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CBSE
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Concept 1 — Multiplying a 2-Digit Number by a 1-Digit Number
```To multiply a 2-digit number by a single digit, we use the column method. We multiply the 1-digit number with the Ones digit first, carry if needed, then multiply with the Tens digit and add any carry.
▶ Steps — 2-Digit × 1-Digit
- Write the 2-digit number on top. Write the 1-digit number below it, aligned to the right.
- Write the × sign to the left. Draw a line.
- Step 1: Multiply the 1-digit number by the Ones digit. If the result ≥ 10, write the units digit below and carry the tens digit.
- Step 2: Multiply the 1-digit number by the Tens digit. Add any carry. Write the result.
- Draw a double line. Read the answer.
▶ Solved Example 1 — No Carry
Find 23 × 3
23
×3
69
O
3 × 3 = 9. Write 9.
T
3 × 2 = 6. Write 6.
23 × 3 = 69
▶ Solved Example 2 — With Carry
Find 47 × 6
4
47
×6
282
O
6 × 7 = 42 → write 2, carry 4.
T
6 × 4 = 24, + carry 4 = 28. Write 28.
- 1Ones: 6 × 7 = 42. Write 2, carry 4.
- 2Tens: 6 × 4 = 24, + 4 (carry) = 28. Write 28.
47 × 6 = 282
▶ Solved Example 3
Find 85 × 9
4
85
×9
765
O
9 × 5 = 45 → write 5, carry 4.
T
9 × 8 = 72, + 4 = 76. Write 76.
85 × 9 = 765
✍ Practice — Concept 1
Q1
Find: (a) 32 × 4 (b) 56 × 7 (c) 74 × 8
▼
Answer
(a) 128 (b) 392 (c) 592
(a)
32
×4
128
(b)
4
56
×7
392
(c)
3
74
×8
592
Concept 2 — Multiplying a 3-Digit Number by a 1-Digit Number
```Same method — but now we have three steps: multiply the 1-digit number by Ones, then Tens, then Hundreds, carrying each time if needed.
▶ Solved Example 4 — No Carry
Find 213 × 3
213
×3
639
O
3 × 3 = 9.
T
3 × 1 = 3.
H
3 × 2 = 6.
213 × 3 = 639
▶ Solved Example 5 — With Carries
Find 348 × 6
2 4
348
×6
2088
O
6 × 8 = 48 → write 8, carry 4.
T
6 × 4 = 24, + 4 = 28 → write 8, carry 2.
H
6 × 3 = 18, + 2 = 20. Write 20.
- 1O: 6×8=48 → write 8, carry 4.
- 2T: 6×4=24, +4=28 → write 8, carry 2.
- 3H: 6×3=18, +2=20. Write 20.
348 × 6 = 2,088
✍ Practice — Concept 2
Q2
Find: (a) 124 × 4 (b) 235 × 7 (c) 469 × 8
▼
Answer
(a) 496 (b) 1,645 (c) 3,752
(a)
1
124
×4
496
(b)
2 3
235
×7
1645
(c)
7 7
469
×8
3752
Concept 3 — Multiplying a 2-Digit Number by a 2-Digit Number
```When the multiplier has two digits, we multiply twice — once by the Ones digit and once by the Tens digit — giving two partial products. The second partial product is shifted one place to the left (or a × placeholder is written). Then we add both partial products to get the final answer.
▶ Steps — 2-Digit × 2-Digit
- Write the larger number on top, the multiplier below it. Draw a line.
- Step 1 (Ones): Multiply the top number by the Ones digit of the multiplier. Write this as the 1st partial product (PP1) — no shift.
- Step 2 (Tens): Multiply the top number by the Tens digit of the multiplier. Write this as the 2nd partial product (PP2) — shifted one place left (write × in the Ones place as a placeholder).
- Draw a second line. Add PP1 and PP2. This is the final product.
⚠ The Shift Rule
The 2nd partial product (from the Tens digit) is always shifted one place to the left. Write a × symbol (or 0) in the Ones column of PP2 as a placeholder. Forgetting this shift is the most common mistake in long multiplication.
▶ Solved Example 6 — 2-Digit × 2-Digit
Find 23 × 12
23
×12
46
23×
276
O
23 × 2 (ones of 12) = 46 → PP1
T
23 × 1 (tens of 12) = 23 → PP2, shifted left (× placeholder)
+
Add: 46 + 230 = 276
- 1Multiply 23 × 2 (ones digit of 12) = 46. Write as PP1.
- 2Multiply 23 × 1 (tens digit of 12) = 23. Write as PP2 with × placeholder (shifted left). PP2 = 23× = 230.
- 3Add: 46 + 230 = 276.
23 × 12 = 276
▶ Solved Example 7 — With Carries in Partial Products
Find 47 × 35
47
×35
235
141×
1645
O
47 × 5 = 235 → PP1
T
47 × 3 = 141 → PP2 (shifted, = 1410)
+
235 + 1410 = 1645
- 1PP1: 47 × 5 = 235. (7×5=35, write 5 carry 3; 4×5=20+3=23)
- 2PP2: 47 × 3 = 141. Write 141× (shifted one left = 1410).
- 3Add: 235 + 1410 = 1645.
47 × 35 = 1,645
▶ Solved Example 8
Find 68 × 74
68
×74
272
476×
5032
O
68 × 4 = 272 → PP1
T
68 × 7 = 476 → PP2 (= 4760)
+
272 + 4760 = 5032
68 × 74 = 5,032
✍ Practice — Concept 3
Q3
Find: (a) 34 × 21 (b) 56 × 43 (c) 78 × 65
▼
Answer
(a) 714 (b) 2,408 (c) 5,070
(a) 34×21
34
×21
34
68×
714
(b) 56×43
56
×43
168
224×
2408
(c) 78×65
78
×65
390
468×
5070
Q4
True or False: In 45 × 23, the 2nd partial product (45×2) should be written as 90, NOT as 90×. Correct if false.
▼
Answer
False. The 2nd partial product must be shifted one place left. It should be written as 90× (representing 900, not 90). Without the shift, the answer will be wrong.
Correct: PP1 = 45×3 = 135. PP2 = 45×2 = 90× = 900. Total = 135 + 900 = 1035.
Correct: PP1 = 45×3 = 135. PP2 = 45×2 = 90× = 900. Total = 135 + 900 = 1035.
Concept 4 — Multiplying a 3-Digit Number by a 2-Digit Number
```Exactly the same method — two partial products, same shift rule. Now the partial products are larger since we are multiplying a 3-digit number each time.
▶ Solved Example 9
Find 124 × 32
124
×32
248
372×
3968
O
124 × 2 = 248 → PP1
T
124 × 3 = 372 → PP2 (= 3720)
+
248 + 3720 = 3968
- 1PP1: 124 × 2 = 248. (4×2=8, 2×2=4, 1×2=2)
- 2PP2: 124 × 3 = 372. Write 372× (= 3720).
- 3Add: 248 + 3720 = 3968.
124 × 32 = 3,968
▶ Solved Example 10 — With Carries
Find 356 × 47
356
×47
2492
1424×
16732
O
356 × 7 = 2492 → PP1
T
356 × 4 = 1424 → PP2 (= 14240)
+
2492 + 14240 = 16732
- 1PP1: 356 × 7:
6×7=42 → write 2, carry 4.
5×7=35, +4=39 → write 9, carry 3.
3×7=21, +3=24 → write 24.
PP1 = 2492. - 2PP2: 356 × 4:
6×4=24 → write 4, carry 2.
5×4=20, +2=22 → write 2, carry 2.
3×4=12, +2=14 → write 14.
PP2 = 1424× = 14240. - 3Add: 2492 + 14240 = 16,732.
356 × 47 = 16,732
▶ Solved Example 11
Find 408 × 56
408
×56
2448
2040×
22848
O
408 × 6 = 2448 → PP1
T
408 × 5 = 2040 → PP2 (= 20400)
+
2448 + 20400 = 22848
408 × 56 = 22,848
✍ Practice — Concept 4
Q5
Find: (a) 213 × 42 (b) 374 × 56
▼
Answer
(a) 8,946 (b) 20,944
(a) 213×42
213
×42
426
852×
8946
(b) 374×56
374
×56
2244
1870×
20944
Q6
A student solves 245 × 36 and writes: PP1 = 1470, PP2 = 735×. Final answer = 1470 + 735 = 2205. Find the correct answer. What mistake was made?
▼
Answer
PP1: 245×6 = 1470 ✓
PP2: 245×3 = 735× = 7350 (not 735!)
Correct addition: 1470 + 7350 = 8820.
Mistake: The student added 1470 + 735 instead of 1470 + 7350. They forgot that PP2× = PP2 shifted left = PP2 × 10.
245
×36
1470
735×
8820
PP2: 245×3 = 735× = 7350 (not 735!)
Correct addition: 1470 + 7350 = 8820.
Mistake: The student added 1470 + 735 instead of 1470 + 7350. They forgot that PP2× = PP2 shifted left = PP2 × 10.
Concept 5 — Word Problems
```For word problems involving multiplication, look for key words like each, every, per, total, altogether, times, groups of. Identify the two factors and multiply using the long multiplication method.
▶ Solved Example 12
A school orders 35 boxes of books. Each box has 48 books. How many books are there in total?
- 1Books per box = 48. Number of boxes = 35. Find 48 × 35.
48
×35
240
144×
1680
O
48×5 = 240
T
48×3 = 144× = 1440
+
240 + 1440 = 1680
There are 1,680 books in total.
▶ Solved Example 13
A factory produces 346 toys every day. How many toys does it produce in 28 days?
346
×28
2768
692×
9688
O
346×8 = 2768
T
346×2 = 692× = 6920
+
2768 + 6920 = 9688
The factory produces 9,688 toys in 28 days.
✍ Practice — Concept 5
Q7
A train has 64 seats per coach. There are 18 coaches. How many seats in total?
▼
Answer
64×8=512 (PP1). 64×1=64× (PP2=640). 512+640=1152.
The train has 1,152 seats in total.
64
×18
512
64×
1152
The train has 1,152 seats in total.
Q8
A farmer plants 425 trees in each field. He has 36 fields. How many trees does he plant altogether?
▼
Answer
425×6=2550 (PP1). 425×3=1275× (=12750). 2550+12750=15300.
The farmer plants 15,300 trees altogether.
425
×36
2550
1275×
15300
The farmer plants 15,300 trees altogether.
Q9
A shop sells 275 packets of biscuits every week. How many packets does it sell in 52 weeks?
▼
Answer
275×2=550 (PP1). 275×5=1375× (=13750). 550+13750=14300.
The shop sells 14,300 packets in 52 weeks.
275
×52
550
1375×
14300
The shop sells 14,300 packets in 52 weeks.
★ Key Points to Remember
- Always multiply Ones first, then Tens, then Hundreds. Carry where needed.
- When multiplying by a 2-digit multiplier, write two partial products: PP1 (ones digit × top) and PP2 (tens digit × top).
- PP2 is always shifted one place left — write a × placeholder in the Ones column of PP2.
- The × placeholder means PP2 is worth 10 times its face value.
- After writing both partial products, add them to get the final answer.
- Always verify: a rough estimate (round both numbers) should be close to your answer.
Exam Style — Class 3, 4 & 5
5 Questions on Multiplying 2–3 Digit Numbers
Q1
Solve using long multiplication: (a) 96 × 7 (b) 75 × 48
▼
Answer
(a) 672 (b) 3,600
(a) 96×7
4
96
×7
672
(b) 75×48
75
×48
600
300×
3600
Q2
Solve using long multiplication showing both partial products: 537 × 64
▼
Answer
PP1: 537×4=2148. PP2: 537×6=3222× (=32220). 2148+32220=34,368
537
×64
2148
3222×
34368
Q3
A student solves 328 × 45 and writes:
PP1 = 328×5 = 1640 PP2 = 328×4 = 1312 Answer = 1640 + 1312 = 2952.
Find the correct answer. What was the mistake? ▼
PP1 = 328×5 = 1640 PP2 = 328×4 = 1312 Answer = 1640 + 1312 = 2952.
Find the correct answer. What was the mistake? ▼
Answer
Correct: PP2 = 1312× = 13120. 1640 + 13120 = 14,760.
Mistake: The student added PP2 as 1312 instead of 13120. They forgot the × shift (forgot to multiply PP2 by 10).
328
×45
1640
1312×
14760
Mistake: The student added PP2 as 1312 instead of 13120. They forgot the × shift (forgot to multiply PP2 by 10).
Q4
A cinema hall has 285 seats. Tickets were sold for 48 shows. How many tickets were sold in total?
▼
Answer
285×8=2280 (PP1). 285×4=1140× (=11400). 2280+11400=13,680.
13,680 tickets were sold in total.
285
×48
2280
1140×
13680
13,680 tickets were sold in total.
Q5
Estimate first, then find the exact answer: 469 × 73.
▼
Answer
Estimate: 469 ≈ 470, 73 ≈ 70. 470 × 70 = 32,900. (Expected range: roughly 30,000–35,000.)
Exact working:
PP1: 469×3=1407. PP2: 469×7=3283× (=32830). 1407+32830=34,237.
Estimate was 32,900 — close to 34,237. ✓
Exact working:
469
×73
1407
3283×
34237
Estimate was 32,900 — close to 34,237. ✓
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