Raksha Bandhan | Preparatory Stage Math Worksheets (NCERT/KVS/CBSE)
Raksha Bandhan (Preparatory Stage Math)
Concepts • Computational Skills • Problem-Solving & Modeling • 10 questions each • 40% Easy, 40% Average, 20% Challenging • One toggle shows Answer + Solution
Worksheet A: Concepts
Easy
Q1. Each Rakhi needs 1 flower; make 5 Rakhis. How many flowers are needed? Also write as repeated addition and as multiplication.
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Answer
5 flowers; 1+1+1+1+1; 5×1.
Solution
Five equal groups of 1 total 5; repeated addition becomes multiplication, 5×1=5.
Easy
Q2. Each Rakhi needs 2 threads; make 5 Rakhis. How many threads? Show as 5 times 2 and as addition.
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Answer
10 threads; 2+2+2+2+2; 5×2=10.
Solution
Five groups of 2 join to 10; multiplication is a quick way to add equal groups.
Easy
Q3. Each Rakhi needs 4 beads; make 5 Rakhis. How many beads? Write the multiplication sentence too.
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Solution
Repeated addition 4+4+4+4+4 equals 20, so 5 groups of 4 give 20 beads.
Easy
Q4. A laddoo box is 3 by 3. How many laddoos are in one full box? Show as “rows × columns.”
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Solution
Three rows with three each make 9 total; array structure supports multiplication meaning.
Average
Q5. Two full boxes of laddoos (each 3×3). How many laddoos in all? Show two ways: 2×9 and 6×3.
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Answer
18 laddoos; 2×9=18; also 6×3=18.
Solution
Either count boxes (2×9) or rows over both boxes (6 rows of 3) to total 18.
Average
Q6. Share 18 laddoos equally among 9 people. How many each? Write the division and a matching multiplication fact.
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Answer
2 each; 18÷9=2; check 9×2=18.
Solution
Equal sharing makes groups of 2; division and multiplication are inverse operations.
Average
Q7. Kaju katlis: 20 pieces shared equally among 5 people. How many each? Show with repeated subtraction groups of 5 briefly in words or marks.
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Solution
Remove sets of 5 four times: 20→15→10→5→0, so each plate gets 4.
Average
Q8. Jalebi tray shows 6 groups of 4. Write two multiplication sentences for the same total using factor order and grouping idea (commutativity mention allowed).
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Solution
Groups-of-4 six times or groups-of-6 four times both reach 24; order changes, product stays.
Challenging
Q9. Garden bed: Dhara says it’s 8×6, Gopal says 6×8. Who is correct and why do both give the same total? State the product and one reason in 1 line each.
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Answer
Both are correct; total 48; order of factors can swap in arrays.
Solution
8 rows×6 plants or 6 rows×8 plants count the same set; commutativity explains equality.
Challenging
Q10. “Times tables by sticks/dots”: Complete the missing step for 5 times: 7×5 = __. Give one pattern in the last digit of 5×n (n=1…10) in one phrase.
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Answer
35; ends in 5,0 alternating.
Solution
Counting dot intersections for 5’s row shows 5,10,15,… with last digits 5 then 0 repeating.
Worksheet B: Computational Skills
Easy
Q1. Fill multiplication from equal groups: 5 jars with 4 cookies each → __ cookies. Also write the addition sentence once only.
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Solution
5 equal groups of 4 total 20; 5×4=20 concisely records the count.
Easy
Q2. Times table via skip-jumps of 3: 0→3→6→9→__→__. Fill next two landings and state the jump size in one word in brackets at end.
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Solution
Each landing adds 3; the reached numbers match 3×n products in order.
Easy
Q3. Division as equal sharing: 15 pedas to 5 children equally. How many each? Also write 15÷5=__ once only.
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Solution
Make 5 equal piles; each pile gets 3 pedas to use up all 15.
Easy
Q4. “How many wheels?” A cycle has 2 wheels. With 12 wheels, how many cycles can be made? State division sentence only once at the end too.
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Solution
Group wheels in sets of 2 to count complete cycles formed.
Average
Q5. “Idli plates” model: 6 plates with 4 idlis each. How many idlis at once? Give 6×4 and 4×6 both briefly in one line after the number only if space permits.
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Answer
24; 6×4=24; 4×6=24.
Solution
Equal groups multiply either by groups or by items per group interchangeably.
Average
Q6. “Rath spokes”: each wheel needs 5 spokes. For 10 wheels, how many spokes? For 20 wheels, how many? Write both as ×5 facts once each after the numbers if space allows.
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Answer
50; 100; (10×5=50; 20×5=100).
Solution
Scale linearly by wheels; doubling wheels doubles total spokes.
Average
Q7. “Spiders’ legs”: 8 legs each. How many legs for 5 spiders? for 10 spiders? for 15 spiders? Write three totals only, then one short observation about ×8 scaling if possible in brackets.
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Answer
40; 80; 120 (×8 grows by 8s).
Solution
5×8=40; 10×8=80; 15×8=120; add 40 then 80 gives 120 by grouping.
Average
Q8. “Auto rickshaw wheels”: one auto has 3 wheels. How many wheels for 18 autos? Also for 34 autos? Give both numbers and one multiplication sentence each if space allows only once after the numbers.
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Answer
54; 102 (18×3=54; 34×3=102).
Solution
Multiply autos by 3 wheels each; or 30×3 + 4×3 = 90+12 = 102.
Challenging
Q9. “Frog jumps”: starting at 0, jumps of 7. What largest number below or equal to 50 can be reached? Also, if starting at 50 and jumping backwards by 7, what last number before stopping? Write both numbers only if space is tight, else short reasoning allowed.
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Answer
49; 1 (or 50→43→36→29→22→15→8→1, cannot jump to −6).
Solution
Multiples of 7 under 50: 7,14,21,28,35,42,49; reverse steps from 50 subtracting 7 stop at 1.
Challenging
Q10. “Envelope puzzle”: cards 1–10, envelope shows product 20. Name two cards inside. Give one more pair for product 45 (use single-digit cards only) in the same line if space permits separated by a semicolon.
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Answer
(4,5); (5,9) for 45.
Solution
Factor 20 from 1–10 gives 4×5; factor 45 gives 5×9 with those cards.
Worksheet C: Problem-Solving & Modeling
Easy
Q1. Make 10 Rakhis: flowers, threads, beads per Rakhi are 1, 2, 4. How many of each are needed in total? State three multiplication facts in one line or as a list quickly under the answer box if space allows.
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Answer
Flowers 10; Threads 20; Beads 40 (10×1, 10×2, 10×4).
Solution
Scale per-Rakhi requirements by 10; proportional reasoning across equal groups.
Easy
Q2. Material limit: With 30 flowers, 30 threads, 30 beads, how many complete Rakhis can be made if each Rakhi needs (1,2,4)? Choose the limiting item and give the count of Rakhis possible once only at end.
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Answer
7 Rakhis (beads limit: 30÷4=7 full sets, remainder 2).
Solution
Flowers allow 30, threads 15, beads 7 full; the minimum governs complete Rakhis.
Easy
Q3. Sea-shell necklaces: 112 shells, each necklace uses 28. How many necklaces, and how many shells left (if any)? Solve by repeated subtraction in words or division once at end too if space permits.
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Answer
4 necklaces; 0 left (112÷28=4).
Solution
112−28=84; −28=56; −28=28; −28=0; four equal takes fit exactly.
Easy
Q4. “Money math”: Rabdi ₹75 a cup. For 5 cups, total cost? If paying with ₹100 notes only, how many notes, and how much change back in rupees (state both counts clearly in one line)?
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Answer
Total ₹375; give 4 notes (₹400); change ₹25.
Solution
5×75=375; nearest 100-multiple above is 400, so return 25.
Average
Q5. “Groups from total”: 45 spokes collected; how many wheels of 5 spokes each can be completed? Any spokes left? Give quotient and remainder in one line if space is tight.
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Answer
9 wheels; 0 left (45÷5=9).
Solution
Group spokes in fives; 9 equal groups use all 45 with none left over.
Average
Q6. “Enough for all?” 24 jalebis for 9 members at home. Can each get 4? If not, how many are needed in total for 4 each, and how many more to buy (two answers only if space is tight)?
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Answer
No; need 36 total; must buy 12 more.
Solution
9×4=36 needed; 36−24=12 short; present gap as “more to buy.”
Average
Q7. “Back-to-100 grouping”: 100 shells into groups of 17 per necklace. How many full necklaces and how many shells left? Solve by subtracting 17s or division with remainder; write both numbers once only at end if short on space.
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Answer
5 necklaces; 15 left (100÷17=5 R15).
Solution
17×5=85; 100−85=15; cannot make another full group from 15.
Average
Q8. “Card pairs sum to a table”: Two tables add to a third (e.g., 2s and 3s make 5s). Check 4×n + 1×n equals 5×n for n=1,2,3 (write three small checks only in one line with equals signs).
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Answer
4+1=5; 8+2=10; 12+3=15 (matches 5×1,5×2,5×3).
Solution
Add row-by-row: “table 4” plus “table 1” equals “table 5” entries.
Challenging
Q9. “Construct a times table”: From skip-jumps 6, write 6×1 to 6×10 as a single sequence of products in order (no commas if short on space; use spaces).
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Answer
6 12 18 24 30 36 42 48 54 60.
Solution
Ten equal jumps of size 6 enumerate the 6 times table to 60.
Challenging
Q10. “Construct a division story”: 127 pebbles shared equally among 3 friends. How many each and how many left? Present by repeated subtraction or by quotient–remainder once at end if short on space.
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Answer
42 each; 1 left (127÷3=42 R1).
Solution
3×42=126; one pebble remains; equal sharing forces remainder 1.
Two best activities
Activity 1: Skip-Jump Tables Race (3, 4, 5, 6)
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Answer
Build 3, 4, 5, 6 times tables on a floor number track using equal jumps and record products.
Solution
Tape a 0–60 track on the floor. Teams draw a card (3/4/5/6) and jump equal steps to reach targets (e.g., flower at 30). They record landings as multiplication sentences (e.g., 5 jumps of 6 = 30). Compare which tables reach 60 exactly and discuss common multiples (e.g., 30 appears in 3s, 5s, 6s). This builds patterning, visualization, and communication.
Activity 2: Equal Groups Lab — Make, Share, Model
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Answer
Use counters to model “make Rakhis” (×) and “share sweets” (÷); write both repeated addition and division sentences.
Solution
Station A: Place counters in 5 trays with 4 each (5×4). Write addition and multiplication. Station B: Share 20 counters into 5 bowls (20÷5). Station C: Tray arrays for 3×3 and 2 boxes (2×9 and 6×3). Learners explain strategies aloud, compare models, and link × with ÷ as inverse, promoting mathematization and reasoning.
