NCERT Solutions for Class 4 Maths Chapter 13 the Transport Museum – step-by-step answers to every exercise and in-text question of the chapter, as per the latest NCERT syllabus for 2027. Read them online or download the free question-answer PDF.
Amala, Raahi and Farzan visit a transport museum. Along the way we build new times-tables by splitting and doubling, multiply by 10s, 100s and 200s, and divide big numbers by taking away equal groups.
NCERT Solutions for Class 4 Maths Chapter 13 the Transport Museum PDF
All the answers are on this page below. The PDF also explains every answer – why it is right and how to check it – so download it for revision.
NCERT Solutions for Class 4 Maths Chapter 13 the Transport Museum – Questions and Answers
All 99 questions of the chapter (textbook pages 184 – 202) are answered below, in the order of the book – in-text questions, activities and the end-of-chapter exercises. Each answer is written the way you would write it in the exam.
Mystery Matrix — Page 184Mystery Matrix
Q1Fill 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
The Mystery Matrix as printed. The yellow boxes are empty; four products are given.
Answer
One answer:
×
4
5
7
2
8
32
40
56
16
6
24
30
42
12
9
36
45
63
18
3
12
15
21
6
One correct filling. The four given products are in black.
21 = 3 × 7 and 42 = 6 × 7 → that column box is 7; its row boxes are 3 and 6. 32 = 8 × 4 → row box 8, column box 4 45 = 9 × 5 → row box 9, column box 5 Last column: no product is given, so any 1-digit number fits. We took 2.
Q2The 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
54
63
48
First grid: row products 56 and 54 (orange), column products 63 and 48 (blue).
42
50
60
35
Second grid: row products 42 and 50 (orange), column products 60 and 35 (blue).
Answer
First grid
56 = 7 × 8 and 63 = 7 × 9, so the top-left box is 7. Then 56 ÷ 7 = 8 and 63 ÷ 7 = 9. Last box: 54 ÷ 9 = 6.
7
8
56
9
6
54
63
48
First grid filled in.
Second grid
42 = 6 × 7 and 60 = 6 × 10, so the top-left box is 6. Then 42 ÷ 6 = 7 and 60 ÷ 6 = 10. Last box: 50 ÷ 10 = 5.
6
7
42
10
5
50
60
35
Second grid filled in.
Times-10 — Page 184Times-10
Q1Match each problem with the appropriate pictorial representation and write the answer. 2 × 10 = 2 Tens = ______The three dashed groups of ten-strips printed beside the question. We have named them A, B and C.
Answer
2 × 10 = 2 Tens 2 Tens = 10 + 10 = 20 Picture: group C, which has 2 strips of ten.
Q25 × 10 = _____ Tens = ______The three dashed groups of ten-strips printed beside the question. We have named them A, B and C.
Answer
5 × 10 = 5 Tens 5 Tens = 10 + 10 + 10 + 10 + 10 = 50 Picture: group A, which has 5 strips of ten.
Q38 × 10 = _____ Tens = ______The three dashed groups of ten-strips printed beside the question. We have named them A, B and C.
Answer
8 × 10 = 8 Tens 8 Tens = 10, 20, 30, 40, 50, 60, 70, 80 Picture: group B, which has 8 strips of ten.
Q1Recall the times-tables 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. (1 × 15 = ___, 2 × 15 = ___, … 10 × 15 = ___)The pebble arrangement printed on page 186 — 15 pebbles across and 10 rows down.
AnswerThe 10 x 15 pebble arrangement, with each row split into 10 and 5.
Take one row at a time. Each row is 10 pebbles and 5 pebbles.
10 × part
5 × part
Answer
1 × 15
10
5
15
2 × 15
20
10
30
3 × 15
30
15
45
4 × 15
40
20
60
5 × 15
50
25
75
6 × 15
60
30
90
7 × 15
70
35
105
8 × 15
80
40
120
9 × 15
90
45
135
10 × 15
100
50
150
So the times-15 table is: 15, 30, 45, 60, 75, 90, 105, 120, 135, 150
Q21. What patterns do you see in this table?The pebble arrangement printed on page 186 — 15 pebbles across and 10 rows down.The times-15 table printed on page 186, with every answer still to be filled in.
Answer
Three patterns are easy to see.
The ones digit goes 5, 0, 5, 0, 5, 0 … again and again.
Each answer is 15 more than the one before it.
Every second answer (30, 60, 90, 120, 150) is in the times-30 table.
Do the same for 11 (10 + 1), 13 (10 + 3), 14 (10 + 4) … up to 20 (10 + 10).
Q24. 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.
Answer
times-1 and times-11
Times-1
Times-11
Difference
1
11
10
2
22
20
3
33
30
4
44
40
5
55
50
6
66
60
7
77
70
8
88
80
9
99
90
10
110
100
What we notice
The difference is always the times-10 table: 10, 20, 30, 40 …
The same thing happens for 2 and 12, for 3 and 13, and so on.
So a number times 12 = that number times 2 + that number times 10. For example, 4 × 12 = 8 + 40 = 48.
Example: 7 × 13 = 7 × 3 + 7 × 10 = 21 + 70 = 91
Q36 × 14 = Double of 6 × 7. Why?Tara’s arrangement of wheels: 6 rows of 14, split into two equal groups of 7.
Q2Raahi 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?
Answer
We have to find 15 × 14. Split 15 as 10 + 5 and 14 as 10 + 4.
The 15 x 14 arrangement split into 10 and 5, and into 10 and 4.
×
10
4
Total
10
100
40
140
5
50
20
70
15 × 14 = 140 + 70 = 210
In 15 coaches, 210 children can be seated.
A Day at the Transport Museum — Page 191Division with a remainder
Q1She wonders how many coaches will be needed for the 324 children from her school. Remember, each coach can seat only 14 children. (324 ÷ 14) Total no. of coaches = _____
Answer
The table in the book takes the children away in groups of coaches:
No. of children
No. of coaches needed
No. of children remaining
324
140
10
324 − 140 = 184
140
10
184 − 140 = 44
14
1
44 − 14 = 30
28
2
2
10 coaches take 140 children → 324 − 140 = 184 left 10 more coaches take 140 → 184 − 140 = 44 left 1 coach takes 14 → 44 − 14 = 30 left 2 coaches take 28 → 30 − 28 = 2 left Total no. of coaches = 10 + 10 + 1 + 2 = 23 Remainder = 2 children
So 23 coaches are filled and 2 children are still standing.
Q2What do we do with the remaining 2 children? Discuss in grade.
Answer
Sample answer. We cannot leave 2 children behind, so we take one more coach for them.
23 full coaches + 1 part-full coach = 24 coaches The 24th coach carries only 2 children.
Other fair ideas the class may give:
Let the 2 children sit with a teacher in the first coach.
Send the 2 children in the next trip of the toy train.
Let Us Solve — Page 192Also, identify remainder (if any) in the division problems.
Q1a) 25 × 34
Answer
Split each number by its places and multiply piece by piece.
×
30
4
Total
20
600
80
680
5
150
20
170
25 × 34 = 680 + 170 = 850
Q2b) 16 × 43
Answer
Split each number by its places and multiply piece by piece.
×
40
3
Total
10
400
30
430
6
240
18
258
16 × 43 = 430 + 258 = 688
Q3c) 68 × 12
Answer
Split each number by its places and multiply piece by piece.
×
10
2
Total
60
600
120
720
8
80
16
96
68 × 12 = 720 + 96 = 816
Q4d) 39 × 13
Answer
Split each number by its places and multiply piece by piece.
×
10
3
Total
30
300
90
390
9
90
27
117
39 × 13 = 390 + 117 = 507
Q5e) 125 ÷ 15
Answer
Take away groups of 15 again and again.
15 × 8 = 120 125 − 120 = 5 Groups taken = 8 Left over = 5
125 ÷ 15 = 8, remainder 5
Check: 15 × 8 + 5 = 125 ✓
Q6f) 94 ÷ 11
Answer
Take away groups of 11 again and again.
11 × 8 = 88 94 − 88 = 6 Groups taken = 8 Left over = 6
94 ÷ 11 = 8, remainder 6
Check: 11 × 8 + 6 = 94 ✓
Q7g) 440 ÷ 22
Answer
Take away groups of 22 again and again.
22 × 20 = 440 440 − 440 = 0 Groups taken = 20 Left over = 0
440 ÷ 22 = 20, remainder 0
The remainder is 0, so 22 divides 440 exactly.
Q8h) 508 ÷ 18
Answer
Take away groups of 18 again and again.
18 × 20 = 360 508 − 360 = 148 18 × 8 = 144 148 − 144 = 4 Groups taken = 20 + 8 = 28 Left over = 4
50 is half of 100. So 80 × 50 is half of 80 × 100. Half of 8000 = 4000
40 is half of 80. So 40 × 50 is half of 80 × 50. Half of 4000 = 2000
In-text Questions — Page 194Multiples of 100
Q1Share what you notice about the answers to these problems. What do you notice about any number times-100s? 11 × 100 = _______ 11 × 200 = _______ 22 × 100 = _______ 22 × 200 = _______
Answer
Problem
Answer
11 × 100
1100
11 × 200
2200
22 × 100
2200
22 × 200
4400
What we notice
Multiplying by 100 simply writes two zeros after the number: 11 → 1100.
Doubling the group size (100 → 200) doubles the answer: 1100 → 2200.
Doubling the number of groups (11 → 22) also doubles the answer.
Doubling both doubles it twice: 1100 → 2200 → 4400.
Q1Amala wonders how many people travelled the first week of this ‘Vande Bharat Mission’. Help her find the answer. (In the first week, 64 flights carried 152 people each.) 64 × 152
Answer
Split 64 as 60 + 4 and 152 as 100 + 50 + 2.
The 64 x 152 rectangle split into 60 and 4, and into 100, 50 and 2.
×
100
50
2
Total
60
6000
3000
120
9120
4
400
200
8
608
64 × 152 = 9120 + 608 = 9,728
So 9,728 people travelled in the first week.
In-text Questions — Page 197The snake boat race
Q1In a particular race, 960 participants volunteered. Each boat is pedalled by 64 people. How many boats will be needed? (960 ÷ 64)
Answer
No. of boats
No. of participants
No. of people remaining
960
10
640
320
5
320
0
10 boats hold 64 × 10 = 640 people 960 − 640 = 320 people left 5 boats hold 64 × 5 = 320 people 320 − 320 = 0 people left Total no. of boats = 10 + 5 = 15 Remainder = 0
15 boats will be needed, and no one is left out.
Let Us Solve — Page 197Also, identify remainder (if any) in the division problems.
Q1a) 237 × 28
Answer
Split each number by its places and multiply piece by piece.
×
20
8
Total
200
4000
1600
5600
30
600
240
840
7
140
56
196
237 × 28 = 5600 + 840 + 196 = 6,636
Q2b) 140 × 16
Answer
Split each number by its places and multiply piece by piece.
×
10
6
Total
100
1000
600
1600
40
400
240
640
140 × 16 = 1600 + 640 = 2,240
Q3c) 389 × 57
Answer
Split each number by its places and multiply piece by piece.
×
50
7
Total
300
15000
2100
17100
80
4000
560
4560
9
450
63
513
389 × 57 = 17100 + 4560 + 513 = 22,173
Q4d) 807 ÷ 24
Answer
Take away groups of 24 again and again.
24 × 30 = 720 807 − 720 = 87 24 × 3 = 72 87 − 72 = 15 Groups taken = 30 + 3 = 33 Left over = 15
807 ÷ 24 = 33, remainder 15
Check: 24 × 33 + 15 = 807 ✓
Q5e) 692 ÷ 33
Answer
Take away groups of 33 again and again.
33 × 20 = 660 692 − 660 = 32 Groups taken = 20 Left over = 32
692 ÷ 33 = 20, remainder 32
Check: 33 × 20 + 32 = 692 ✓
Q6f) 996 ÷ 45
Answer
Take away groups of 45 again and again.
45 × 20 = 900 996 − 900 = 96 45 × 2 = 90 96 − 90 = 6 Groups taken = 20 + 2 = 22 Left over = 6
996 ÷ 45 = 22, remainder 6
Check: 45 × 22 + 6 = 996 ✓
Dividing by 10 and 100 — Page 197A farmer packs his rice in sacks of 10 kg each.
Q1a) If he has 60 kg of rice, how many sacks does he need? 60 ÷ 10 = _______
Each time the group is ten times bigger, so the answer is ten times smaller: 200, 20, 2.
Q41. Manku the monkey sees 870 bananas in the market. Each bunch has 10 bananas. How many bunches are there in the market? _______ (Write the division statement.)
Answer
Division statement: 870 ÷ 10 = 87 870 = 87 Tens Each bunch is one ten.
There are 87 bunches in the market.
Q52. Rukhma Bi wants to distribute ₹1000/- equally among her 10 grandchildren on the occasion of Eid. How much money will each of them get? ________
Answer
Division statement: ₹1000 ÷ 10 = ₹100 1000 = 10 Hundreds Give 1 Hundred to each child.
Each grandchild gets ₹100.
Let Us Solve — Page 198The Punjab Mail
Q11. 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. It had 6 coaches: 3 carrying 96 passengers and 3 for goods. a) How many people travelled in each coach on the first journey?
Answer
Only 3 coaches carried people, and they carried 96 in all. 96 ÷ 3 = ? 3 × 30 = 90 → 96 − 90 = 6 3 × 2 = 6 → 6 − 6 = 0 96 ÷ 3 = 30 + 2 = 32
32 people travelled in each passenger coach.
Q2b) 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?
Answer
We have to find 24 × 72.
×
70
2
Total
20
1400
40
1440
4
280
8
288
24 × 72 = 1440 + 288 = 1,728
1,728 people can travel on this train.
Let Us Solve — Page 199The school trip, the brick kiln and Chilika lake
Q12. Amala and her 35 classmates, along with 6 teachers, are going on a school trip to Goa. 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?
Answer
Children = Amala + 35 classmates = 36 Teachers = 6 Total people = 36 + 6 = 42
1
3
6
+
6
4
2
2 people sit on one seat. 42 ÷ 2 = 21 seats needed
Seats in the bus = 15 + 10 = 25 25 is more than 21.
They need 21 seats, and yes, there are enough seats — 4 seats are still empty.
Q2b) Find the total cost of the tickets for all children.The ticket-price board printed beside the question.
Answer
Number of children = 36 Cost of one child’s ticket = ₹359 We have to find 36 × 359.
×
300
50
9
Total
30
9000
1500
270
10770
6
1800
300
54
2154
36 × 359 = 10770 + 2154 = 12,924
The tickets for all the children cost ₹12,924.
Q3c) What is the cost of the tickets for all teachers?The ticket-price board printed beside the question.
Answer
Number of teachers = 6 An adult ticket costs ₹899 We have to find 6 × 899.
×
800
90
9
Total
6
4800
540
54
5394
6 × 899 = 5,394
The tickets for all the teachers cost ₹5,394.
Q43. Kedar works in a brick kiln. a) The kiln makes 125 bricks in a day. How many bricks can be made in a month?
Answer
Bricks in one day = 125 Days in a month = 30 (we take a 30-day month) We have to find 125 × 30.
Q5b) Each brick is sold in the market for ₹ 9. How much money can they earn in a month?
Answer
Bricks in a month = 3,750 Price of one brick = ₹9 We have to find 3750 × 9.
×
9
Total
3000
27000
27000
700
6300
6300
50
450
450
3750 × 9 = 27000 + 6300 + 450 = 33,750
They can earn ₹33,750 in a month.
Q64. Chilika lake in Odisha is the largest saltwater lake in India. It is famous for the Irrawaddy dolphins. Boats can be hired to go 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?
Answer
First make the bill for all 8 people.
Item
Working
Amount
Bus tickets, Puri to Satapada
8 × ₹60
₹480
Two-hour boat ride for 8 people
given
₹1200
Total
₹480 + ₹1200
₹1680
The two amounts are added in columns:
₹
4
8
0
+
₹
1
2
0
0
₹
1
6
8
0
Now share the total between 8 people. ₹1680 ÷ 8 = ? 8 × 200 = 1600 → 1680 − 1600 = 80 8 × 10 = 80 → 80 − 80 = 0 ₹1680 ÷ 8 = ₹210
We need to spend ₹210 on each person.
Let Us Solve — Page 200Find the multiplication and division sentences
Q15. Find the multiplication and division sentences below. Shade the sentences. How many can you find? Some are done for you. (250 × 4 = 1000, 50 × 20 = 1000, 525 ÷ 5 = 105)
134
52
30
31
931
10
93
55
5
20
15
15
250
3
33
101
22
1040
450
0
4
26
104
5555
110
100
50
20
1000
60
16
99
44
104
19
3
6
22
132
7
20
6
950
6
6000
30
200
693
808
624
31
14
1200
8
16
24
35
9
525
5
105
62
3200
78
The number square as printed in the book. Three sentences are already shaded in the book.
Answer
15 sentences in all: the 3 done in the book and these 12 more.
No.
Sentence
Where
1
52 × 20 = 1040
down, column 2
2
30 × 15 = 450
down, column 3
3
55 × 101 = 5555
down, column 8
4
5 × 22 = 110
down, column 1
5
4 × 26 = 104
across, row 3
6
50 × 19 = 950
down, column 3
7
1000 × 6 = 6000
down, column 5
8
99 × 7 = 693
down, column 8
9
104 × 6 = 624
down, column 2
10
6 × 22 = 132
across, row 5
11
6000 ÷ 30 = 200
across, row 6
12
200 × 16 = 3200
down, column 7
134
52
30
31
931
10
93
55
5
20
15
15
250
3
33
101
22
1040
450
0
4
26
104
5555
110
100
50
20
1000
60
16
99
44
104
19
3
6
22
132
7
20
6
950
6
6000
30
200
693
808
624
31
14
1200
8
16
24
35
9
525
5
105
62
3200
78
All fifteen sentences shaded.
Solve — Page 2006. Solve
Q1a) 35 × 76
Answer
Split each number by its places and multiply piece by piece.
×
70
6
Total
30
2100
180
2280
5
350
30
380
35 × 76 = 2280 + 380 = 2,660
Q2b) 267 × 38
Answer
Split each number by its places and multiply piece by piece.
×
30
8
Total
200
6000
1600
7600
60
1800
480
2280
7
210
56
266
267 × 38 = 7600 + 2280 + 266 = 10,146
Q3c) 498 × 9
Answer
Split each number by its places and multiply piece by piece.
×
9
Total
400
3600
3600
90
810
810
8
72
72
498 × 9 = 3600 + 810 + 72 = 4,482
Q4d) 89 × 42
Answer
Split each number by its places and multiply piece by piece.
×
40
2
Total
80
3200
160
3360
9
360
18
378
89 × 42 = 3360 + 378 = 3,738
Q5e) 55 × 23
Answer
Split each number by its places and multiply piece by piece.
×
20
3
Total
50
1000
150
1150
5
100
15
115
55 × 23 = 1150 + 115 = 1,265
Q6f) 345 × 17
Answer
Split each number by its places and multiply piece by piece.
×
10
7
Total
300
3000
2100
5100
40
400
280
680
5
50
35
85
345 × 17 = 5100 + 680 + 85 = 5,865
Q7g) 66 × 22
Answer
Split each number by its places and multiply piece by piece.
×
20
2
Total
60
1200
120
1320
6
120
12
132
66 × 22 = 1320 + 132 = 1,452
Q8h) 704 × 11
Answer
Split each number by its places and multiply piece by piece.
×
10
1
Total
700
7000
700
7700
4
40
4
44
704 × 11 = 7700 + 44 = 7,744
Q9i) 319 × 26
Answer
Split each number by its places and multiply piece by piece.
18 × 40 = 720 774 − 720 = 54 18 × 3 = 54 54 − 54 = 0 Groups taken = 40 + 3 = 43 Left over = 0
774 ÷ 18 = 43, remainder 0
The remainder is 0, so 18 divides 774 exactly.
Q12l) 864 ÷ 26
Answer
Take away groups of 26 again and again.
26 × 30 = 780 864 − 780 = 84 26 × 3 = 78 84 − 78 = 6 Groups taken = 30 + 3 = 33 Left over = 6
864 ÷ 26 = 33, remainder 6
Check: 26 × 33 + 6 = 864 ✓
Q13m) 304 ÷ 12
Answer
Take away groups of 12 again and again.
12 × 20 = 240 304 − 240 = 64 12 × 5 = 60 64 − 60 = 4 Groups taken = 20 + 5 = 25 Left over = 4
304 ÷ 12 = 25, remainder 4
Check: 12 × 25 + 4 = 304 ✓
Q14n) 670 ÷ 9
Answer
Take away groups of 9 again and again.
9 × 70 = 630 670 − 630 = 40 9 × 4 = 36 40 − 36 = 4 Groups taken = 70 + 4 = 74 Left over = 4
670 ÷ 9 = 74, remainder 4
Check: 9 × 74 + 4 = 670 ✓
Q15o) 584 ÷ 25
Answer
Take away groups of 25 again and again.
25 × 20 = 500 584 − 500 = 84 25 × 3 = 75 84 − 75 = 9 Groups taken = 20 + 3 = 23 Left over = 9
584 ÷ 25 = 23, remainder 9
Check: 25 × 23 + 9 = 584 ✓
Q16p) 900 ÷ 15
Answer
Take away groups of 15 again and again.
15 × 60 = 900 900 − 900 = 0 Groups taken = 60 Left over = 0
900 ÷ 15 = 60, remainder 0
The remainder is 0, so 15 divides 900 exactly.
Q17q) 658 ÷ 32
Answer
Take away groups of 32 again and again.
32 × 20 = 640 658 − 640 = 18 Groups taken = 20 Left over = 18
658 ÷ 32 = 20, remainder 18
Check: 32 × 20 + 18 = 658 ✓
Q18r) 974 ÷ 9
Answer
Take away groups of 9 again and again.
9 × 100 = 900 974 − 900 = 74 9 × 8 = 72 74 − 72 = 2 Groups taken = 100 + 8 = 108 Left over = 2
974 ÷ 9 = 108, remainder 2
Check: 9 × 108 + 2 = 974 ✓
Chinnu’s Coins — Page 201Chinnu’s Coins
Q11. 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 has 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.The price board at the amusement park ticket counter.
Answer
Child
Note / coin
Working
How many
Bujji
₹200 notes
3 notes = ₹600, 4 notes = ₹800
4 notes
Munna
₹50 notes
750 ÷ 50 = 15
15 notes
Balu
₹20 notes
37 notes = ₹740, 38 notes = ₹760
38 notes
Chinnu
₹5 coins
750 ÷ 5 = 150
150 coins
Sansu
₹2 coins
750 ÷ 2 = 375
375 coins
Bujji: 750 ÷ 200 = 3, and ₹150 is still left. So 3 notes are not enough. He must bring 4 notes (₹800). Balu: 750 ÷ 20 = 37, and ₹10 is still left. So he must bring 38 notes (₹760).
Q2b) Which of these children will not receive any change from the cashier?
Answer
Munna, Chinnu and Sansu will not get any change.
Child
Money given
Ticket
Change
Bujji
₹800
₹750
₹50
Munna
₹750
₹750
nil
Balu
₹760
₹750
₹10
Chinnu
₹750
₹750
nil
Sansu
₹750
₹750
nil
Bujji’s change:
7
10
₹
8
0
0
−
₹
7
5
0
₹
5
0
Balu’s change:
₹
7
6
0
−
₹
7
5
0
₹
1
0
Q3c) How long would the cashier take to count Chinnu’s coins?
Answer
Chinnu brings 150 coins of ₹5.
Sample answer. Say the cashier counts 1 coin every second.
If the cashier counts 2 coins a second, it takes 75 seconds — about 1 minute 15 seconds.
Q42. Observe the following multiplications. The answers have been provided. (12 × 13 = 156, 11 × 14 = 154, 13 × 13 = 169, 11 × 12 = 132) In each case, do you see any pattern in the two numbers and their product? (Hint: Look at the coloured digits!)The four multiplications printed in the book, with the coloured digits.
Answer
Both numbers start with 1. Look only at the ones digits.
Problem
Ones digits
Their sum
Their product
Answer
12 × 13
2 and 3
5
6
156
11 × 14
1 and 4
5
4
154
13 × 13
3 and 3
6
9
169
11 × 12
1 and 2
3
2
132
The pattern Hundreds digit of the answer = 1 Tens digit = sum of the two ones digits Ones digit = product of the two ones digits
Example: 12 × 13 2 + 3 = 5 and 2 × 3 = 6 So the answer is 1 5 6 = 156
Q5For what other multiplication problems will this pattern hold? Find 5 such examples.The four multiplications printed in the book, with the coloured digits.
Answer
Five examples (both numbers between 11 and 19, with small ones digits):
Problem
Sum of ones digits
Product of ones digits
Answer
11 × 13
1 + 3 = 4
1 × 3 = 3
143
12 × 12
2 + 2 = 4
2 × 2 = 4
144
11 × 15
1 + 5 = 6
1 × 5 = 5
165
12 × 14
2 + 4 = 6
2 × 4 = 8
168
11 × 17
1 + 7 = 8
1 × 7 = 7
187
Where it breaks: 14 × 14 4 + 4 = 8 but 4 × 4 = 16, which is not one digit. The true answer is 196, not 1–8–16.
Chinnu’s Coins — Page 202Matching vehicles with the number of people
Q13. Assume each vehicle is travelling with full capacity. How many people can travel in each of these vehicles? Match them up.
Vehicles
Number of people
75 Cycles
400
52 Autos
75
103 Cars
4560
20 Minibus
156
30 Aeroplanes
864
15 Train sleeper coaches
412
The two columns printed on page 202. Beside the vehicles the book shows pictures seen from above: a cycle; an auto with a back seat for 3; a car with a driver and 4 more seats; a minibus with 20 passenger seats; an aeroplane with many rows of seats; and a sleeper coach marked S1–S72.
Answer
People in one vehicle: cycle 1; auto 3, car 4 and minibus 20 (not counting the driver).
Vehicles
People in one
Working
Matches
75 Cycles
1
75 × 1
75
52 Autos
3
52 × 3
156
103 Cars
4
103 × 4
412
20 Minibus
20
20 × 20
400
30 Aeroplanes
152
30 × 152
4560
15 Train sleeper coaches
—
the box left over
864
Aeroplanes: of 4560 and 864, only 4560 can be shared among 30 exactly (4560 ÷ 30 = 152 seats). Only 864 is left, so it goes with the 15 train sleeper coaches.