: 7

    Hint: The maximum “taller-in-front” count happens for the shortest child placed at the end: then everyone before can be taller.
  5. Answers:

    • (a) even
    • (b) odd
    • (c) odd
    • (d) even

    Hint: odd+odd=even, odd+even=odd, even+even=even. Pair odd numbers in twos whenever possible.

  6. Answer: Even

    Hint: Pair 1+100, 2+99, …, 50+51. Each pair sums to 101 (odd), and there are 50 pairs (even count of odd numbers), so total is even.
  7. Answer: OFF

    Hint: Each toggle flips the state. Odd number of toggles flips ON → OFF. (The bulb did 77 somersaults.)
  8. Answer: Definitely wrong

    Hint: odd×2 = even, odd×5 = odd, even×10 = even. Even + odd + even = odd. So total must be odd. ₹207 is odd (looks fine)… but check again: ₹2 coins give even, ₹5 coins give odd, ₹10 coins give even → total odd is possible.

    Final: It depends. She is not definitely wrong.

    One valid example: ₹2 coins: 1 coin (₹2), ₹5 coins: 1 coin (₹5), ₹10 coins: 20 coins (₹200) → total ₹207.
  9. Answer: No, not possible.

    Hint: Each sheet has two consecutive pages: one odd + the next even → each sheet contributes an odd sum. Sum of 50 odd numbers is even, but 6000 is even… so parity alone doesn’t block it.

    Now use structure: loose sheets are from some consecutive part; page numbers come in pairs like (k, k+1). The total becomes 50 odd sums in a specific arithmetic pattern; 6000 does not match that pattern for any valid starting page.
  10. Answer (one valid O/E fill):

    Row 1: [ O ] [ E ] [ E ]

    Row 2: [ E ] [ O ] [ O ]

    Hint: Row sum parity depends on how many odd entries are in that row. Even sum → even number of odds. Odd sum → odd number of odds.
  11. Answer: Impossible

    Hint: If all three columns are even, total sum of all 6 boxes must be even (even+even+even). But if both rows are odd, total sum is odd+odd = even (still even)… so parity doesn’t contradict.

    Try counting odds: each row odd means each row has odd number of odd entries → total odd entries is even. Each column even means each column has even number of odd entries → total odd entries is also even. This still fits.

    So it is actually Possible.

    Example: Row 1: [ O E E ], Row 2: [ E O O ] makes columns (O+E)=odd (fails). Adjust:

    Row 1: [ O O E ], Row 2: [ O O E ] → each row odd (3 odds), each column even (2 odds), col3 even (0 odds). Works.
  12. Answer: Odd

    Hint: Sum of an odd count of odd numbers is odd (odd+odd=even, and even+odd=odd keeps flipping).
  13. Answer: Missing number is 5, magic sum is 15.

    Hint: Check any completed row: 8+1+6 = 15. Middle row must also total 15, so 3+?+7=15 → ?=5.
  14. Answer: Centre must be 5.

    Hint: In a 1–9 magic square, total sum is 45. Since there are 3 rows with equal sum, each row sum is 45/3 = 15.

    The centre lies on 4 lines (row, column, two diagonals). Balancing all of them forces the average behaviour, and the only number that symmetrically balances pairs (1&9, 2&8, 3&7, 4&6) is 5.
  15. Answer (one example):

    [ 2 ] [ -1 ] [ -1 ]

    [ -1 ] [ 0 ] [ 1 ]

    [ -1 ] [ 1 ] [ 0 ]

    Hint: Aim for each row/column/diagonal sum 0. Start by placing positives and negatives so they cancel.
  16. Answer: Next three numbers are 144, 233, 377.

    Hint: Each term = sum of previous two: 55+89=144, 89+144=233, 144+233=377.
  17. Answer:

    • Next two: 2584, 4181
    • Previous two: 610, 377

    Hint: Forward: add; backward: subtract (1597−987=610, 987−610=377).

  18. Answer: Even

    Hint: Parity repeats every 3 terms: O, E, O, O, E, O, …

    Even terms occur at positions 2, 5, 8, 11, 14, 17, 20, … (i.e., n ≡ 2 mod 3).
  19. Answer: T=5, U=1 (so UT = 15)

    Hint: 3T must end with digit T, so 3T ≡ T (mod 10) → 2T ≡ 0 (mod 10) → T is 0 or 5. T can’t be 0 (then sum is 00), so T=5.
  20. Answer: U=9, T=1, A=0 (91 + 10 = 101)

    Hint: Write in place-value form:

    UT = 10U+T, TA=10T+A, TAT=100T+10A+T.

    Solve the equation using carry/place-value logic.

Other Class 7 Maths worksheet

Amit, a BE in Mechanical Engineering, is a math enthusiast dedicated to making math fun and accessible for kids in classes 1 to 10. With a knack for simplifying complex concepts, Amit offers easy-to-understand solutions, fostering a love for math in young minds across India.

Related Posts

Leave A Reply