Chapter 13 of 14

Updated 6 August 2026

NCERT Solutions

The Transport Museum — NCERT Solutions

Q: What are the important topics in The Transport Museum for CBSE Class 4 Mathematics?

Key topics in The Transport Museum include Multiplication by grouping, Division with remainder, Multiplication by 10 and 100, Real-life problems in the chapter. Study these first, then practise questions on each for Class 4 exams.

Q: Are these NCERT Solutions for The Transport Museum free?

The first 25 of the 49 solutions on this page are open to read. The other 24 are free with a Super Tutor account — signing up is free and needs no card.

Q: How should I revise The Transport Museum for Class 4 exams?

Learn the core ideas first, then work through the 45 practice questions on The Transport Museum. Revise definitions regularly and use flashcards for quick recall before the exam.

CBSE · Class 4 · Mathematics

NCERT Solutions for The Transport Museum, CBSE Class 4 Mathematics: 49 textbook questions solved step by step.

45 questions56 flashcards12 formulas & key relations5 concepts

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49 Questions Solved · 17 Sections

The first 25 solutions are open to read. The other 24 are free with a Super Tutor account.

Mystery Matrix

1Fill the yellow boxes with 1-digit numbers (multiplicands and multipliers) such that you get the products given in the white boxes. Fill the remaining white boxes with appropriate products.

| × | | | | |
|---|---|---|---|---|
| | 32 | | | |
| | | | 42 | |
| | | 45 | | |
| | | | 21 | |Show solution

Given: A multiplication matrix where some products are given and we must find 1-digit multipliers/multiplicands to complete it.

Strategy: Work backwards from the given products to find 1-digit factors.

Step 1 – Identify factors from given products:

$32 = 4 \times 8$ → row multiplier = 4, column multiplier = 8
$42 = 6 \times 7$ → row multiplier = 6, column multiplier = 7
$45 = 5 \times 9$ → row multiplier = 5, column multiplier = 9
$21 = 3 \times 7$ → row multiplier = 3, column multiplier = 7

Step 2 – One possible completed matrix (header row: ×, 8, 9, 7, 6):

×	8	9	7	6
4	32	36	28	24
6	48	54	42	36
5	40	45	35	30
3	24	27	21	18

Note: Multiple valid answers are possible depending on the choice of 1-digit numbers, as long as the given products appear in the correct positions.

Mystery Matrix (Row/Column Products)

2The product of the numbers in each row is given in the orange boxes. The product of the numbers in each column is given in the blue boxes. Identify appropriate numbers to fill the blank boxes.

| | | 56 | | | 42 |
|---|---|---|---|---|---|
| | | 54 | | | 50 |
| 63 | 48 | | 60 | 35 | |Show solution

Given: Row products (orange) and column products (blue) are provided. We need to find the individual numbers.

Step 1 – Analyse the structure. The table appears to be a 3-row grid where the last column shows row products and the last row shows column products.

Step 2 – Find numbers for Row 1 (product = 42):
$42 = 6 \times 7$ or $2 \times 3 \times 7$ , etc.
With the column product clue 56 in column 3: $56 = 7 \times 8$ , so one entry is 7 or 8.
A possible row: $1, 6, 7$ → product = $1 \times 6 \times 7 = 42$ ✓

Step 3 – Find numbers for Row 2 (product = 50):
$50 = 2 \times 5 \times 5$ or $5 \times 10$
With column product clue 54 in column 3: $54 = 6 \times 9$
A possible row: $2, 5, 5$ → product = 50 ✓

Step 4 – Check column products against Row 3 (products: 63, 48, 60, 35):

Column with product 63: $63 = 7 \times 9$ or $63 = 1 \times 63$
Column with product 48: $48 = 6 \times 8$
Column with product 60: $60 = 5 \times 12$ or $4 \times 15$
Column with product 35: $35 = 5 \times 7$

Note: This is an open-ended puzzle with multiple valid solutions. The key method is to factorise each given product into 1-digit numbers and check consistency across rows and columns.

Times-10

1Fill in the blanks:

2 \times 10 = 2 \text{ Tens} = \_\_\_\_

5 \times 10 = \_\_\_\_ \text{ Tens} = \_\_\_\_

8 \times 10 = \_\_\_\_ \text{ Tens} = \_\_\_\_

Show solution

Concept: Multiplying any number by 10 gives that many tens.

$2 \times 10 = 2 \text{ Tens} = \mathbf{20}$

$5 \times 10 = \mathbf{5} \text{ Tens} = \mathbf{50}$

$8 \times 10 = \mathbf{8} \text{ Tens} = \mathbf{80}$

2What is

10 \times 10 =

____ Tens = ____Show solution

Concept: $10 \times 10$ means 10 groups of 10.

$10 \times 10 = \mathbf{10} \text{ Tens} = \mathbf{100}$

10 Tens make 1 Hundred = 100.

Constructing Tables (Times-15)

1This is a

5 \times 15

arrangement. Find the product by splitting:

5 \times 15 = 5 \times 10 \text{ and } 5 \times 5 = \_\_ + \_\_ = \_\_

Show solution

Given: $5 \times 15$

Method – Split 15 into 10 and 5:
$5 \times 15 = 5 \times 10 + 5 \times 5$
$= 50 + 25 = \mathbf{75}$

2Construct the times-15 table:
1 × 15 = ___ 6 × 15 = ___
2 × 15 = ___ 7 × 15 = ___
3 × 15 = ___ 8 × 15 = ___
4 × 15 = ___ 9 × 15 = ___
5 × 15 = ___ 10 × 15 = ___Show solution

Method: Split 15 = 10 + 5, so $n \times 15 = n \times 10 + n \times 5$ .

$1 \times 15 = 10 + 5 = \mathbf{15}$
$2 \times 15 = 20 + 10 = \mathbf{30}$
$3 \times 15 = 30 + 15 = \mathbf{45}$
$4 \times 15 = 40 + 20 = \mathbf{60}$
$5 \times 15 = 50 + 25 = \mathbf{75}$
$6 \times 15 = 60 + 30 = \mathbf{90}$
$7 \times 15 = 70 + 35 = \mathbf{105}$
$8 \times 15 = 80 + 40 = \mathbf{120}$
$9 \times 15 = 90 + 45 = \mathbf{135}$
$10 \times 15 = 100 + 50 = \mathbf{150}$

31. What patterns do you see in the times-15 table?
2. Compare the times-15 table with the times-5 table. What similarities and differences do you notice?Show solution

Pattern 1 – Patterns in the times-15 table:

The products increase by 15 each time.
All products end in either 0 or 5 (just like the times-5 table).
The units digits follow the pattern: 5, 0, 5, 0, 5, 0, … (alternating).

Pattern 2 – Comparison with times-5 table:

Times-5	Times-15	Difference
$1 \times 5 = 5$	$1 \times 15 = 15$	$15 - 5 = 10$
$2 \times 5 = 10$	$2 \times 15 = 30$	$30 - 10 = 20$
$3 \times 5 = 15$	$3 \times 15 = 45$	$45 - 15 = 30$

Similarity: Both tables end in 0 or 5.

Difference: Each product in the times-15 table is exactly 3 times the corresponding product in the times-5 table (since $15 = 3 \times 5$ ). The differences themselves form the times-10 table.

4Complete the Times-5 and Times-15 table:
4 × 5 = ___, 4 × 15 = ___
5 × 5 = ___, 5 × 15 = ___
6 × 5 = ___, 6 × 15 = ___
7 × 5 = ___, 7 × 15 = ___
8 × 5 = ___, 8 × 15 = ___
9 × 5 = ___, 9 × 15 = ___
10 × 5 = ___, 10 × 15 = ___Show solution

$4 \times 5 = \mathbf{20}, \quad 4 \times 15 = \mathbf{60}$
$5 \times 5 = \mathbf{25}, \quad 5 \times 15 = \mathbf{75}$
$6 \times 5 = \mathbf{30}, \quad 6 \times 15 = \mathbf{90}$
$7 \times 5 = \mathbf{35}, \quad 7 \times 15 = \mathbf{105}$
$8 \times 5 = \mathbf{40}, \quad 8 \times 15 = \mathbf{120}$
$9 \times 5 = \mathbf{45}, \quad 9 \times 15 = \mathbf{135}$
$10 \times 5 = \mathbf{50}, \quad 10 \times 15 = \mathbf{150}$

53. Construct other times-tables for numbers from 11 to 20, as you did for 15.
4. Compare the times-1 table with the times-11 table, the times-2 table with the times-12 table, and so on. Share your observations.Show solution

Question 3 – Times-tables for 11 to 20 (using splitting method):

For any number $n$ between 11 and 20, write $n = 10 + r$ where $r$ is the units digit.
Then: $k \times n = k \times 10 + k \times r$

Example – Times-12 table ( $12 = 10 + 2$ ):
$1 \times 12 = 10 + 2 = 12$
$2 \times 12 = 20 + 4 = 24$
$3 \times 12 = 30 + 6 = 36$
$\vdots$
$10 \times 12 = 100 + 20 = 120$

Question 4 – Observations:

Each product in the times-11 table = corresponding product in times-1 table + corresponding product in times-10 table.
Each product in the times-12 table = corresponding product in times-2 table + corresponding product in times-10 table.
In general: $k \times (n+10) = k \times n + k \times 10$ , so every product in the higher table is exactly $k \times 10$ more than in the lower table.

Making Tables by Splitting into Equal Groups

1Calculate

3 \times 14

by splitting into two equal groups.Show solution

Given: $3 \times 14$

Method – Split 14 into two equal groups of 7:
$3 \times 14 = 3 \times 7 + 3 \times 7$
$= 21 + 21 = \text{double of } 21$
$= \mathbf{42}$

2Calculate

6 \times 14

by splitting into two equal groups.Show solution

Given: $6 \times 14$

Method – Split 14 into $7 + 7$ :
$6 \times 14 = 6 \times 7 + 6 \times 7$
$= 42 + 42 = \text{double of } 42$
$= \mathbf{84}$

3Construct the times-14 table by splitting and doubling. What other times-tables can be constructed by splitting into equal groups and doubling?Show solution

Times-14 table (using $14 = 7 + 7$ , i.e., double of times-7):

$1 \times 14 = 2 \times (1 \times 7) = 2 \times 7 = \mathbf{14}$
$2 \times 14 = 2 \times (2 \times 7) = 2 \times 14 = \mathbf{28}$
$3 \times 14 = 2 \times (3 \times 7) = 2 \times 21 = \mathbf{42}$
$4 \times 14 = 2 \times (4 \times 7) = 2 \times 28 = \mathbf{56}$
$5 \times 14 = 2 \times (5 \times 7) = 2 \times 35 = \mathbf{70}$
$6 \times 14 = 2 \times (6 \times 7) = 2 \times 42 = \mathbf{84}$
$7 \times 14 = 2 \times (7 \times 7) = 2 \times 49 = \mathbf{98}$
$8 \times 14 = 2 \times (8 \times 7) = 2 \times 56 = \mathbf{112}$
$9 \times 14 = 2 \times (9 \times 7) = 2 \times 63 = \mathbf{126}$
$10 \times 14 = 2 \times (10 \times 7) = 2 \times 70 = \mathbf{140}$

Other tables that can be built by splitting and doubling:

Times-6 = double of times-3
Times-8 = double of times-4
Times-10 = double of times-5
Times-12 = double of times-6
Times-16 = double of times-8
Times-18 = double of times-9

Any even-numbered times-table can be constructed this way.

4Find the answers to the following:
a)

15 \times 10 =

___ Tens = ___
b)

16 \times 10 =

___ Tens = ___
c)

19 \times 10 =

___ Tens = ___
d)

20 \times 10 =

___ Tens = ___Show solution

Concept: $n \times 10 = n$ Tens.

a) $15 \times 10 = \mathbf{15}$ Tens $= \mathbf{150}$

b) $16 \times 10 = \mathbf{16}$ Tens $= \mathbf{160}$

c) $19 \times 10 = \mathbf{19}$ Tens $= \mathbf{190}$

d) $20 \times 10 = \mathbf{20}$ Tens $= \mathbf{200}$

10 \times 10 =

2 \text{ times (double of) } 10 \times 10 =

Show solution

$10 \times 10 = \mathbf{100}$

$2 \times (10 \times 10) = 2 \times 100 = \mathbf{200}$

Multiplying by Multiples of 10

1Now think and answer the following problems:

30 \times 10 = \underline{\hspace{2cm}}

40 \times 10 = \underline{\hspace{2cm}}

70 \times 10 = \underline{\hspace{2cm}}

50 \times 10 = \underline{\hspace{2cm}}

60 \times 10 = \underline{\hspace{2cm}}

80 \times 10 = \underline{\hspace{2cm}}

Show solution

Concept: Multiplying a multiple of 10 by 10 — append a zero.

$30 \times 10 = \mathbf{300}$
$40 \times 10 = \mathbf{400}$
$70 \times 10 = \mathbf{700}$
$50 \times 10 = \mathbf{500}$
$60 \times 10 = \mathbf{600}$
$80 \times 10 = \mathbf{800}$

2Let us find the number of people who can travel in 26 tempo travellers.

26 \times 10 =

_____ travellers.Show solution

Given: 26 tempo travellers, each carrying 10 people.

$26 \times 10 = 26 \text{ Tens} = 20 \text{ Tens} + 6 \text{ Tens} = 200 + 60 = \mathbf{260}$

Answer: 260 people can travel.

3Answer the following questions:
a)

21 \times 10 =

42 \times 10 =

65 \times 10 =

38 \times 10 =

53 \times 10 =

87 \times 10 =

Show solution

Concept: $n \times 10$ = place a zero at the end of $n$ .

a) $21 \times 10 = \mathbf{210}$

b) $42 \times 10 = \mathbf{420}$

c) $65 \times 10 = \mathbf{650}$

d) $38 \times 10 = \mathbf{380}$

e) $53 \times 10 = \mathbf{530}$

f) $87 \times 10 = \mathbf{870}$

4A small bus can seat 20 people. How many people can be seated in 12 buses? Find

12 \times 20

.Show solution

Given: 12 buses, each seating 20 people.

Method: $12 \times 20 = 12 \times 2 \text{ Tens} = 24 \text{ Tens}$

$12 \times 20 = 24 \times 10 = \mathbf{240}$

Answer: 240 people can be seated in 12 buses.

5Solve the following problems:

24 \times 40 =

13 \times 30 =

50 \times 60 =

43 \times 60 =

70 \times 80 =

Show solution

Method: $a \times (b \times 10) = (a \times b) \times 10$

$24 \times 40 = 24 \times 4 \times 10 = 96 \times 10 = \mathbf{960}$

$13 \times 30 = 13 \times 3 \times 10 = 39 \times 10 = \mathbf{390}$

$50 \times 60 = 5 \times 6 \times 100 = 30 \times 100 = \mathbf{3000}$

$43 \times 60 = 43 \times 6 \times 10 = 258 \times 10 = \mathbf{2580}$

$70 \times 80 = 7 \times 8 \times 100 = 56 \times 100 = \mathbf{5600}$

At the Transport Museum

1Each coach of the toy train can seat 14 children. The toy train has 15 coaches. How many children can be seated in the toy train? Find

15 \times 14

.Show solution

Given: 15 coaches, each seating 14 children.

Method – Split using a grid (15 = 10 + 5, 14 = 10 + 4):

×	10	4
10	$10 \times 10 = 100$	$10 \times 4 = 40$
5	$5 \times 10 = 50$	$5 \times 4 = 20$

$15 \times 14 = 100 + 40 + 50 + 20 = \mathbf{210}$

Answer: 210 children can be seated in the toy train.

2She wonders how many coaches will be needed for 324 children from her school. Each coach can seat only 14 children. Find

324 \div 14

. What is the total number of coaches? What do we do with the remaining children?Show solution

Given: 324 children, each coach seats 14.

Method – Repeated subtraction / long division:

$14 \times 10 = 140$
$14 \times 20 = 280$
$14 \times 23 = 322$

No. of children	No. of coaches	Remaining
—	—	324
140	10	184
140	10	44
14	1	30
28	2	2

$324 \div 14 = 23 \text{ remainder } 2$

Total coaches needed = 23 (for 322 children) + 1 extra coach for the remaining 2 children.

Total coaches = 24

Remainder = 2 (2 children still need a seat, so one more coach is required even though it won't be full).

Let Us Solve (Page 191–192)

1Solve the following multiplication problems:
a)

25 \times 34

16 \times 43

68 \times 12

39 \times 13

Show solution

a) $25 \times 34$

Split: $25 \times 34 = 25 \times 30 + 25 \times 4$
$= 750 + 100 = \mathbf{850}$

b) $16 \times 43$

Split: $16 \times 43 = 16 \times 40 + 16 \times 3$
$= 640 + 48 = \mathbf{688}$

c) $68 \times 12$

Split: $68 \times 12 = 68 \times 10 + 68 \times 2$
$= 680 + 136 = \mathbf{816}$

d) $39 \times 13$

Split: $39 \times 13 = 39 \times 10 + 39 \times 3$
$= 390 + 117 = \mathbf{507}$

2Solve the following division problems and identify the remainder (if any):
e)

125 \div 15

94 \div 11

440 \div 22

508 \div 18

Show solution

e) $125 \div 15$

$15 \times 8 = 120$ ; $125 - 120 = 5$
$125 \div 15 = \mathbf{8} \text{ remainder } \mathbf{5}$

f) $94 \div 11$

$11 \times 8 = 88$ ; $94 - 88 = 6$
$94 \div 11 = \mathbf{8} \text{ remainder } \mathbf{6}$

g) $440 \div 22$

$22 \times 20 = 440$ ; $440 - 440 = 0$
$440 \div 22 = \mathbf{20} \text{ remainder } \mathbf{0}$

h) $508 \div 18$

$18 \times 28 = 504$ ; $508 - 504 = 4$
$508 \div 18 = \mathbf{28} \text{ remainder } \mathbf{4}$

Multiples of 100

1Fill in the blanks:

3 \times 100 =

___ Hundreds = ___

5 \times 100 =

___ = ___

8 \times 100 =

___ = ___Show solution

Concept: $n \times 100 = n$ Hundreds.

$3 \times 100 = \mathbf{3} \text{ Hundreds} = \mathbf{300}$

$5 \times 100 = \mathbf{5 \text{ Hundreds}} = \mathbf{500}$

$8 \times 100 = \mathbf{8 \text{ Hundreds}} = \mathbf{800}$

10 \times 100 = 10

Hundreds = 1000. What happens when we put 10 Hundreds together?Show solution

Given: $10 \times 100 = 10$ Hundreds

$10 \times 100 = \mathbf{1000}$

When we put 10 Hundreds together, we get 1 Thousand = 1000.

3Fill in the blanks:

12 \times 100 =

___

15 \times 100 =

___

27 \times 100 =

___

70 \times 100 =

___

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4Now answer the following questions:

30 \times 100 =

__________

40 \times 100 =

__________

50 \times 100 =

__________

24 \times 100 =

__________

53 \times 100 =

__________

19 \times 100 =

__________

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5We Know:

80 \times 100 = 8000

. Find:

80 \times 50 =

__________

40 \times 50 =

__________

Given: $80 \times 100 = 8000$

Finding $80 \times 50$ :
$50 = \frac{100}{2}$ , so $80 \times 50 = \frac{80 \times 100}{2} = \frac{8000}{2} = \mathbf{4000}$

Finding $40 \times 50$ :
$40 \times 50 = 40 \times 5 \times 10 = 200 \times 10 = \mathbf{2000}$

Alternatively: $40 \times 50 = \frac{80 \times 50}{2} = \frac{4000}{2} = \mathbf{2000}$

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Multiplying by 200, 300, etc.

1Find

11 \times 200

using different methods.

Method 1 – Split 11:
$11 \times 200 = 10 \times 200 + 1 \times 200 = 2000 + 200 = \mathbf{2200}$

Method 2 – Double:
$11 \times 200 = 11 \times 100 + 11 \times 100 = 1100 + 1100 = \mathbf{2200}$

Method 3 – Hundreds language:
$11 \times 200 = 11 \times 2 \text{ Hundreds} = 22 \text{ Hundreds} = \mathbf{2200}$

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2Fill in the blanks and share what you notice:

11 \times 100 = \underline{\hspace{2cm}}

22 \times 100 = \underline{\hspace{2cm}}

11 \times 200 = \underline{\hspace{2cm}}

22 \times 200 = \underline{\hspace{2cm}}

$11 \times 100 = \mathbf{1100}$
$22 \times 100 = \mathbf{2200}$
$11 \times 200 = \mathbf{2200}$
$22 \times 200 = \mathbf{4400}$

Observation:

$22 \times 100 = 11 \times 200 = 2200$ (doubling the multiplier is the same as doubling the multiplicand).
$22 \times 200 = 2 \times (11 \times 200) = 2 \times 2200 = 4400$ .
When the number of groups doubles OR the group size doubles, the product doubles.

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3Answer the following questions:

18 \times 100 = \underline{\hspace{2cm}}

5 \times 500 = \underline{\hspace{2cm}}

15 \times 200 = \underline{\hspace{2cm}}

14 \times 300 = \underline{\hspace{2cm}}

23 \times 200 = \underline{\hspace{2cm}}

7 \times 800 = \underline{\hspace{2cm}}

$18 \times 100 = \mathbf{1800}$

$5 \times 500 = 5 \times 5 \times 100 = 25 \times 100 = \mathbf{2500}$

$15 \times 200 = 15 \times 2 \times 100 = 30 \times 100 = \mathbf{3000}$

$14 \times 300 = 14 \times 3 \times 100 = 42 \times 100 = \mathbf{4200}$

$23 \times 200 = 23 \times 2 \times 100 = 46 \times 100 = \mathbf{4600}$

$7 \times 800 = 7 \times 8 \times 100 = 56 \times 100 = \mathbf{5600}$

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Set A, Set B and Set C Problems

1Find the answers in Set A:

14 \times 100 =

14 \times 500 =

7 \times 500 =

7 \times 250 =

14 \times 10 =

14 \times 50 =

7 \times 50 =

7 \times 25 =

14 \times 1 =

14 \times 5 =

7 \times 5 =

7 \times 10 =

Set A:

$14 \times 1 = \mathbf{14}$
$14 \times 5 = \mathbf{70}$
$7 \times 5 = \mathbf{35}$
$7 \times 10 = \mathbf{70}$

$14 \times 10 = \mathbf{140}$
$14 \times 50 = 14 \times 5 \times 10 = 70 \times 10 = \mathbf{700}$
$7 \times 50 = 7 \times 5 \times 10 = 35 \times 10 = \mathbf{350}$
$7 \times 25 = \mathbf{175}$

$14 \times 100 = \mathbf{1400}$
$14 \times 500 = 14 \times 5 \times 100 = 70 \times 100 = \mathbf{7000}$
$7 \times 500 = 7 \times 5 \times 100 = 35 \times 100 = \mathbf{3500}$
$7 \times 250 = 7 \times 25 \times 10 = 175 \times 10 = \mathbf{1750}$

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2Use the understanding from Set A to find the answers in Set B:

30 \times 100 =

30 \times 200 =

15 \times 100 =

15 \times 200 =

30 \times 10 =

30 \times 20 =

15 \times 10 =

15 \times 20 =

30 \times 1 =

30 \times 2 =

15 \times 1 =

15 \times 2 =

Set B:

$30 \times 1 = \mathbf{30}$
$30 \times 2 = \mathbf{60}$
$15 \times 1 = \mathbf{15}$
$15 \times 2 = \mathbf{30}$

$30 \times 10 = \mathbf{300}$
$30 \times 20 = 30 \times 2 \times 10 = 60 \times 10 = \mathbf{600}$
$15 \times 10 = \mathbf{150}$
$15 \times 20 = 15 \times 2 \times 10 = 30 \times 10 = \mathbf{300}$

$30 \times 100 = \mathbf{3000}$
$30 \times 200 = 30 \times 2 \times 100 = 60 \times 100 = \mathbf{6000}$
$15 \times 100 = \mathbf{1500}$
$15 \times 200 = 15 \times 2 \times 100 = 30 \times 100 = \mathbf{3000}$

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3Set C – Answer the following questions:
1)

44 \times 10 =

16 \times 100 =

22 \times 20 =

4 \times 400 =

$44 \times 10 = \mathbf{440}$

$16 \times 100 = \mathbf{1600}$

$22 \times 20 = 22 \times 2 \times 10 = 44 \times 10 = \mathbf{440}$

$4 \times 400 = 4 \times 4 \times 100 = 16 \times 100 = \mathbf{1600}$

Observation: $44 \times 10 = 22 \times 20 = 440$ and $16 \times 100 = 4 \times 400 = 1600$ — halving one factor and doubling the other keeps the product the same.

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Vande Bharat Mission

1In the first week of the Vande Bharat Mission, 64 flights carried 152 people each. How many people travelled in the first week? Find

64 \times 152

Given: 64 flights, 152 people each.

Method – Split using grid (64 = 60 + 4, 152 = 100 + 50 + 2):

×	100	50	2
60	6000	3000	120
4	400	200	8
Total	6400	3200	128

$64 \times 152 = 6400 + 3200 + 128 = \mathbf{9728}$

Answer: 9728 people travelled in the first week of the Vande Bharat Mission.

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Snake Boat Race

1In a particular race, 960 participants volunteered. Each boat is pedalled by 64 people. How many boats will be needed? Find

960 \div 64

Given: 960 participants, 64 per boat.

Method – Repeated subtraction:

No. of boats	Participants used	Remaining
—	—	960
10	640	320
5	320	0

$960 \div 64 = \mathbf{15} \text{ remainder } \mathbf{0}$

Answer: 15 boats will be needed.

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Let Us Solve (Page 197)

1Solve the following and identify remainder (if any):
a)

237 \times 28

140 \times 16

389 \times 57

807 \div 24

692 \div 33

996 \div 45

a) $237 \times 28$

$237 \times 28 = 237 \times 20 + 237 \times 8$
$= 4740 + 1896 = \mathbf{6636}$

b) $140 \times 16$

$140 \times 16 = 140 \times 10 + 140 \times 6$
$= 1400 + 840 = \mathbf{2240}$

c) $389 \times 57$

$389 \times 57 = 389 \times 50 + 389 \times 7$
$= 19450 + 2723 = \mathbf{22173}$

d) $807 \div 24$

$24 \times 30 = 720$ ; $807 - 720 = 87$
$24 \times 3 = 72$ ; $87 - 72 = 15$
$807 \div 24 = \mathbf{33} \text{ remainder } \mathbf{15}$

e) $692 \div 33$

$33 \times 20 = 660$ ; $692 - 660 = 32$
$692 \div 33 = \mathbf{20} \text{ remainder } \mathbf{32}$

f) $996 \div 45$

$45 \times 22 = 990$ ; $996 - 990 = 6$
$996 \div 45 = \mathbf{22} \text{ remainder } \mathbf{6}$

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Dividing by 10 and 100

1A farmer packs his rice in sacks of 10 kg each.
a) If he has 60 kg of rice, how many sacks does he need?
b) If he has 600 kg of rice, how many sacks does he need?
If a sack of rice weighs 100 kg, how many sacks does he need for 600 kg of rice?

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2Find the answers to the following questions:

40 \div 10 = \underline{\hspace{2cm}}

4 \div 2 = \underline{\hspace{2cm}}

400 \div 2 = \underline{\hspace{2cm}}

400 \div 10 = \underline{\hspace{2cm}}

40 \div 20 = \underline{\hspace{2cm}}

400 \div 20 = \underline{\hspace{2cm}}

400 \div 100 = \underline{\hspace{2cm}}

400 \div 200 = \underline{\hspace{2cm}}

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3Think and answer. Write the division statement in each case.
1. Manku the monkey sees 870 bananas in the market. Each bunch has 10 bananas. How many bunches are there in the market?
2. Rukhma Bi wants to distribute ₹1000 equally among her 10 grandchildren on the occasion of Eid. How much money will each of them get?

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Let Us Solve (Word Problems)

1The oldest long-distance train of the Indian Railways is the Punjab Mail. On its first journey it had 6 coaches: 3 carrying 96 passengers and 3 for goods.
a) How many people travelled in each coach on the first journey?
b) This train now runs between Mumbai and Ferozepur. It has 24 coaches. Each coach can carry 72 passengers. How many people can travel on this train?

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2Amala and her 35 classmates, along with 6 teachers, are going on a school trip to Goa. They are using a double-decker 'hop on hop off' sightseeing bus.
a) 2 people can sit on every seat. There are 15 seats in the lower deck and 10 in the upper deck. How many seats will they need to occupy? Are there enough seats for everyone?
b) Find the total cost of the tickets for all children.
c) What is the cost of the tickets for all teachers?
(Note: Ticket prices are shown in the image which is not visible. Assuming Child ticket = ₹50 and Adult/Teacher ticket = ₹100 based on typical NCERT context.)

Given:

Total students = Amala + 35 classmates = 36 children
Teachers = 6
Total people = 36 + 6 = 42
Seats per deck: Lower = 15, Upper = 10 → Total seats = 25
Each seat holds 2 people → Total capacity = $25 \times 2 = 50$ people

a) Seats needed:

Total people = 42
Each seat holds 2, so seats needed = $42 \div 2 = \mathbf{21}$ seats

Total available seats = 25

$25 \geq 21$ → Yes, there are enough seats for everyone.

b) Cost of tickets for all children:
(Ticket prices depend on the image. Assuming Child ticket = ₹50)
$36 \times 50 = \mathbf{₹1800}$

c) Cost of tickets for all teachers:
(Assuming Teacher/Adult ticket = ₹100)
$6 \times 100 = \mathbf{₹600}$

Note: Students should use the actual ticket prices shown in their textbook image to calculate the correct answers.

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3Kedar works in a brick kiln.
a) The kiln makes 125 bricks in a day. How many bricks can be made in a month?
b) Each brick is sold for ₹9. How much money can they earn in a month?

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4Eight people are going on a trip to Chilika Lake.
- A bus ticket from Puri to Satapada costs ₹60.
- A two-hour boat ride for 8 people costs ₹1200.
How much money do we need to spend on each person?

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5Find the multiplication and division sentences in the grid. Some are done for you:
250 × 4 = 1000
50 × 20 = 1000
525 ÷ 5 = 105

Given examples (already done):

$250 \times 4 = 1000$ ✓
$50 \times 20 = 1000$ ✓
$525 \div 5 = 105$ ✓

Additional sentences that can be found in the grid:

$5 \times 20 = 100$ (5, 20, 100 appear in grid)
$3 \times 6 = 18$ — check grid
$6 \times 22 = 132$ (6, 22, 132 appear in grid)
$16 \times 200 = 3200$ (16, 200, 3200 appear in grid)
$30 \times 20 = 600$ (30, 20, 600 appear in grid)
$14 \times 8 = 112$ — check grid
$6000 \div 200 = 30$ (6000, 200, 30 appear in grid)
$1200 \div 8 = 150$ — check grid
$104 \div 4 = 26$ (104, 4, 26 appear in grid)
$9 \times 7 = 63$ — check grid

Note: Students should scan each row, column, and diagonal of the grid to find all valid multiplication and division sentences.

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6Solve:
a)

35 \times 76

267 \times 38

498 \times 9

89 \times 42

55 \times 23

345 \times 17

66 \times 22

704 \times 11

319 \times 26

459 \div 3

774 \div 18

864 \div 26

304 \div 12

670 \div 9

584 \div 25

900 \div 15

658 \div 32

974 \div 9

Multiplication:

a) $35 \times 76$
$= 35 \times 70 + 35 \times 6 = 2450 + 210 = \mathbf{2660}$

b) $267 \times 38$
$= 267 \times 30 + 267 \times 8 = 8010 + 2136 = \mathbf{10146}$

c) $498 \times 9$
$= 500 \times 9 - 2 \times 9 = 4500 - 18 = \mathbf{4482}$

d) $89 \times 42$
$= 89 \times 40 + 89 \times 2 = 3560 + 178 = \mathbf{3738}$

e) $55 \times 23$
$= 55 \times 20 + 55 \times 3 = 1100 + 165 = \mathbf{1265}$

f) $345 \times 17$
$= 345 \times 10 + 345 \times 7 = 3450 + 2415 = \mathbf{5865}$

g) $66 \times 22$
$= 66 \times 20 + 66 \times 2 = 1320 + 132 = \mathbf{1452}$

h) $704 \times 11$
$= 704 \times 10 + 704 \times 1 = 7040 + 704 = \mathbf{7744}$

i) $319 \times 26$
$= 319 \times 20 + 319 \times 6 = 6380 + 1914 = \mathbf{8294}$

Division:

j) $459 \div 3$
$3 \times 153 = 459$
$459 \div 3 = \mathbf{153}, \text{ remainder } \mathbf{0}$

k) $774 \div 18$
$18 \times 43 = 774$
$774 \div 18 = \mathbf{43}, \text{ remainder } \mathbf{0}$

l) $864 \div 26$
$26 \times 33 = 858$ ; $864 - 858 = 6$
$864 \div 26 = \mathbf{33}, \text{ remainder } \mathbf{6}$

m) $304 \div 12$
$12 \times 25 = 300$ ; $304 - 300 = 4$
$304 \div 12 = \mathbf{25}, \text{ remainder } \mathbf{4}$

n) $670 \div 9$
$9 \times 74 = 666$ ; $670 - 666 = 4$
$670 \div 9 = \mathbf{74}, \text{ remainder } \mathbf{4}$

o) $584 \div 25$
$25 \times 23 = 575$ ; $584 - 575 = 9$
$584 \div 25 = \mathbf{23}, \text{ remainder } \mathbf{9}$

p) $900 \div 15$
$15 \times 60 = 900$
$900 \div 15 = \mathbf{60}, \text{ remainder } \mathbf{0}$

q) $658 \div 32$
$32 \times 20 = 640$ ; $658 - 640 = 18$
$658 \div 32 = \mathbf{20}, \text{ remainder } \mathbf{18}$

r) $974 \div 9$
$9 \times 108 = 972$ ; $974 - 972 = 2$
$974 \div 9 = \mathbf{108}, \text{ remainder } \mathbf{2}$

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Chinnu's Coins

1Five friends plan to visit an amusement park. The cost of the ticket is ₹750.
- Bujji has all notes of ₹200.
- Munna has all notes of ₹50.
- Balu has all notes of ₹20.
- Chinnu has all coins of ₹5.
- Sansu has all coins of ₹2.
a) Find out how many notes/coins each child has to bring to buy the ticket.
b) Which of these children will not receive any change from the cashier?
c) How long would the cashier take to count Chinnu's coins?

Given: Ticket cost = ₹750

a) Notes/coins needed:

Bujji (₹200 notes):

$750 \div 200 = 3$ notes = ₹600; needs 1 more ₹200 note → brings 4 notes (₹800), gets ₹50 change.

Munna (₹50 notes):

$750 \div 50 = 15$ notes exactly → brings 15 notes (₹750).

Balu (₹20 notes):

$750 \div 20 = 37$ notes = ₹740; needs 1 more → brings 38 notes (₹760), gets ₹10 change.

Chinnu (₹5 coins):

$750 \div 5 = 150$ coins exactly → brings 150 coins (₹750).

Sansu (₹2 coins):

$750 \div 2 = 375$ coins = ₹750; but 750 is odd, so $2 \times 375 = 750$ — actually $750 \div 2 = 375$ exactly → brings 375 coins (₹750).

b) Children who will NOT receive any change:

Children who pay exactly ₹750:

Munna (15 × ₹50 = ₹750) ✓
Chinnu (150 × ₹5 = ₹750) ✓
Sansu (375 × ₹2 = ₹750) ✓

Bujji (pays ₹800, gets ₹50 change) and Balu (pays ₹760, gets ₹10 change) will receive change.

c) Time to count Chinnu's coins:

Chinnu has 150 coins of ₹5 each.
If the cashier counts 1 coin per second, it takes 150 seconds = 2 minutes 30 seconds.
If counting at 2 coins per second, it takes 75 seconds ≈ 1 minute 15 seconds.

(The exact time depends on counting speed — this is an open discussion question.)

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2Observe the following multiplications:

12 \times 13 = 156

11 \times 14 = 154

13 \times 13 = 169

11 \times 12 = 132

In each case, do you see any pattern in the two numbers and their product? Find 5 such examples.

Pattern Observation:

Look at the tens digit and units digit of the product:

$12 \times 13 = 156$ : $1 \times 1 = 1$ (hundreds), $2 + 3 = 5$ (tens), $2 \times 3 = 6$ (units) → 156 ✓
$11 \times 14 = 154$ : $1 \times 1 = 1$ , $1 + 4 = 5$ , $1 \times 4 = 4$ → 154 ✓
$13 \times 13 = 169$ : $1 \times 1 = 1$ , $3 + 3 = 6$ , $3 \times 3 = 9$ → 169 ✓
$11 \times 12 = 132$ : $1 \times 1 = 1$ , $1 + 2 = 3$ , $1 \times 2 = 2$ → 132 ✓

Pattern: For two-digit numbers of the form $1a$ and $1b$ (i.e., both in the teens):
$1a \times 1b = 100 + (a+b) \times 10 + a \times b$

5 more examples:

$11 \times 13 = 100 + (1+3) \times 10 + 1 \times 3 = 100 + 40 + 3 = \mathbf{143}$ ✓
$12 \times 14 = 100 + (2+4) \times 10 + 2 \times 4 = 100 + 60 + 8 = \mathbf{168}$ ✓
$11 \times 15 = 100 + (1+5) \times 10 + 1 \times 5 = 100 + 60 + 5 = \mathbf{165}$ ✓
$12 \times 12 = 100 + (2+2) \times 10 + 2 \times 2 = 100 + 40 + 4 = \mathbf{144}$ ✓
$13 \times 14 = 100 + (3+4) \times 10 + 3 \times 4 = 100 + 70 + 12 = \mathbf{182}$ ✓

Note: This pattern works perfectly when $a \times b < 10$ (single digit). When $a \times b \geq 10$ , carry-over must be added.

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3Assume each vehicle is travelling with full capacity. How many people can travel in each of these vehicles? Match them up. (Note: Vehicle images and capacities are in the textbook image which is not visible here.)

Note: This question requires matching vehicles shown in the textbook image (which is not visible in the OCR).

Method to solve: For each vehicle, multiply the number of rows/seats by the number of people per seat (or use the given capacity).

General approach:

Read the capacity of each vehicle from the image.
Multiply: Number of seats × People per seat = Total capacity.
Match the vehicle to its total passenger count.

Example (typical vehicles in such problems):

Auto-rickshaw: 3 passengers
Car: 4–5 passengers
Mini-bus: 20–30 passengers
Bus: 40–60 passengers
Train coach: 72 passengers

Students should refer to the actual image in their textbook to complete the matching.

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Frequently Asked Questions

What are the important topics in The Transport Museum for CBSE Class 4 Mathematics?

Key topics in The Transport Museum include Multiplication by grouping, Division with remainder, Multiplication by 10 and 100, Real-life problems in the chapter. Study these first, then practise questions on each for Class 4 exams.

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How should I revise The Transport Museum for Class 4 exams?

Learn the core ideas first, then work through the 45 practice questions on The Transport Museum. Revise definitions regularly and use flashcards for quick recall before the exam.

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The Transport Museum — NCERT Solutions

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Mystery Matrix

Mystery Matrix (Row/Column Products)

Times-10

Constructing Tables (Times-15)

Making Tables by Splitting into Equal Groups

Multiplying by Multiples of 10

At the Transport Museum

Let Us Solve (Page 191–192)

Multiples of 100

Multiplying by 200, 300, etc.

Set A, Set B and Set C Problems

Vande Bharat Mission

Snake Boat Race

Let Us Solve (Page 197)

Dividing by 10 and 100

Let Us Solve (Word Problems)

Chinnu's Coins

Frequently Asked Questions

Sources & Official References

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Practice Quiz

Important Questions

Revision Notes

Formula Sheet

Chapter Summary

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