Class 4 Maths Chapter 13 The Transport Museum – Stepwise NCERT Solutions

The Transport Museum Class 4 Solutions Question Answer Maths Chapter 13

NCERT Textbook Page 184 Mystery Matrix

Fill 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.

Solution:

The 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.  

Solution:

NCERT Textbook Pages 184-185 Times-10

Match each problem with the appropriate pictorial representation and write the answer.

What is 10 × 10 = ____ Tens = _______

Solution:

What is 10 × 10 = __10__ Tens = ___1Hundred____

NCERT Textbook Pages 185-187 Constructing Tables

How many pebbles are there in this arrangement?

Solution:
This arrangement contains 75 pebbles, organized in a 5 × 15 pattern. An easy way to calculate this product is by dividing the arrangement into smaller parts.   

5 × 15 = 5 × 10 and 5 × 5 = 50 + 25 = 75

Recall the times-table that we created in Grade 3. Now construct a times-15 table. You may use the arrangement given below and split the columns into 10 and 5 for ease of counting, as shown on the previous page.

Solution:
1 × 15 = 1 × 10 + 1 × 5 = 10 + 5 = 15
2 × 15 = 2 × 10 + 2 × 5 = 20 + 10 = 30
3 × 15 = 3 × 10 + 3 × 5 = 30 + 15 = 45
4 × 15 = 4 × 10 + 4 × 5 = 40 + 20 = 60
5 × 15 = 5 × 10 + 5 × 5 = 50 + 25 = 75
6 × 15 = 6 × 10 + 6 × 5 = 60 + 30 = 90
7 × 15 = 7 × 10 + 7 × 5 = 70 + 35 = 105
8 × 15 = 8 × 10 + 8 × 5 = 80 + 40 = 120
9 × 15 = 9 × 10 + 9 × 5 = 90 + 45 = 135
10 × 15 = 10 × 10 + 10 × 5 = 100 + 50 = 150

Question 1.
What patterns do you see in this table?

Solution:

1 × 15 = 15, which is the number of pebbles in the first row.
2 × 15 = 15 + 15 = 30, representing the total number of pebbles in the first and second rows.
3 × 15 = 15 + 15 + 15 = 45, which is the total number of pebbles in the first three rows.
4 × 15 = 15 + 15 + 15 + 15 = 60, the number of pebbles in the first four rows, and so on.

Thus, the values in the times-15 table increase by 15 each time, forming a pattern of 15, 30, 45, 60, and so on.

Question 2.

Compare the times-15 table with the times-5 table. What similarities and differences do you notice? It is time-10 table.

Solution:

Similarity: Both tables show a pattern where the results alternate between odd and even numbers. For instance, in the times-5 table, the sequence 5, 10, 15 alternates between odd and even, and similarly, in the times-15 table, the sequence 15, 30, 45 follows the same pattern.

Difference: The main difference lies in the base number being multiplied. The times-5 table uses 5 as the multiplier, whereas the times-15 table uses 15. Additionally, each number in the times-15 table is three times larger than the corresponding number in the times-5 table.

Question 3.
Construct other times-tables for numbers from 11 to 20, as you did for 15.

Solution:

Question 4.
As you compared the times-5 table with the times-15 table, 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.

Solution:

Each successive number in the times-11 table is eleven times greater than the corresponding number in the times-1 table.
Similarly, each successive number in the times-12 table is six times greater than the corresponding number in the times-2 table, and so on.

NCERT Textbook Pages 187-188

Making tables by splitting into equal groups

Here is an arrangement of wheels. To count the total number of wheels, Tara splits them into two equal groups.

3 × 14 = 3 × 7 and 3 × 7

= 21 + 21 = double of 21

= 42

Similarly, 6 × 14 can be obtained by splitting the arrangement into two equal groups.

6 × 14 = 6 × 7 and 6 × 7

= 42 + 42 = double of 42

= 84

We have seen how to calculate 3 × 14 and 6 × 14 by splitting and doubling. Can we construct the times-14 table by splitting and doubling? Try!

Solution:

Yes, we can construct the times-14 table by splitting and doubling.

1 × 14 = 1 × , 1 7 and 1 × 7
= 1 × 7 + 1 × 7 = 7 + 7 = 14

2 × 14 = 2 × 7 and 2 × 7
= 2 × 7 + 2 × 7 = 14 + 14 = 28

3 × 14 = 3 × 7 and 3 × 7
= 3 × 7 + 3 × 7 = 21 + 21 = 42

4 × 14 = 4 × 7 and 4 × 7
= 4 × 7 + 4 × 7 = 28 + 28 = 56

5 × 14 = 5 × 7 and 5 × 7
= 5 × 7 + 5 × 7 = 35 + 35 = 70

6 × 14 = 6 × 7 and 6 × 7
= 6 × 7 + 6 × 7 = 42 + 42 = 84

7 × 14 = 7 × 7 and 7 × 7
= 7 × 7 + 7 × 7 = 49 + 49 = 98

8 × 14 = 8 × 7 and 8 × 7
= 8 × 7 + 8 × 7 = 56 + 56 = 112

9 × 14 = 9 × 7 and 9 × 7
= 9 × 7 + 9 × 7 = 63 + 63 = 126

10 × 14 = 10 × 7 and 10 × 7
= 10 × 7 + 10 × 7 = 70 + 70 = 140

What other times tables can be constructed by splitting into equal groups and doubling? Give examples.

Solution: We can create the 8, 10, 12, 16, 18, 20, and other times tables by dividing them into equal groups and then doubling.

NCERT Textbook Page 188 Multiples of 10

Find the answers to the following:

(a) 15 × 10 = ____ Tens = _______.
(b) 16 × 10 = ____ Tens = _______.
(c) 19 × 10 = ____ Tens = _______.
(d) 20 × 10 = ____ Tens = _______.

Solution:
(a) 15 × 10 = 15 Tens = 10 Tens + 5 Tens
= 100 + 50 = 150

(b) 16 × 10 = 16 Tens = 10 Tens + 6 Tens
= 100 + 60 = 160

(c) 19 × 10 = 19 Tens = 10 Tens + 9 Tens
= 100 + 90
= 190

(d) 20 × 10 = 20 Tens = 10 Tens + 10 Tens
= 100 + 100 = 200

10 × 10 = 10 Tens = 100 2 times (double)
10 × 10 = 10 Tens + 10 Tens
= 100 + 100 = 200

Discuss in grade what happens when we take several groups of 10.

Solution:
Students should do it by themselves.

NCERT Textbook Pages 189-190

Now think and answer the following Problems.

30 × 10 = _____
40 × 10 = _____
50 × 10 = _____
60 × 10 = _____
70 × 10 = _____
80 × 10 = _____

Solution:
30 × 10 = 30 Tens = 300
40 × 10 = 40 Tens = 400
50 × 10 = 50 Tens = 500
60 × 10 = 60 Tens = 600
70 × 10 = 70 Tens = 700
80 × 10 = 80 Tens = 800

Answer the following questions. Share your thoughts.

(a) 21 × 10 = _______

Solution:
21 × 10 = 21 Tens = 20 Tens + 1 Ten
= 200 + 10
= 210

(b) 42 × 10 = _______

Solution:
42 × 10 = 42 Tens = 40 Tens + 2 Tens
= 400 + 20
= 420

(c) 65 × 10 = _______

Solution:
65 × 10 = 65 Tens = 60 Tens + 5 Tens
= 600 + 50
= 650

(d) 38 × 10 = _______

Solution:
38 × 10 = 38 Tens = 30 Tens + 8 Tens
= 300 + 80
= 380

(e) 53 × 10 = _______

Solution:
53 × 10 = 53 Tens = 50 Tens + 8 Tens
= 500 + 30
= 530

(f) 87 × 10 = _______

Solution:
87 × 10 = 87 Tens = 80 Tens + 7 Tens
= 800 + 70
= 870

Solve the following problems. Share your thoughts.

24 × 40 = _______
50 × 60 = _______
13 × 30 = _______
43 × 60 = _______
70 × 80 = _______

Solution:
24 × 40 = 20 × 40 and 4 × 40
= 20 × 4 Tens + 4 × 4 Tens
= 800 + 160
= 960

50 × 60 = 50 × 6 Tens
= 300 Tens
= 3000

13 × 30 = 13 × 3 Tens
= 39 Tens
= 390

43 × 60 = 40 × 60 and 3 × 60
= 40 × 6 Tens + 3 × 6 Tens
= 240 Tens + 18 Tens
= 2400 + 180
= 2580

70 × 80 = 8 × 7 × 10 × 10
= 56 × 10 Tens
= 560 Tens
= 5600

NCERT Textbook Pages 190-191 – A Day at the Transport Museum

Amala, Raahi and Farzan are visiting the “Transport Museum”. This museum has a collection of different modes of transport used by people in India. It includes several vehicles from the olden days. Raahi spots a toy train. She figures out that each coach can seat 14 children. The toy train has 15 coaches.

How many children can be seated in the toy train?

We have to find 15 × 14.

15 × 14 = 100 + 40 + 50 + 20

= 210

In 15 coaches, 210 children can be seated.

She wonders how many coaches will be needed for the 324 children from her school. Remember, each coach can seat only 14 children.

We have to find 324 ÷ 14

Total no. of coaches: 10 + 10 + 1 + 2 = 23

What do we do with the remaining 2 children? Discuss in grade.   

Solution:
Students should do it by themselves.

NCERT Textbook Page 192 Let Us Solve

Also, identify remainder (if any) in the division problems.

(a) 25 × 34

Solution:

(b) 16 × 43

Solution:

(c) 68 × 12

Solution:

(d) 39 × 13

Solution:

(e) 125 ÷ 15

Solution:

Therefore, when 125 is divided by 15, the result is (5 + 3) = 8 with a remainder of 5.

(f) 94 ÷ 11

Solution:

Therefore, when 94 is divided by 11, the result is (2 + 3 + 3) = 8, with a remainder of 6.

(g) 440 ÷ 22

Solution:

(h) 508 ÷ 18

Solution:

Therefore, when 508 is divided by 18, the result is (10 + 10 + 5 + 3) = 28, with a remainder of 4.

NCERT Textbook Pages 192-197 Multiples of 100

2 × 100 = 2 Hundreds = 200
3 × 100 = _____ Hundreds = ______
5 × 100 = _______ = __________
8 × 100 = ______ = _________
10 × 100 = 10 Hundreds = 1000

Solution:
2 × 100 = 2 Hundreds = 200
3 × 100 = 3 Hundreds = 300
5 × 100 = 5 Hundreds = 500
8 × 100 = 8 Hundreds = 800

What happens when we put 10 hundreds together?

11 × 100 = 11 Hundreds
= 10 Hundreds + 1 Hundred
= 1000+ 100 = 1100
12 × 100 = _______
15 × 100 = _______
20 × 100 = 20 Hundreds = 2000
27 × 100 = _______
70 × 100 = _______

Solution:
10 × 1 Hundred = 1000

12 × 100 = 12 Hundreds

= 10 Hundreds + 2 Hundreds
= 1000 + 200
= 1200

15 × 100 = 15 Hundreds
= 10 Hundreds + 5 Hundreds
= 1000 + 500
= 1500

27 × 100 = 27 Hundreds
= 20 Hundreds + 7 Hundreds
= 2000 + 700
= 2700

70 × 100 = 70 Hundreds
= 7000

Now answer the following questions. Share your thoughts.

30 × 100 = ________
40 × 100 = ________
50 × 100 = ________
24 × 100 = ________
53 × 100 = ________
19 × 100 = ________

We Know
80 × 100 = 8000

Find
80 × 50 = ________
40 × 50 = ______

Solution:
30 × 100 = 3 × 1000 = 3000
40 × 100 = 4 × 1000 = 4000
50 × 100 = 5 × 1000 = 5000
24 × 100 = 24 × 1 Hundred = 2400
53 × 100 = 53 × 1 Hundred = 5300
19 × 100 = 19 × 1 Hundred = 1900

80 × 50 = 4000
40 × 50 = 2000

Share what you notice about the answers to these problems.

11 × 100 = _______
11 × 200 = _______
22 × 100 = _______
22 × 200 = _______

Solution:
11 × 100 = 1100

11 × 200 = 11 × 100 + 11 × 100
= 1100 + 1100
= 2200

22 × 100 = 2200
22 × 200 = 22 × 2 Hundreds
= 44 Hundreds
= 4400

Answer the following questions. Share your thoughts.

18 × 100 = _______
5 × 500 = _______
15 × 200 = _______
14 × 300 = _______
23 × 200 = _______
7 × 800 = _______

Solution:
18 × 100 = 1800

5 × 500 = 5 × 5 Hundreds
= 25 Hundreds
= 2500

15 × 200 = 15 × 2 Hundreds
= 30 Hundreds
= 3000

14 × 300 = 14 × 3 Hundreds
= 42 Hundreds
= 4200

23 × 200 = 23 × 2 Hundreds
= 46 Hundreds
= 4600

7 × 800 = 7 × 8 Hundreds
= 56 Hundreds
= 5600

Find the answers in Set A. Examine the relationships between the problems and the answers in Set A carefully. Then use this understanding to find the answers in Set B.

Solution:

C. Answer the following questions:

1. 44 × 10 = 440
22 × 20 = 22 × 10 + 22 × 10
= 220 + 220
= 440

2. 16 × 100 = 1600
4 × 400 = 4 × 100 + 4 × 100 + 4 × 100 + 4 × 100
= 400 + 400 + 400 + 400
= 1600

NCERT Textbook Page 197 – Let Us Solve

Also, identify remainder (if any) in the division problems,

(a) 237 × 28

Solution:

(b) 140 × 16

Solution:

(c) 389 × 57

Solution:

d) 807 ÷ 24

Solution:

When 807 is divided by 24, the result is 10 + 10 + 10 + 2 + 1 = 33, with a remainder of 15.

e) 692 ÷ 33

Solution:

f) 996 ÷ 45

Solution:

NCERT Textbook Pages 197-198

Dividing by 10 and 100

A farmer packs his rice in sacks of 10 kg each.

(а) If he has 60 kg of rice, how many sacks does he need?

Solution:

The weight of rice to be packed in each sack is 10 kg, and the total weight of rice is 60 kg.

The number of sacks needed is 600 + 10 = 60.

Therefore, the total number of sacks required is 6.

(b) If he has 600 kg of rice, how many sacks does he need?

Solution: Each sack is to be packed with 100 kg of rice, and the total weight of rice is 600 kg.

The number of sacks needed is 600 ÷ 10 = 60.

Therefore, the total number of sacks required is 60.

If a sack of rice weighs loo kg then how many sacks does he need for 600 kg of rice?

Solution: Each sack is to be packed with 100 kg of rice, and the total weight of rice is 600 kg.
The number of sacks required is 600 ÷ 100 = 6.

Find the answers to the following questions. Share your thoughts in class.

Solution:
40 ÷ 10 = 4
4 ÷ 2 = 2
400 ÷ 2 = 200
400 ÷ 10 = 40
40 ÷ 20 = 2
400 ÷ 20 = 20
400 ÷ 100 = 4
400 ÷ 200 = 2
400 ÷ 200 = 2

Think and answer. Write the division statement in each case.

Question 1.

Manku the monkey sees 870 bananas in the market. Each bunch has 10 bananas. How many bunches are there in the market?

Solution: Total number of bananas in the market is 870, with 10 bananas in each bunch.
The number of bunches in the market is 870 ÷ 10 = 87.

Division statement: 870 ÷ 10 = 87.

Question 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?

Solution: Rukhma Bi has 10 grandchildren, and she distributes ₹1000 among them.

Each grandchild receives ₹100 from the total amount.

Division statement: 1000 ÷ 10 = 100.

NCERT Textbook Pages 198-200 Let Us Solve

Question 1.
The oldest long-distance train of the Indian Railways is the Punjab Mail which ran between Mumbai and Peshawar. Its first journey was on 12 October 1912. Do you I know how many coaches it had 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 has been running for 106 years now. It runs between Mumbai, Maharashtra and Ferozepur, Punjab. It has 24 coaches. Each coach can carry 72 passengers. How many people can travel on this train?

Solution: (a) The train has 3 coaches carrying passengers, and there are 96 people in total.
The number of people traveling in each coach is 96 ÷ 3.

96 ÷ 3 = 10 + 10 + 10 + 2 = 32,

So, 32 people traveled in each coach.

(b) The train has 24 coaches, with 72 passengers in each coach.

The total number of people traveling in the train is 24 × 72.

Thus, 1728 passengers can travel on this train.

Question 2.

Amala and her 35 classmates, along with 6 teachers, are going on a school trip to Goa. They are using the double-decker “hop on hop off” sightseeing bus to explore the city.

(a) 2 people can sit on every seat of the bus. 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?

Solution:

(a) The total number of people on the bus is the sum of students and teachers: 36 + 6 = 42.
Since each seat can accommodate 2 people, the required number of seats occupied by the people is 42 ÷ 2.

(b) The ticket price for each child is ₹359, and the total cost of the tickets for all the children is 36 × ₹359.   

Total coat of tickets for all children = ₹ 12924

(c) Ticket price for adult = ₹ 899
Total cost of the tickets for all teachers = 6 × ₹ 899

Total cost of the tickets forn all teachers = ₹ 5394

Question 3.

Kedar 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 in the market for ₹ 9. How much money can they earn in a month?

Solution:
(a) Number of bricks made in a day = 125

So, number of pricks can be made in a month = 30 × 125

Therefore, 3750 brecks can be made in a month.

(b) The price of each brick is ₹9, and the total money earned in a month is ₹9 × 3750.   

Money they earn in a month = ₹ 33750

Question 4.
Chilika lake in Odisha is the largest saltwater lake in India. It is famous for the Irrawaddy dolphins. Boats can be hired to go to see the dolphins. The trip from Puri includes a bus ride followed by a boat ride. Eight people will be going on the trip.

• 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?

Solution:
Boat ride costs for each person = ₹ 1200 ÷ 8

The cost of the boat ride for each person is 100 + 50 = ₹150.

The price of the bus ticket is ₹60.

Therefore, the total amount of money needed for each person is ₹60 + ₹150 = ₹210.

Question 5.

Find the multiplication and division sentences below. Shade the sentences. How many can you find? Some are done for you.

Solution:

250 × 4 = 1000

50 × 20 = 1000   

5 × 22 = 110

52 × 20 = 1040

104 × 6 = 624

30 × 15 = 450

50 × 19 = 950

1000 × 6 = 6000

55 × 101 = 5555

99 × 7 = 693

200 × 16 = 3200

35 × 9 = 315

931 ÷ 10 = 93

4 × 26 = 104

6 × 22 = 132

6000 ÷ 30 = 200

Question 6.
Solve

a) 35 × 76
Solution:

b) 267 × 38

Solution:

c) 498 × 9

Solution:

d) 89 × 42

Solution:

e) 55 × 23

Solution:

f) 345 × 17

Solution:

g) 66 × 22

Solution:

h) 704 × 11

Solution:

i) 319 × 26

Solution:

j) 459 ÷ 3

Solution:

k) 774 ÷ 18

Solution:

l) 864 ÷ 26

Solution:

m) 304 ÷ 12

Solution:

n) 670 ÷ 9

Solution:

o) 584 ÷ 25

Solution:

p) 900 ÷ 15

Solution:

q) 658 ÷ 32

Solution:

r) 974 ÷ 9

Solution:

NCERT Textbook Pages 201 -202 Chinnu’s Coins

Question 1.
Five friends plan to visit an amusement park nearby. Each of them uses different notes and coins to buy the ticket. The cost of the ticket is ?750.

• Bujji has brought all notes of ₹ 200.

• And Munna has brought all notes of ₹ 50.

• Whereas Balu has brought all notes of ₹ 20.

• And guess what, Chinnu got all coins of ₹ 5.

• And 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?

Solution:

(a) Notes brought by Bujji = 750 ÷ 200

Notes brought by Balu = 750 ÷ 20

So, Balu brought 10 + 10 + 10 + 5 + 2 = 37 notes of ₹20. For the remaining ₹10, he gave a ₹20 note and received a ₹10 note in return from the cashier, making the total 38 ₹20 notes.

Coins brought by Chinnu = 750 ÷ 5

So, Chinnu brought 100 + 20 + 20 + 10 = 150 coins.

Coins brought by Sansu= 750 ÷ 2

Thus, Sansu brought 100 + 100 + 100 + 50 + 20 + 5 = 375 coins of ₹20.

(b) Munna, Chinnu, and Sansu will not receive any change from the cashier.

(c) Chinnu has all coins of ₹ 5. And 750 + 5 = 150. So, the cashier counted 150 times to count Chinnu’s coins.

Question 2.
Observe the following multiplications. The answers have been provided.

In each case, do you notice any pattern between the two numbers and their product? (Hint: Pay attention to the coloured digits!)

For what other multiplication problems will this pattern hold? Find 5 such examples.

Solution: We observe that, the ones digit of the product is the product of ones digits of the multiplicand and multiplier, and the tens digit of the product is the sum of ones digit of the multiplicand and multiplier.

Other such examples are

Question 3.

Assume each vehicle is travelling with full capacity. How many people can travel in each of these vehicles? Match them up.

Solution:

NCERT Solutions Class 4 Maths Chapter 13 – The Transport Museum

In NCERT Solutions Class 4 Maths Chapter 13 The Transport Museum, children explore multiplication and division through vehicles, tickets, and passengers. These 2025-26 solutions make ideas like times-10, hundreds and division by 10 and 100 easy with stepwise, child-friendly explanations.

The chapter builds confidence in tables, multiples of 10 and 100, and word problems on travel. By revising every solved example, students strengthen mental maths and learn to quickly connect numbers with real-life situations like fares, seats, and coins.

While practicing Class 4 Maths Chapter 13, try to first guess answers using estimation, then check with written steps. Regular revision of these NCERT-based questions helps you score higher marks and makes multiplication–division completely hassle-free.