NCERT Solutions for Class 5 Maths Chapter 1 We the Travellers I – 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.
Chapter 1 of Math Mela takes you on a journey — bullock carts, cycles, trains, ships and spacecraft — and uses that journey to open the door to five-digit numbers. You will learn to read, write, compare and round numbers up to 99,999, and finish with two clever puzzles: Mira’s subtraction trick and the King’s 20 horses.
NCERT Solutions for Class 5 Maths Chapter 1 We the Travellers I 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.
| Chapter | Download | |
|---|---|---|
| Solutions PDF (English medium) | We the Travellers I | View PDF |
| Solutions PDF (Hindi medium) | हम हैं यात्री – 1 | View PDF |
NCERT Solutions for Class 5 Maths Chapter 1 We the Travellers I – Questions and Answers
All 96 questions of the chapter (textbook pages 1 – 16) 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.
In-text Questions — Page 1Opening page and Reading and writing large numbers
This is about your own trip, so the answer is your own. Write five things: when, where, how you travelled, how long it took and how far it was.
Nobody can guess this correctly — it has to be looked up. The count runs into lakhs and crores, so it is a very big number.
We group the objects into thousands. Then we write how many thousands there are, followed by what is left over. For example, 7 thousands and 209 more are written as 7,209.
Each new box is one thousand more than the box before it. The six empty boxes are:
We write it as 10,000.
It has five digits — a 1 followed by four zeros.
In-text Questions — Page 2The place value chart and the Indian place value system
The pattern is simple: each column is ten times the column on its right.
10 Tens = 1 Hundred = 100
10 Hundreds = 1 Thousand = 1,000
10 Thousands = 1 Ten Thousand = 10,000
So the chart now has five columns:
| TTh | Th | H | T | O |
|---|---|---|---|---|
| 1 | ||||
| 1 | 0 | |||
| 1 | 0 | 0 | ||
| 1 | 0 | 0 | 0 | |
| 1 | 0 | 0 | 0 | 0 |
Yes. We use only the ten digits 0–9. The place of a digit gives its value.
3 Hundreds = 300
8 Tens = 80
0 Ones = 0
1,000 + 300 + 80 + 0 = 1,380
1 Hundred = 100
2 Tens = 20
3 Ones = 3
9,000 + 100 + 20 + 3 = 9,123
In-text Questions — Pages 3–4Writing and naming numbers beyond 10,000 (token table)
The number is 10,033.
Step 1 — add up the tokens.
Three 10 tokens = 30
Three 1 tokens = 3
10,000 + 30 + 3 = 10,033
Step 2 — put each digit in its column.
| Number | TTh | Th | H | T | O |
|---|---|---|---|---|---|
| 10,033 | 1 | 0 | 0 | 3 | 3 |
Read the number from the left, one digit to one column.
| Number | TTh | Th | H | T | O |
|---|---|---|---|---|---|
| 10,458 | 1 | 0 | 4 | 5 | 8 |
The number is 11,214, read as eleven thousand two hundred fourteen.
Step 1 — add up the tokens.
+ 1,000
+ 100 + 100 = 200
+ 10
+ 1 + 1 + 1 + 1 = 4
Total = 11,214
Step 2 — say the name. Split at the comma: 11 | 214.
| Number | TTh | Th | H | T | O |
|---|---|---|---|---|---|
| 13,520 | 1 | 3 | 5 | 2 | 0 |
Check with the tokens:
Five 100 tokens = 500
Two 10 tokens = 20
13,000 + 500 + 20 = 13,520 ✓
| Number | TTh | Th | H | T | O |
|---|---|---|---|---|---|
| 20,000 | 2 | 0 | 0 | 0 | 0 |
| Number | TTh | Th | H | T | O |
|---|---|---|---|---|---|
| 45,867 | 4 | 5 | 8 | 6 | 7 |
Check with the tokens shown in the book:
Five 1,000 tokens = 5,000
Eight 100 tokens = 800
Six 10 tokens = 60
Seven 1 tokens = 7
40,000 + 5,000 + 800 + 60 + 7 = 45,867 ✓
Let Us Do — Page 5Question 1 — continuing number patterns
Step 1 — find the jump. Take one number away from the next one.
678 – 567 = 111
So the rule is: add 111 each time.
The full chain:
Step 1 — find the jump. Use two numbers that are next to each other.
So the rule is: add 1,050 each time.
Step 1 — find the jump.
7,301 – 6,401 = 900
Rule: add 900 each time.
Step 1 — find the jump.
Top row: 10,100 → 10,200 → 10,300 → 10,400 → 10,500 → 10,600 → 10,700
Bottom row (from the right box to the left box): 10,800 → 10,900 → 11,000
Step 1 — find the jump.
Top row: 10,105 → 10,125 → 10,145 → 10,165 → 10,185 → 10,205 → 10,225
Bottom row (from the right box to the left box): 10,245 → 10,265 → 10,285
Step 1 — find the jump.
Top row: 10,992 → 10,993 → 10,994 → 10,995 → 10,996 → 10,997 → 10,998
Bottom row (from the right box to the left box): 10,999 → 11,000 → 11,001
Step 1 — find the jump.
10,798 – 10,796 = 2 → rule: add 2 each time.
Top row: 10,794 → 10,796 → 10,798 → 10,800 → 10,802 → 10,804 → 10,806
Bottom row (from the right box to the left box): 10,808 → 10,810 → 10,812
This chain goes down, not up.
Step 1 — find the jump.
Top row: 73,005 → 72,004 → 71,003 → 70,002 → 69,001 → 68,000 → 66,999
Bottom row (from the right box to the left box): 65,998 → 64,997 → 63,996
Step 1 — find the jump.
Top row: 82,350 → 83,350 → 84,350 → 85,350 → 86,350 → 87,350 → 88,350
Bottom row (from the right box to the left box): 89,350 → 90,350 → 91,350
Let Us Do — Pages 6–7Questions 2 to 5 — number names, order, comparing and digit swap
| Number | Number Name |
|---|---|
| 8,045 | Eight thousand forty-five |
| 7,209 | |
| 10,599 | |
| Ten thousand seven hundred forty-three | |
| 20,869 | Twenty thousand eight hundred sixty-nine |
| 13,579 | |
| Ten thousand ten | |
| Fifty-six thousand four hundred ninety-one | |
| 45,045 | |
| 39,593 | |
| 50,005 | |
| 26,050 | |
| 81,200 | |
| Ninety thousand nine | |
| Twenty-three thousand two hundred thirty | |
| Thirty-six thousand one |
Here is the whole table filled in. The orange cells are the ones you had to write.
| Number | Number Name |
|---|---|
| 8,045 | Eight thousand forty-five |
| 7,209 | Seven thousand two hundred nine |
| 10,599 | Ten thousand five hundred ninety-nine |
| 10,743 | Ten thousand seven hundred forty-three |
| 20,869 | Twenty thousand eight hundred sixty-nine |
| 13,579 | Thirteen thousand five hundred seventy-nine |
| 10,010 | Ten thousand ten |
| 56,491 | Fifty-six thousand four hundred ninety-one |
| 45,045 | Forty-five thousand forty-five |
| 39,593 | Thirty-nine thousand five hundred ninety-three |
| 50,005 | Fifty thousand five |
| 26,050 | Twenty-six thousand fifty |
| 81,200 | Eighty-one thousand two hundred |
| 90,009 | Ninety thousand nine |
| 23,230 | Twenty-three thousand two hundred thirty |
| 36,001 | Thirty-six thousand one |
Increasing order means smallest first.
No, the student is not correct. 49,014 is the bigger number.
Step 1 — count the digits.
49,014 has 5 digits
A 5-digit number is always bigger than a 4-digit number.
Step 2 — line them up in the place value chart.
| TTh | Th | H | T | O | |
|---|---|---|---|---|---|
| 9,990 | 0 | 9 | 9 | 9 | 0 |
| 49,014 | 4 | 9 | 0 | 1 | 4 |
In the TTh column, 9,990 has nothing (0) and 49,014 has 4. So 49,014 wins straight away.
So each box goes up by 5,000.
Where the two numbers sit:
- 9,990 sits just before the 10,000 mark — only 10 short of it.
- 49,014 sits just after the 45,000 mark, close to 50,000.
Two swaps work. Either one is a correct answer.
Swap the 1 and the 8 : 1,478 → 8,471
The digits of 10,593, in order, are 1 · 0 · 5 · 9 · 3.
(i) Between 11,000 and 15,000 → 13,590
(ii) More than 35,000 → 50,193 (or 90,513)
The digits of 48,247, in order, are 4 · 8 · 2 · 4 · 7.
(i) As small as possible → 28,447
(ii) As big as possible → 84,247
Nearest Tens (10s), Hundreds (100s), and Thousands (1,000s) — Pages 7–8Nearest Tens (10s), Hundreds (100s), and Thousands (1,000s)
The rabbit should go to 2,350.
Step 1 — name the two neighbouring tens. They are 2,340 and 2,350.
Step 2 — measure the two distances.
2,350 – 2,346 = 4 steps
Step 3 — pick the shorter one. 4 is less than 6, so the rabbit goes forward to 2,350, needing 4 jumps.
2,300 is the nearest hundred of 2,346. It will need 46 jumps to reach 2,300.
Step 1 — name the two neighbouring hundreds. 2,300 and 2,400.
Step 2 — measure both distances.
2,400 – 2,346 = 54
Step 3 — pick the shorter one. 46 < 54, so the rabbit goes back to 2,300.
2,000 is the nearest thousand of 2,346. It will need 346 jumps to reach 2,000.
Step 1 — name the two neighbouring thousands. 2,000 and 3,000.
Step 2 — measure both distances.
3,000 – 2,346 = 654
Step 3 — pick the shorter one. 346 < 654, so the rabbit goes back to 2,000.
| Number | Nearest Tens | Nearest Hundreds | Nearest Thousands |
|---|---|---|---|
| 3,176 | 3,180 | 3,200 | 3,000 |
| 4,017 | 4,020 | 4,000 | 4,000 |
| 5,789 | 5,790 | 5,800 | 6,000 |
| 8,203 | 8,200 | 8,200 | 8,000 |
Let Us Think — Pages 8–9Rounding off, and numbers that share the same neighbour
Circle 7,030 and 6,999.
| Number | Vijay: nearest hundred | Suma: nearest thousand | Same? |
|---|---|---|---|
| 7,126 | 7,100 | 7,000 | No |
| 7,835 | 7,800 | 8,000 | No |
| 7,030 | 7,000 | 7,000 | Yes ✓ |
| 6,999 | 7,000 | 7,000 | Yes ✓ |
Sample answer: 71 and 73. Both have nearest ten 70.
73 → ones digit 3, which is less than 5 → nearest ten is 70
Sample answer: 512 and 549. Both have nearest hundred 500.
549 → last two digits 49, less than 50 → nearest hundred is 500
Sample answer: 4,120 and 4,480. Both have nearest thousand 4,000.
4,480 → last three digits 480, less than 500 → nearest thousand is 4,000
Sample answer: 302.
Nearest hundred of 302 → last two digits 02, less than 50 → 300
Both are 300, so they match ✓
Sample answer: 4,020.
Nearest thousand of 4,020 → last three digits 020, less than 500 → 4,000
Both are 4,000 ✓
Sample answer: 5,002. All three roundings give 5,000.
Nearest hundred → last two digits 02, less than 50 → 5,000
Nearest thousand → last three digits 002, less than 500 → 5,000
Let Us Do — Page 10Travelling, Now and Then — speed and how many vehicles
She will cover 60 km.
Distance in 4 hours = 4 × distance in 1 hour
Step 1 — find the total number of children.
461 + 439 = 900 children
Step 2 — one bicycle carries 2 people. Share 900 into groups of 2.
Answer: 450 bicycles are needed.
Total children = 900. One autorickshaw carries 3.
Answer: 300 autorickshaws.
Total children = 900. One car carries 4.
4 × 200 = 800, leaving 100
4 × 25 = 100, leaving 0
200 + 25 = 225
Answer: 225 cars.
Total children = 900. One big car carries 6.
6 × 100 = 600, leaving 300
6 × 50 = 300, leaving 0
100 + 50 = 150
Answer: 150 big cars.
Total children = 900. One tempo traveller carries 10.
Answer: 90 tempo travellers.
Total children = 900. One boat carries 20.
20 × 40 = 800, leaving 100
20 × 5 = 100, leaving 0
40 + 5 = 45
Answer: 45 boats.
Total children = 900. One minibus carries 25.
25 × 20 = 500, leaving 400
25 × 16 = 400, leaving 0
20 + 16 = 36
Answer: 36 minibuses.
Total children = 900. One aeroplane carries 180.
180 × 5 = 900, leaving 0
So the answer is 5
Answer: 5 aeroplanes.
Finding Large Numbers Around Us — Pages 10–11Finding Large Numbers Around Us
A 4-digit number is anything from 1,000 to 9,999. Do not count one by one — estimate, which means make a careful guess using multiplication.
One page has about 150 words. The chapter has 16 pages.
150 × 16 = 2,400 words — a 4-digit number.
This is about your own school, so go and count.
Count the children in your own classroom.
Count all the books in the room — textbooks, notebooks and library books.
456 + 44 books in the cupboard = 500 books
Take a full-grown mango tree in the village.
| What we count | Rough count | How many digits |
|---|---|---|
| Mangoes on the tree in one season | about 1,500 | 4-digit |
| Leaves on the tree | about 25,000 | 5-digit |
| Seeds dropped in a year | about 1,500 | 4-digit |
| Days the tree has been alive (about 15 years) | about 5,475 | 4-digit |
Choose your own village or town and fill in real figures where you can. Here is a filled-in example.
| What we count | Rough count | How many digits |
|---|---|---|
| People living in the village | about 4,800 | 4-digit |
| Houses in the village | about 1,100 | 4-digit |
| People living in a small town | about 45,000 | 5-digit |
| Two-wheelers in the town | about 12,000 | 5-digit |
| Trees along the town roads | about 3,000 | 4-digit |
| Students in all the schools of the town | about 10,500 | 5-digit |
Pastime Mathematics — Pages 11–13Pastime Mathematics — Sanju and Mira’s puzzles on the train
It can be done in 7 trips. The trick is that the boatman is allowed to bring something back.
| Trip | What the boatman does | Left behind on this side | Waiting on the far side |
|---|---|---|---|
| 1 | Takes the sheep across | lion, grass | sheep |
| 2 | Comes back alone | lion, grass | sheep |
| 3 | Takes the grass across | lion | sheep, grass |
| 4 | Brings the sheep back | lion, sheep | grass |
| 5 | Takes the lion across | sheep | lion, grass |
| 6 | Comes back alone | sheep | lion, grass |
| 7 | Takes the sheep across | — | lion, sheep, grass |
Check every stage. The lion and the sheep are never left alone together. The sheep and the grass are never left alone together. So nothing gets eaten.
Play second, and always make the two piles equal again. Do that every turn and you will win.
Whatever they take from one pile, take exactly the same number from the other pile.
The piles are equal after every one of your turns.
Mira knew because every difference you can ever get is a number in the 9 times table, and the chain of 9s always ends at 9 itself.
Here: 7 – 3 = 4, and 9 × 4 = 36 ✓
The differences in the book’s example are 36, 27, 45, 9. Two things are true of every one of them.
- Each one is in the 9 times table.
- The digits of each one add up to 9.
| Difference | Is it 9 × something? | Digits added |
|---|---|---|
| 36 | 9 × 4 | 3 + 6 = 9 |
| 27 | 9 × 3 | 2 + 7 = 9 |
| 45 | 9 × 5 | 4 + 5 = 9 |
| 9 | 9 × 1 | 9 |
Take the digits 2 and 9.
Step 2 : 63 – 36 = 27
Step 3 : 72 – 27 = 45
Step 4 : 54 – 45 = 9
What is common: every difference is a multiple of 9 — 63, 27, 45, 9. And you always end at 9, however many steps it takes.
Choose two digits that are next-door neighbours — digits that differ by 1.
2 and 3 → 32 – 23 = 9
4 and 5 → 54 – 45 = 9
6 and 7 → 76 – 67 = 9
8 and 9 → 98 – 89 = 9
The pattern: the two digits are consecutive — one is exactly 1 more than the other.
Choose two digits that are 3 apart.
27 = 9 × 3, so the digits must differ by 3.
All the pairs that work:
| Digits | The two numbers | Subtraction |
|---|---|---|
| 1, 4 | 41 and 14 | 41 – 14 = 27 |
| 2, 5 | 52 and 25 | 52 – 25 = 27 |
| 3, 6 | 63 and 36 | 63 – 36 = 27 |
| 4, 7 | 74 and 47 | 74 – 47 = 27 |
| 5, 8 | 85 and 58 | 85 – 58 = 27 |
| 6, 9 | 96 and 69 | 96 – 69 = 27 |
| Digits | Differences in digits | Difference in numbers formed by the digits |
|---|---|---|
| 3, 7 | 7 – 3 = 4 | 73 – 37 = 36 |
| 1, 9 | 9 – 1 = 8 | 91 – 19 = 72 |
| 2, 8 | 8 – 2 = 6 | 82 – 28 = 54 |
| 4, 5 | 5 – 4 = 1 | 54 – 45 = 9 |
Yes. The relationship is:
Check it on every row of Mira’s table.
| Digits | Difference in digits | 9 × that | Difference in numbers |
|---|---|---|---|
| 3, 7 | 4 | 9 × 4 = 36 | 73 – 37 = 36 ✓ |
| 1, 9 | 8 | 9 × 8 = 72 | 91 – 19 = 72 ✓ |
| 2, 8 | 6 | 9 × 6 = 54 | 82 – 28 = 54 ✓ |
| 4, 5 | 1 | 9 × 1 = 9 | 54 – 45 = 9 ✓ |
| Digits | Differences in digits | Difference in numbers formed by the digits |
|---|---|---|
| 3, 7 | 7 – 3 = 4 | 73 – 37 = 36 |
| 1, 9 | 9 – 1 = 8 | 91 – 19 = 72 |
| 2, 8 | 8 – 2 = 6 | 82 – 28 = 54 |
| 4, 5 | 5 – 4 = 1 | 54 – 45 = 9 |
Mira’s table already covers digit differences 4, 8, 6 and 1. The four missing ones are 2, 3, 5 and 7. Here they are.
| Digits | Difference in digits | Difference in the numbers formed |
|---|---|---|
| 3, 5 | 5 – 3 = 2 | 53 – 35 = 18 |
| 1, 4 | 4 – 1 = 3 | 41 – 14 = 27 |
| 2, 7 | 7 – 2 = 5 | 72 – 27 = 45 |
| 1, 8 | 8 – 1 = 7 | 81 – 18 = 63 |
The difference between the two digits tells you which multiple of 9 you will get. Multiply it by 9.
| Difference in digits | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Difference in numbers | 9 | 18 | 27 | 36 | 45 | 54 | 63 | 72 |
So the digit difference is like a code number. Once you know it, you know the answer without subtracting at all.
You reach 9 on the third subtraction when your two digits are 3 apart or 8 apart.
The numbers are: 41 and 14 · 52 and 25 · 63 and 36 · 74 and 47 · 85 and 58 · 96 and 69.
The numbers are: 91 and 19.
The longest possible chain takes 5 subtractions. It happens when the two digits are exactly 2 apart.
The pairs: (1, 3) · (2, 4) · (3, 5) · (4, 6) · (5, 7) · (6, 8) · (7, 9).
Let Us Do — Pages 13–15Questions 1 to 9 — writing numbers in between, ordering, expanded form and counting notes
Sample answer: 23,600 · 23,750 · 23,900 · 24,000 · 24,150
Sample answer: 38,200 · 38,300 · 38,400 · 38,500 · 38,550
Sheetal’s car has been driven more.
Step 1 — count the digits. Both numbers have 5 digits, so we must compare place by place.
Step 2 — start from the left.
| TTh | Th | H | T | O | |
|---|---|---|---|---|---|
| Ravi 56,987 | 5 | 6 | 9 | 8 | 7 |
| Sheetal 67,543 | 6 | 7 | 5 | 4 | 3 |
So 67,543 > 56,987 → Sheetal’s car
Ascending order means cheapest first.
| Town | Population |
|---|---|
| Town 1 | 65,232 |
| Town 2 | 53,231 |
| Town 3 | 56,380 |
| Town 4 | 51,336 |
| Town 5 | 45,858 |
| Town 6 | 66,540 |
Descending order means biggest first.
| Place | Town | Population |
|---|---|---|
| 1st (largest) | Town 6 | 66,540 |
| 2nd | Town 1 | 65,232 |
| 3rd | Town 3 | 56,380 |
| 4th | Town 2 | 53,231 |
| 5th | Town 4 | 51,336 |
| 6th (smallest) | Town 5 | 45,858 |
There are 11 such numbers:
8,062 = 8,000 + 60 + 2
| Th | H | T | O |
|---|---|---|---|
| 8 | 0 | 6 | 2 |
0 in the hundreds place = 0
6 in the tens place = 60
2 in the ones place = 2
9,980 = 9,000 + 900 + 80
9 hundreds = 900
8 tens = 80
0 ones = 0
9,000 + 900 + 80 = 9,980 ✓
10,304 = 10,000 + 300 + 4
| TTh | Th | H | T | O |
|---|---|---|---|---|
| 1 | 0 | 3 | 0 | 4 |
0 thousands = 0
3 hundreds = 300
0 tens = 0
4 ones = 4
10,000 + 300 + 4 = 10,304 ✓
23,004 = 20,000 + 3,000 + 4
3 thousands = 3,000
0 hundreds = 0
0 tens = 0
4 ones = 4
20,000 + 3,000 + 4 = 23,004 ✓
70,405 = 70,000 + 400 + 5
| TTh | Th | H | T | O |
|---|---|---|---|---|
| 7 | 0 | 4 | 0 | 5 |
4 hundreds = 400
5 ones = 5
70,000 + 400 + 5 = 70,405 ✓
68 = 5 Tens + 18 Ones
Check: 5 Tens + 18 Ones = 50 + 18 = 68 ✓
607 = 4 Hundreds + 207 Ones
Check: 400 + 207 = 607 ✓
5,621 = 4 Thousand + 16 Hundreds + 2 Tens + 1 One
2 Tens = 20
5,621 – 4,000 – 20 = 1,601
1,600 = 16 Hundreds
1 = 1 One
Check: 4,000 + 1,600 + 20 + 1 = 5,621 ✓
7,069 = 5 Thousand + 20 Hundreds + 69 Ones
5,000 = 5 Thousand
69 = 69 Ones
Check: 5,000 + 2,000 + 69 = 7,069 ✓
37,608 = 2 Ten Thousand + 17 Thousand + 6 Hundreds + 8 Ones
8 Ones = 8
20,000 = 2 Ten Thousand
600 = 6 Hundreds
Check: 20,000 + 17,000 + 600 + 8 = 37,608 ✓
43,001 = 3 Ten Thousand + 13 Thousand + 0 Hundreds + 1 One
1 One = 1
Thousands blank = 13, Hundreds blank = 0
Check: 30,000 + 13,000 + 0 + 1 = 43,001 ✓
79 notes of ₹100.
7,900 ÷ 100 = 79, and ₹34 is left over
So there are 79 notes of ₹100, with ₹34 left over.
7 thousands.
7,000 ÷ 1,000 = 7, and 934 is left over
So there are 7 thousands, with 934 left over.
15 notes of ₹500 (with ₹434 left over).
Use the hint. Part (c) told us that ₹7,934 holds 7 thousands.
7 thousands → 7 × 2 = 14 notes of ₹500 (that is ₹7,000)
934 is more than 500, so one more ₹500 note fits
14 + 1 = 15 notes
6,534 notes of ₹10 (with ₹2 left over).
65,340 ÷ 10 = 6,534, and ₹2 is left over
The quick way: Cover the last digit of 65,342. What remains is 6,534.
653 notes of ₹100 (with ₹42 left over).
65,300 ÷ 100 = 653, and ₹42 is left over
The quick way: Cover the last two digits of 65,342. What remains is 653.
65 thousands (with 342 left over).
65,000 ÷ 1,000 = 65, and 342 is left over
The quick way: Cover the last three digits of 65,342. What remains is 65.
130 notes of ₹500 (with ₹342 left over).
Use part (g). ₹65,342 holds 65 thousands.
342 is less than 500, so no extra note fits
King’s Horses — Pages 15–16King’s Horses
No. There were only 19 horses. One had already been stolen.
Count the rooms one by one:
Left wall (middle rooms) : 1 + 3 = 4
Right wall (middle rooms) : 2 + 3 = 5
Bottom wall : 0 + 2 + 3 + 0 = 5
5 + 4 + 5 + 5 = 19 horses
The caretaker counted the corner rooms twice.
But the corner horses got counted twice.
The corner rooms hold 1 + 0 + 0 + 0 = 1 horse.
20 – 1 = 19 horses ✓
He moved more horses into the corner rooms. The more horses sit in corners, the more often they get counted twice, and the fewer horses he needs in total.
Here is one arrangement of 18 horses with 5 on every side.
8 more horses. The caretaker can keep up the trick until only 10 horses are left.
Here is that arrangement.
Fewest possible = 10
18 – 10 = 8 more horses can be stolen
Chapter 1 of 15 – Maths
- Next: Chapter 2 Fractions
- All chapters: NCERT Solutions Class 5 Maths
Also for This Chapter
NCERT Solutions Class 5 Maths Chapter 1 We the Travellers I – Overview
| Class | Class 5 |
| Subject | Maths |
| Chapter | 1 – We the Travellers I |
| Textbook pages | 1 – 16 |
| Questions solved | 96 |
| Session | 2027 |
| Format | Full answers on this page + PDF |
NCERT Solutions Class 5 Maths Chapter 1 – FAQs
Where can I download NCERT Solutions Class 5 Maths Chapter 1 We the Travellers I PDF?
Read every answer on this page, or download the free PDF from the table at the top.
Is this for the 2027 session?
Yes, it follows the latest NCERT Class 5 Maths book for 2027.
How many questions are solved in NCERT Class 5 Maths Chapter 1?
All 96 questions of the chapter (textbook pages 1 – 16) – the in-text questions, activities and the end-of-chapter exercises.
How should I use these solutions?
Read the chapter in the NCERT book first, attempt the questions yourself, then check your answers here.