NCERT Solutions Class 5 Maths Chapter 4 We the Travellers Ii

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 5 · M AT H S

NCERT Solutions

Chapter 4: We the Travellers—II

NCERT Textbook — Math Mela

BOOK PAGES SECTIONS QUESTIONS MEDIUM

42 – 56 18 63 English

Solutions, notes, sample papers & more at 60 pages

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

CLASS 5 · MATHS · MATH MELA

NCERT Solutions — Chapter 4: We the Travellers—II
Chapter 4 of Math Mela takes the journey of Chapter 1 further. This time we travel with big numbers — litres of
fuel in a tank, kilometres on the North–South Corridor, rupees spent on a train trip — and learn to add and
subtract numbers up to five digits, to check an answer the other way round, and to find quick, clever
shortcuts instead of always writing columns.

TEXTBOOK BOOK PAGES

Math Mela (Class 5) 42 – 56

SECTIONS QUESTIONS

18 63

MEDIUM

English

Making Sums Equal — Page 42
Making Sums Equal

TRY THIS

Q1 Think what will happen to the sums if we interchange 2 and 5? Try interchanging
other pairs of numbers and find the one that will make the sums equal.

If we swap 2 and 5, the two sums move further apart, not closer.

Left group now: 1 + 5 + 7 + 9 = 22

Right group now: 3 + 4 + 2 + 9 = 18

Before the swap the sums were 19 and 21. Now they are 22 and 18.

Why it happens: When you move a small number out and a bigger number in, the
left group gains the difference between them. 5 − 2 = 3, so the left sum went up by 3
(19 → 22) and the right sum came down by 3 (21 → 18). The gap grew from 2 to 4.

So we must swap the other way round — send a number out and bring in one that is only a
little bigger.
Here is how to find the right pair in 3 steps.

1. Add both sums: 19 + 21 = 40. Share it equally: 40 ÷ 2 = 20. Each group must reach 20.

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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

2. The left group must gain 20 − 19 = 1. So the number coming in must be exactly 1 more than
the number going out.
3. Look for such a pair: 2 is in the left group and 3 is in the right group, and 3 − 2 = 1.

Interchange 2 and 3.

Left: 1 + 3 + 7 + 9 = 20

Right: 2 + 4 + 5 + 9 = 20

Swap the 2 and the 3

1 3
2 4
7 5
9 9

20 20

One move is enough: 2 goes right, 3 comes left, and both sums become 20.

Tip: Always halve the total first. That tells you the target sum, and then you know
exactly how much the smaller group has to gain.

Q2 (a) Interchange pairs of numbers between the two groups to make their sums
equal, using the least number of moves. Left group: 1, 2, 7, 9 (sum 19). Right group:
3, 4, 5, 9 (sum 21).

Interchange 2 and 3. One move is enough.

1. Total of both groups: 19 + 21 = 40.
2. Each group must have 40 ÷ 2 = 20.
3. The left group needs 20 − 19 = 1 more, so bring in a number that is 1 bigger than the one
you send out.
4. 2 goes out, 3 comes in.

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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Left: 1 + 3 + 7 + 9 = 20

Right: 2 + 4 + 5 + 9 = 20

Check: 20 + 20 = 40, the same total we started with ✓

Q3 (b) Interchange pairs of numbers between the two groups to make their sums
equal, using the least number of moves. Left group: 5, 7, 12, 15 (sum 39). Right
group: 9, 11, 13, 14 (sum 47).

Interchange 5 and 9. One move is enough.

1. Total: 39 + 47 = 86.
2. Each group must have 86 ÷ 2 = 43.
3. The left group needs 43 − 39 = 4 more. So the number coming in must be 4 bigger than the
one going out.
4. 5 goes out and 9 comes in, because 9 − 5 = 4.

Left: 9 + 7 + 12 + 15 = 43

Right: 5 + 11 + 13 + 14 = 43

Another correct answer: Swapping 7 and 11 also works, because 11 − 7 = 4 as well.
Left: 5 + 11 + 12 + 15 = 43. Right: 9 + 7 + 13 + 14 = 43.

Q4 (c) Interchange pairs of numbers between the two groups to make their sums
equal, using the least number of moves. Left group: 11, 15, 19, 23 (sum 68). Right
group: 13, 17, 21, 25 (sum 76).

Here one move is not possible. You need two moves: interchange 11 and 13, and interchange
15 and 17.

1. Total: 68 + 76 = 144. Each group must have 144 ÷ 2 = 72.
2. The left group needs 72 − 68 = 4 more.
3. Try one swap. Look for a right-hand number that is exactly 4 bigger than a left-hand number:
11 needs 15, 15 needs 19, 19 needs 23, 23 needs 27. None of these is in the right group. So
one move cannot work.

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Class 5 Maths Chapter 4 We the Travellers—II
a g l AglaSem · NCERT Solutions

4. Now use two swaps. Send out 11 and 15 (that is 26) and bring in 13 and 17 (that is 30). The
co m
left group gains 30 − 26 = 4.
e m.
m l as
m .co a g
l a se
Left: 13 + 17 + 19 + 23 = 72
g
aRight: 11 + 15 + 21 + 25 = 72

co m
ag
Check: 72 + 72 = 144 ✓
m .
as e
a g l
Why one move can never work here: Each left number leaves a remainder of 3
when divided by 4 (11, 15, 19, 23), and each right number leaves a remainder of 1
co m
m.
(13, 17, 21, 25). So the difference between any right number and any left number is

m as e
.co l
2, 6, 10 … — always 2 more than a multiple of 4. It can never be exactly 4.
a g
a s em
a gl Other correct answers: Swapping 11 with 13 and 19 with 21 also gives 72 and 72.
Any two swaps whose numbers gain the left group exactly 4 will do.
m a s
m .co agl
l a se
Q5 a g
(d) Interchange pairs of numbers between the two groups to make their sums
equal, using the least number of moves. Left group: 77, 78, 79, 80 (sum 314). Right

co m
.
group: 81, 82, 83, 84 (sum 330).

e m
m l as
.co a g
em two moves are needed: interchange 77 and 81, and interchange 78 and 82.

a s
gl
Again
a
m
1. Total: 314 + 330 = 644. Each group must have 644 ÷ 2 = 322.

a se
com l
2. The left group needs 322 − 314 = 8 more.
. a g
e m
3. For one swap the incoming number would have to be 8 bigger than the outgoing one. But

g l
the biggest possible gap is 84as− 77 = 7. So one move is impossible.
4. Send out 77 and 78 (thatais 155) and bring in 81 and 82 (that is 163). The gain is 163 − 155 =

m
8.

. co
em
m l as
.co g
Left: 81 + 82 + 79 + 80 = 322
m Right: 77 + 78 + 83 + 84 = 322 a
l a se
ag Check: 322 + 322 = 644 ✓
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
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.co


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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Other correct answers: 79 ↔ 83 with 80 ↔ 84 also works, and so does 77 ↔ 81 with
78 ↔ 82 in any order. Each pair of swaps must together add exactly 8 to the left
group.

Fuel Arithmetic — Page 43
Fuel Arithmetic

Q1 A lorry has 28 litres of fuel in its tank. An additional 75 litres is filled. What is the
total quantity of fuel in the lorry?

The lorry now has 103 litres of fuel.
The tank already had 28 l. Another 75 l was poured in. Fuel is being put together, so we add.

Total = 28 l + 75 l

Add column by column, starting from the ones.

1. Ones: 8 + 5 = 13. That is 1 ten and 3 ones. Write 3 in the ones place and carry the 1 ten.
2. Tens: 2 + 7 = 9, and the carried 1 makes 10. That is 1 hundred and 0 tens. Write 0 in the tens
place and carry the 1 hundred.
3. Hundreds: only the carried 1. Write 1.

H T O

1 (carried) 1 (carried)

2 8

+ 7 5

1 0 3

Total fuel = 103 litres

Check it yourself: 103 − 75 = 28, which is what the tank had at the start. So the
answer is right.

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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Q2 Do you remember how to add two numbers using place value of numbers?

Yes. Give every digit its own column, then add one column at a time from the right.

1. Write the numbers one under the other. Ones must sit under ones, tens under tens,
hundreds under hundreds.
2. Start with the ones column, on the far right.
3. If a column adds up to 9 or less, just write the answer below the line.
4. If a column adds up to 10 or more, keep only the ones digit there and regroup — carry the
ten to the next column on the left.
5. Move left, one column at a time, until every column is done.

Why we regroup: A place can only hold the digits 0 to 9. The moment ten ones
collect together, they become one ten, and a ten has to live in the tens column. That
is all a carry is — ten small things swapped for one bigger thing.

28 + 75

Ones: 8 + 5 = 13 → 1 ten and 3 ones

Tens: 2 + 7 + 1 = 10 → 1 hundred and 0 tens

Hundreds: 1

Answer: 103

Q3 Find the sum of 49 and 89.

138

1. Ones: 9 + 9 = 18. Write 8 in the ones place and carry 1 ten.
2. Tens: 4 + 8 = 12, and the carried 1 makes 13. Write 3 in the tens place and carry 1 hundred.
3. Hundreds: only the carried 1. Write 1.

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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

H T O

1 (carried) 1 (carried)

4 9

+ 8 9

1 3 8

49 + 89 = 138

Check it yourself: 138 − 89 = 49 ✓. An easier route: 49 + 89 is the same as 49 + 90 −
1 = 139 − 1 = 138.

Let Us Solve — Page 43
Fuel Arithmetic

LET US SOLVE

Q1 Add the following numbers. Wherever possible, find easier ways to add the pairs of
numbers. (a) 15 + 79 (b) 46 + 99 (c) 38 + 35 (d) 5 + 89 (e) 76 + 28 (f) 69 + 20

All six answers first, then the easy way for each.

SUM EASY WAY ANSWER

(a) 15 + 79 79 + 1 = 80, then 80 + 14 94

(b) 46 + 99 46 + 100 = 146, then take 1 away 145

(c) 38 + 35 38 + 2 = 40, then 40 + 33 73

(d) 5 + 89 89 + 1 = 90, then 90 + 4 94

(e) 76 + 28 76 + 4 = 80, then 80 + 24 104

(f) 69 + 20 only tens change: 6 tens + 2 tens = 8 tens 89

(a) 15 + 79 = 94

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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

1. 79 is close to 80. Give it 1 from the 15: 79 + 1 = 80.
2. The 15 is now 14.
3. 80 + 14 = 94.

(b) 46 + 99 = 145

1. 99 is 1 less than 100. Add 100 instead: 46 + 100 = 146.
2. You added 1 too much, so take 1 back: 146 − 1 = 145.

(c) 38 + 35 = 73

1. 38 needs 2 to reach 40. Take 2 from the 35: 38 + 2 = 40.
2. The 35 is now 33.
3. 40 + 33 = 73.

(d) 5 + 89 = 94

1. 89 needs 1 to reach 90. Take 1 from the 5: 89 + 1 = 90.
2. The 5 is now 4.
3. 90 + 4 = 94.

(e) 76 + 28 = 104

1. 76 needs 4 to reach 80. Take 4 from the 28: 76 + 4 = 80.
2. The 28 is now 24.
3. 80 + 24 = 104.

(f) 69 + 20 = 89

1. 20 is 2 whole tens, so only the tens digit changes.
2. 6 tens + 2 tens = 8 tens; the 9 ones stay as they are.
3. So the answer is 89.

Why these shortcuts are safe: You are only moving a few ones from one number to
the other. Nothing is lost and nothing is added, so the total stays exactly the same —
but one of the numbers becomes a friendly 10, 20, 40 or 80, and those are easy to
add in your head.

Check it yourself: Do (e) by columns as well. Ones: 6 + 8 = 14, write 4 carry 1. Tens: 7
+ 2 + 1 = 10, write 0 carry 1. Hundreds: 1. That gives 104 — the same answer.

Relationship Between Addition and Subtraction — Page 44

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Class 5 Maths Chapter 4 We the Travellers—II
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Relationship Between Addition and Subtraction
co m
em.
m l as
.co a
Find the relationship between the numbers in the given statements and fill in the
g
em appropriately. (a) If 46 + 21 = 67, then 67 – 21 = ___ and 67 – 46 = ___. (b) If 198 –
Q1

a s
blanks

a gl 98 = 100, then 100 + ___ = 198 and 198 – ___ = 98. (c) If 189 + 98 = 287, then 287 – 98 = ___
and 287 – 189 = ___. (d) If 872 – 672 = 200, then 200 + ___ = 872 and 872 – ___ = 672.

com
e m . ag
as

a g l
Every one of these blanks can be filled without any working. The three numbers in each family
are the same three numbers, only rearranged.

co m
GIVEN FACT FIRST BLANK SECOND BLANK
e m.
c=o67m as
67 − 46g=l21
. a
em
(a) 46 + 21 67 − 21 = 46

a s
agl(b) 198 − 98 = 100 100 + 98 = 198 198 − 100 = 98

a s
com agl
(c) 189 + 98 = 287 287 − 98 = 189 287 − 189 = 98

m .
gl ase
(d) 872 − 672 = 200 200 + 672 = 872 872 − 200 = 672

a
Take (a) as the picture for all four.

co m
.
1. 46 and 21 are the two parts. 67 is the whole.
2. Take one part away from the whole and the other part is left: 67 − 21e=m
m a s 46.

. cothe other part away and the first part is left: 67 − 46 = 21.agl
sem
3. Take

a
agl Why it happens: Adding puts two parts together to make a whole. Subtracting takes
one part out of the whole. They undo each other, exactly like walking 5 steps forward
se m
o m g l a
m .chides an addition fact.
and then 5 steps back. So one addition fact always hides two subtraction facts inside
a
a se
it, and one subtraction fact always
l
ag
m
Check it yourself: For (d), 200 + 672 = 872 ✓ and 872 − 200 = 672 ✓. Both blanks use
. co
m
only the numbers already printed in the question — you never need a new number.

m as e
.co a g l
se m
g l a
a In each of the following, write the subtraction and addition sentences that follow
c
.
Q2
from the given sentence. (a) If 78 + 164 = 242, then ___ . (b) If 462 + 839 = 1301, then
s e m
m a
. co agl
___ . (c) If 921 – 137 = 784, then ___ . (d) If 824 – 234 = 590, then ___ .

e m
g l as
a
Each box has two blank lines, so write two sentences in each.

com
m .
m ase
.co


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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

GIVEN FIRST SENTENCE SECOND SENTENCE

(a) 78 + 164 = 242 242 − 164 = 78 242 − 78 = 164

(b) 462 + 839 = 1301 1301 − 839 = 462 1301 − 462 = 839

(c) 921 − 137 = 784 784 + 137 = 921 921 − 784 = 137

(d) 824 − 234 = 590 590 + 234 = 824 824 − 590 = 234

How to build the sentences in 3 steps.

1. Find the biggest of the three numbers. That is the whole. In (a) it is 242.
2. The other two are the parts: 78 and 164.
3. Now write: whole − part = the other part. Do it once for each part.

If you are given a subtraction sentence, as in (c) and (d), the answer of the subtraction is itself a
part.

1. In (c) the whole is 921. The parts are 137 and 784.
2. Put the two parts together: 784 + 137 = 921.
3. Take the other part away from the whole: 921 − 784 = 137.

Check it yourself: 462 + 839 = 1301 ✓, so 1301 − 462 must give 839 — and it does.
Whenever your three numbers are the same three, you have not made a mistake.

More Fuel Arithmetic — Page 44
More Fuel Arithmetic

Q1 A minibus has 18 l of fuel left. After refuelling, the fuel meter indicates 65 l. How
much fuel has been filled in the fuel tank of the minibus?

47 litres of fuel was filled.
The tank had 18 l. After filling, the meter shows 65 l. We want the part that was added, so we
take the old amount away from the new amount.

Fuel filled = 65 l − 18 l

1. Ones: 5 is smaller than 8, so we cannot subtract yet. Regroup — take 1 ten from the 6 tens
and break it into 10 ones.

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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

2. The tens are now 5 and the ones are 5 + 10 = 15.
3. Ones: 15 − 8 = 7.
4. Tens: 5 − 1 = 4.

T O

5 (was 6) 15 (was 5)

−1 8

4 7

Fuel filled = 47 litres

Q2 Check if 18 + 47 = 65?

Yes. 18 + 47 = 65, so the answer 47 l is correct.

1. Ones: 8 + 7 = 15. Write 5 and carry 1 ten.
2. Tens: 1 + 4 = 5, and the carried 1 makes 6.
3. The answer is 65 — exactly the reading on the fuel meter.

18 l (already in the tank) + 47 l (filled) = 65 l (meter reading) ✓

Why this check works: Subtraction pulls a part out of the whole. Adding the part
back must return you to the whole. If it does not, something went wrong in the
subtraction — most often a forgotten regrouping.

Make it a habit: After every subtraction, add your answer to the number you took
away. If you land back on the number you started with, you are right.

Let Us Solve — Page 45

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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

More Fuel Arithmetic

LET US SOLVE

Q1 What is the difference between 82 and 37? Check your answer. Is 37 + ____ = 82?

The difference is 45.
The word difference means how much bigger one number is than the other. So we subtract the
smaller from the bigger.

82 − 37

1. Ones: 2 is smaller than 7. Regroup — take 1 ten from the 8 tens and break it into 10 ones.
2. The tens are now 7 and the ones are 2 + 10 = 12.
3. Ones: 12 − 7 = 5.
4. Tens: 7 − 3 = 4.

T O

7 (was 8) 12 (was 2)

−3 7

4 5

82 − 37 = 45

Check: 37 + 45 = 82.

1. Ones: 7 + 5 = 12. Write 2, carry 1 ten.
2. Tens: 3 + 4 + 1 = 8.
3. That gives 82 ✓

Q2 57 – 11 = ___

46

1. Ones: 7 − 1 = 6. No regrouping is needed.

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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

2. Tens: 5 − 1 = 4.

57 − 11 = 46

Check: 11 + 46 = 57 ✓

Q3 23 – 19 = ___

4

1. Ones: 3 is smaller than 9, so regroup. Take 1 ten from the 2 tens.
2. The tens are now 1 and the ones are 13.
3. Ones: 13 − 9 = 4.
4. Tens: 1 − 1 = 0.

23 − 19 = 4

Easier way: 19 is 1 less than 20. 23 − 20 = 3, then put the 1 back: 3 + 1 = 4. Same
answer. Check: 19 + 4 = 23 ✓

Q4 49 – 21 = ___

28

1. Ones: 9 − 1 = 8.
2. Tens: 4 − 2 = 2.

49 − 21 = 28

Check: 21 + 28 = 49 ✓

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Class 5 Maths Chapter 4 We the Travellers—II
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co m
m.
56 – 18 = ___
e
Q5

m l as
m .co a g
ase

38 gl
a
1. Ones: 6 is smaller than 8, so regroup. Take 1 ten from the 5 tens.

co m
2. The tens are now 4 and the ones are 16.
e m . ag
3. Ones: 16 − 8 = 8.
g l as
4. Tens: 4 − 1 = 3. a

co m
m.
56 − 18 = 38

m as e
.co a g l
a s em 18 + 38 = 56 ✓
gl
Check:
a
m a s
m .co agl
se
Q6 93 – 35 = ___

g l a
a

co m
.
58

e m
m are now 8 and the ones are 13.
cotens
1. Ones: 3 is smaller than 5, so regroup. Take 1 ten from the 9 tens.
g l as
. a
s3.em
2. The

a
agl 4. Tens: 8 − 3 = 5.
Ones: 13 − 5 = 8.

se m
com g l a
m . a
ase
93 − 35 = 58

agl
Check: 35 + 58 = 93 ✓

co m
m .
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.co a g l
a s emQ7 84 – 23 = ___

agl
.c
s e m
m a
61
e m . co agl
1. Ones: 4 − 3 = 1.
g l as
2. Tens: 8 − 2 = 6. a

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m .
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.co


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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

84 − 23 = 61

Check: 23 + 61 = 84 ✓

Q8 70 – 43 = ___

27

1. Ones: 0 is smaller than 3, so regroup. Take 1 ten from the 7 tens.
2. The tens are now 6 and the ones are 10.
3. Ones: 10 − 3 = 7.
4. Tens: 6 − 4 = 2.

70 − 43 = 27

Check: 43 + 27 = 70 ✓

Q9 65 – 47 = ___

18

1. Ones: 5 is smaller than 7, so regroup. Take 1 ten from the 6 tens.
2. The tens are now 5 and the ones are 15.
3. Ones: 15 − 7 = 8.
4. Tens: 5 − 4 = 1.

65 − 47 = 18

Check: 47 + 18 = 65 ✓

Sums of Consecutive Numbers — Page 45

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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Sums of Consecutive Numbers

MATH TALK

Q1 In each of the boxes above, state whether the sums are even or odd. Explain why
this is happening.

Box of 2 consecutive numbers — every sum is odd. Box of 3 — the sums are odd and even turn
by turn. Box of 4 — every sum is even.

BOX SUMS PRINTED EVEN OR ODD

2 consecutive numbers 3, 5, 7, 9 all odd

3 consecutive numbers 6, 9, 12, 15 even, odd, even, odd

4 consecutive numbers 10, 14, 18, 22 all even

Why the box of 2 gives only odd sums.

1. Two consecutive numbers are always one even and one odd — 3 and 4, or 4 and 5.
2. The even one splits into pairs completely. The odd one leaves one counter alone.
3. That single counter has no partner, so the total is odd.

Why the box of 4 gives only even sums.

1. Four consecutive numbers are two even and two odd — for example 3, 4, 5, 6.
2. The two even numbers pair off with nothing left over.
3. Each odd number leaves one counter alone, and those two lonely counters join to make one
more pair.
4. Nothing is left over, so the total is even.

Why the box of 3 keeps changing.

1. The sum of 3 consecutive numbers is 3 times the middle number: 1 + 2 + 3 = 3 × 2 = 6.
2. The middle numbers are 2, 3, 4, 5 — even, odd, even, odd, one after the other.
3. 3 × an even number is even; 3 × an odd number is odd.
4. So the sums go even, odd, even, odd.

Try This: Write the sums of 5 consecutive numbers starting from 1, 2, 3 and 4. You
will get 15, 20, 25, 30 — odd and even turn by turn again, for the same reason as the
box of 3.

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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Q2 What is the difference between two successive sums in each box? Is it the same
throughout?

Yes, it is the same all the way down each box. The jump equals the count of numbers being
added.

BOX SUMS JUMP EACH TIME

2 consecutive numbers 3→5→7→9 2

3 consecutive numbers 6 → 9 → 12 → 15 3

4 consecutive numbers 10 → 14 → 18 → 22 4

Why it happens: Look at the box of 4. The first row is 1 + 2 + 3 + 4 and the next row
is 2 + 3 + 4 + 5. Every number has slid up by exactly 1. There are 4 numbers, so 4 lots
of 1 have been added — the sum grows by 4. Nothing else changes, so the jump is 4
every single time.

1 + 2 + 3 + 4 = 10

2 + 3 + 4 + 5 = 14 (each number 1 bigger, 4 numbers, so +4)
3 + 4 + 5 + 6 = 18 (+4 again)

Q3 What will be the difference between two successive sums for — (a) 5 consecutive
numbers (b) 6 consecutive numbers

(a) 5 (b) 6
The rule from Q2 is: the jump is the same as how many numbers you are adding.
(a) Five consecutive numbers

1. 1 + 2 + 3 + 4 + 5 = 15.
2. 2 + 3 + 4 + 5 + 6 = 20.
3. The jump is 20 − 15 = 5, because 5 numbers each grew by 1.

(b) Six consecutive numbers

1. 1 + 2 + 3 + 4 + 5 + 6 = 21.

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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

2. 2 + 3 + 4 + 5 + 6 + 7 = 27.
3. The jump is 27 − 21 = 6, because 6 numbers each grew by 1.

Check it yourself: Carry on with 3 + 4 + 5 + 6 + 7 + 8 = 33. And 33 − 27 = 6 ✓ The
jump stays 6 for ever.

In-text Questions — Page 46
Sums of Consecutive Numbers

Q1 Notice how the sums of 3, 4, and 5 consecutive numbers are related to the numbers
being added. Use your understanding to find the following sums without adding
the numbers directly. (a) 67 + 68 + 69 (b) 24 + 25 + 26 + 27 (c) 48 + 49 + 50 + 51 + 52 (d)
237 + 238 + 239 + 240 + 241 + 242

The red arrows in the book pair the outside numbers together. That is the whole secret.

67 + 68 + 69 = 3 × 68
= 204

67 + 69 = 136
136 + 68 = 204

Two outside numbers make twice the middle one, so three consecutive numbers make three times
the middle one.

Two rules to use

Odd count (3 numbers, 5 numbers): sum = count × middle number.
Even count (4 numbers, 6 numbers): pair the outside numbers. Every pair adds to the same
total, so sum = (count ÷ 2) × (first + last).

(a) 67 + 68 + 69 = 204

1. Three numbers, so the middle one is 68.
2. Sum = 3 × 68.
3. 3 × 68 = 204.

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Class 5 Maths Chapter 4 We the Travellers—II
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(b) 24 + 25 + 26 + 27 = 102
co m
e m.
m 25 + 26 = 51 — both pairs give the same total. as
1. Four numbers, so make 2 pairs.

. coand a g l
3. Sums=em
2. 24 + 27 = 51

l a 2 × 51 = 102.
ag48 + 49 + 50 + 51 + 52 = 250
(c)

co m
ag
1. Five numbers, so the middle one is 50.

m .
2. Sum = 5 × 50.
ase
3. 5 × 50 = 250.
a g l
(d) 237 + 238 + 239 + 240 + 241 + 242 = 1,437

co m
m.
1. Six numbers, so make 3 pairs.

m as e
3. Sum =.c3o× 479.
l
2. 237 + 242 = 479, 238 + 241 = 479, 239 + 240 = 479.
a g
em
4.a3s× 479 = 1,437.
agl
m
Why the pairs are always equal: Move one step in from the left and the number
a s
m .co
goes up by 1. Move one step in from the right and the number goes down by 1. The
agl
l a se
gain and the loss cancel out, so every pair keeps the same total.
ag
Check it yourself: Add (b) the long way — 24 + 25 = 49, 49 + 26 = 75, 75 + 27 = 102 ✓
co m
m .
e
The shortcut gave the same answer far faster.
m l as
m .co a g
l a se
g
a The Longest Land Route — Adding Large Numbers — Pages 46–47
se m
The Longest Land Route — Adding Large Numbers
com g l a
m . a
e
asNorth–South Corridor was 1,855 km from Srinagar and
Q1
a g l
One of the places on the
1,862 km from Kanniyakumari. What was the total length of the North–South

m
Corridor in 2019?

. co
em
m l as
.co g

em a
s
The North–South Corridor was 3,717 km long.
a
l The place marked A on the map splits the whole road into two parts. One part reaches back to
ag c
m .
e
Srinagar, the other goes on to Kanniyakumari. Put the two parts together to get the whole road.

m a s
.
emkm
c o agl
a s
Total length = 1,855 km + 1,862

agl
1. Ones: 5 + 2 = 7. Write 7.

co m
m .
m ase
.co


a g l Page 19 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

2. Tens: 5 + 6 = 11. Write 1 and carry 1 hundred.
3. Hundreds: 8 + 8 = 16, and the carried 1 makes 17. Write 7 and carry 1 thousand.
4. Thousands: 1 + 1 = 2, and the carried 1 makes 3. Write 3.

TH H T O

1 (carried) 1 (carried)

1 8 5 5

+1 8 6 2

3 7 1 7

Total length of the North–South Corridor = 3,717 km

One thing to notice in the book: The map beside the paragraph prints 1885 km for
the northern part, but the words in the paragraph say 1,855 km, and the worked
sum on page 47 also uses 1,855 km. Use 1,855 km — the map label is a printing slip.

Q2 Do you remember how to add large numbers?

Yes — a big number is added in exactly the same way as a small one. Only the number of
columns changes.

1. Write the numbers one under the other so that ones sit under ones, tens under tens,
hundreds under hundreds and thousands under thousands.
2. Add the ones column first.
3. If any column reaches 10 or more, keep the ones digit there and carry 1 to the column on its
left.
4. Work leftwards, column by column, until you have done the last one.

Why the method does not change: Ten ones always make one ten, ten tens always
make one hundred, and ten hundreds always make one thousand. The rule for
swapping ten small units for one bigger unit is the same at every place, so the same
method carries on for as long as the number is.

Page 20 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

1,855 + 1,862

Ones: 5 + 2 = 7

Tens: 5 + 6 = 11 → write 1, carry 1

Hundreds: 8 + 8 + 1 = 17 → write 7, carry 1

Thousands: 1 + 1 + 1 = 3

3,717

Q3 Mahesh and his family decide to drive from Srinagar to Kanniyakumari. He spends
₹21,880 on fuel and toll tax, and ₹38,900 on other expenses during this journey. How
much did he spend in total?

Mahesh spent ₹60,780 in all.
There are two amounts of money, and we want them put together. So we add.

Total spent = ₹21,880 + ₹38,900

1. Ones: 0 + 0 = 0.
2. Tens: 8 + 0 = 8.
3. Hundreds: 8 + 9 = 17. Write 7 and carry 1 thousand.
4. Thousands: 1 + 8 = 9, and the carried 1 makes 10. Write 0 and carry 1 ten-thousand.
5. Ten-thousands: 2 + 3 = 5, and the carried 1 makes 6. Write 6.

TTH TH H T O

1 (carried) 1 (carried)

2 1 8 8 0

+3 8 9 0 0

6 0 7 8 0

Total money spent = ₹60,780

Page 21 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Check it yourself: 60,780 − 38,900 = 21,880, which is the fuel and toll amount we
started with ✓

Q4 Mentally track the positions of the digits as you add. (Example: 267 + 54)

Once the digits are lined up properly you no longer need to write TTh, Th, H, T and O at the top.
Your eye can see the columns.

267

+ 54

———
321

1. Line up the numbers from the right. The 7 and the 4 must be in the same column, and the 6
and the 5 in the same column. The 2 stands alone.
2. Ones: 7 + 4 = 11. Write 1 and carry 1 ten.
3. Tens: 6 + 5 = 11, and the carried 1 makes 12. Write 2 and carry 1 hundred.
4. Hundreds: 2 and the carried 1 make 3. Write 3.

267 + 54 = 321

Why lining up from the right matters: A digit means nothing on its own — 5 can
stand for five, or fifty, or five hundred. The column tells you which. If the 54 slipped
one column to the left, you would be adding 540 by mistake.

Tip: Numbers of different lengths are the easy place to slip. Always start writing
from the ones end, never from the left.

Let Us Solve — Page 48

Page 22 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

The Longest Land Route — Adding Large Numbers

LET US SOLVE

Q1 Find the following sums. Try not to write TTh, Th, H, T, and O at the top. Just align
the digits properly, at least for the smaller numbers. (a) 238 + 367 (b) 1,234 + 12,345
(c) 12 + 123 (d) 46,120 + 12,890 (e) 878 + 8,789 (f) 1,749 + 17,490

SUM ANSWER

(a) 238 + 367 605

(b) 1,234 + 12,345 13,579

(c) 12 + 123 135

(d) 46,120 + 12,890 59,010

(e) 878 + 8,789 9,667

(f) 1,749 + 17,490 19,239

(a) 238 + 367 = 605

1. Ones: 8 + 7 = 15. Write 5, carry 1.
2. Tens: 3 + 6 + 1 = 10. Write 0, carry 1.
3. Hundreds: 2 + 3 + 1 = 6.

(b) 1,234 + 12,345 = 13,579

1. Line up from the right: the 4 under the 5, the 3 under the 4, and so on. The 1 of 12,345 has
no partner.
2. Ones: 4 + 5 = 9. Tens: 3 + 4 = 7. Hundreds: 2 + 3 = 5. Thousands: 1 + 2 = 3. Ten-thousands: 1.
3. Nothing had to be carried at all.

(c) 12 + 123 = 135

1. Ones: 2 + 3 = 5.
2. Tens: 1 + 2 = 3.
3. Hundreds: 1.

(d) 46,120 + 12,890 = 59,010

1. Ones: 0 + 0 = 0.
2. Tens: 2 + 9 = 11. Write 1, carry 1.
3. Hundreds: 1 + 8 + 1 = 10. Write 0, carry 1.
4. Thousands: 6 + 2 + 1 = 9.

Page 23 of 60

as e
Class 5 Maths Chapter 4 We the Travellers—II
a g l AglaSem · NCERT Solutions

5. Ten-thousands: 4 + 1 = 5.
co m
e m.
om as
(e) 878 + 8,789 = 9,667

1. Ones: 8 +.9c = 17. Write 7, carry 1. a g l
a s e7m+ 8 + 1 = 16. Write 6, carry 1.
l
agHundreds:
2. Tens:
3. 8 + 7 + 1 = 16. Write 6, carry 1.

m
4. Thousands: 8 + 1 = 9.
. co ag
e m
as
(f) 1,749 + 17,490 = 19,239

1. Ones: 9 + 0 = 9.
a g l
2. Tens: 4 + 9 = 13. Write 3, carry 1.

co m
m.
3. Hundreds: 7 + 4 + 1 = 12. Write 2, carry 1.
4. Thousands: 1 + 7 + 1 = 9.
m as e
.co a g l
em
5. Ten-thousands: 1.

a s
a gl Check it yourself: For (d), subtract back — 59,010 − 12,890 = 46,120 ✓
m a s
m .co agl
l a se
a g

co m
m .
m as e
.co a g l
se m
g l a
a
se m
com g l a
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ase
agl

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Q2 The great Indian road trip! Nazrana and her friends planned a road trip across India,
starting from Delhi. They first drove to Mumbai, then Goa, then Hyderabad, and
finally Puri. Look at the distances marked on the map and help them find the total
distance travelled.

N

Delhi
Delhi

1,600 km
1,600 km

Mumbai
Mumbai
Puri
Puri
Hyderabad
Hyderabad
590 km
590 km
Goa
Goa 1,055 km
1,055 km
670 km
670 km

*Map not to scale

The road trip: Delhi → Mumbai → Goa → Hyderabad → Puri, with the distance marked
on each part of the drive.

They travelled 3,915 km in all.
Read the four distances off the map, in the order they drove.

Page 25 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

PART OF THE JOURNEY DISTANCE

Delhi → Mumbai 1,600 km

Mumbai → Goa 590 km

Goa → Hyderabad 670 km

Hyderabad → Puri 1,055 km

Now add them, two at a time, so the numbers stay small.

1. 1,600 + 590 = 2,190 km (Delhi to Goa)
2. 2,190 + 670 = 2,860 km (Delhi to Hyderabad)
3. 2,860 + 1,055 = 3,915 km (Delhi to Puri)

Total distance travelled = 3,915 km

Check it yourself: Add in a different order and you must still get 3,915. Try 1,600 +
1,055 = 2,655, then 590 + 670 = 1,260, then 2,655 + 1,260 = 3,915 ✓

Q3 Find 2 numbers among 5,205, 6,220, 7,095, 8,455, and 4,840 whose sum is closest to
the following. (a) 10,000 (b) 15,000 (c) 13,000 (d) 16,000

TARGET THE TWO NUMBERS THEIR SUM HOW FAR OFF

(a) 10,000 5,205 and 4,840 10,045 45 more

(b) 15,000 6,220 and 8,455 14,675 325 less

(c) 13,000 8,455 and 4,840 13,295 295 more

(d) 16,000 7,095 and 8,455 15,550 450 less

How to search without adding every pair.

1. Round each number to the nearest thousand: 5,205 → 5,000; 6,220 → 6,000; 7,095 → 7,000;
8,455 → 8,000; 4,840 → 5,000.
2. For a target, look for two rounded numbers that make it. For 10,000, try 5,000 + 5,000 — that
points at 5,205 and 4,840.
3. Now do the real addition to check, and try the one or two nearest rivals as well.

Page 26 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

(a) 5,205 + 4,840 = 10,045 — only 45 away from 10,000. The next best is 6,220 + 4,840 = 11,060,
which is 1,060 away.
(b) 6,220 + 8,455 = 14,675 — 325 away. Its rival 7,095 + 8,455 = 15,550 is 550 away, so 14,675
wins.
(c) 8,455 + 4,840 = 13,295 — 295 away. Its close rival 6,220 + 7,095 = 13,315 is 315 away, so
13,295 wins by 20.
(d) 7,095 + 8,455 = 15,550 — 450 away. This is the biggest pair you can make, so nothing gets
closer to 16,000.

Tip: When two answers look equally close, work out both differences and compare
them. In part (c) the gap between the two rivals was only 20.

Subtracting Large Numbers — Pages 49–50
Subtracting Large Numbers

Q1 The total distance of the sea route from Mumbai to Chennai is 2,700 km. A ship
starting from Mumbai first reaches the Cochin port, travelling 1,083 km by sea. How
much more distance does it have to travel to reach the Chennai port?

The ship still has to travel 1,617 km.
The whole sea route is 2,700 km. The ship has already covered 1,083 km of it. We want the part
still left, so we subtract.

Distance left = 2,700 km − 1,083 km

1. Ones: 0 is smaller than 3. The tens are 0 too, so first take 1 hundred from the 7 hundreds
and turn it into 10 tens.
2. Now the hundreds are 6 and the tens are 10. Take 1 ten from those 10 tens and turn it into
10 ones.
3. Now the tens are 9 and the ones are 10.
4. Ones: 10 − 3 = 7.
5. Tens: 9 − 8 = 1.
6. Hundreds: 6 − 0 = 6.
7. Thousands: 2 − 1 = 1.

Page 27 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

TH H T O

2 6 (was 7) 9 (was 0) 10 (was 0)

−1 0 8 3

1 6 1 7

Distance still to be travelled = 1,617 km

Q2 Do you remember how to subtract numbers using place value?

Yes. Line the numbers up in columns and take away one column at a time, starting at the ones.

1. Write the bigger number on top and the smaller one below it, ones under ones.
2. Start at the ones column.
3. If the top digit is big enough, subtract straight away.
4. If the top digit is too small, regroup — take 1 from the column on the left and turn it into 10
in this column. Remember that the left column now has 1 less.
5. Move left and repeat until every column is done.

Why regrouping is allowed: You are not changing the number, only the way it is
written. 2,700 can be seen as 2 thousands, 7 hundreds, 0 tens, 0 ones — or as 2
thousands, 6 hundreds, 9 tens, 10 ones. Both are 2,700. The second way just
happens to be easier to subtract from.

Tip: When there are zeros in the middle, as in 2,700, borrow one step at a time —
hundred to tens first, then tens to ones. Do not try to jump two columns at once.

Q3 Check if the solution is correct.

Yes, it is correct. Add the answer back to the distance already sailed and you land on the full
route.

Page 28 of 60

as e
Class 5 Maths Chapter 4 We the Travellers—II
a g l AglaSem · NCERT Solutions

co m
e m.
1,617 km (still to sail) + 1,083 km (already sailed) = 2,700 km ✓
m l as
.co a g
a s em
1. Ones: 7 + 3 = 10. Write 0, carry 1 ten.
2. g l 1 + 8 + 1 = 10. Write 0, carry 1 hundred.
a Tens:
3. Hundreds: 6 + 0 + 1 = 7.

com ag
4. Thousands: 1 + 1 = 2.
.
a s em to Chennai sea route. So 1,617 km is right.
That is 2,700 — exactly the length of the Mumbai

agl

. com
As you learnt earlier, the longest land route is 15,150 km between Talon (Russia) and
emconnecting the
Q4

m a s
glmuch longer is the land
Sagres (Portugal). The longest highway in Africa is 10,228 km long,

. co of Cairo, in Egypt and Cape Town, in South Africa. How
a
sem
cities

a
agl
route between Talon and Sagres compared to the highway between Cairo and Cape
Town?

m a s
.co agl

se m
The Talon–Sagres route is 4,922 km longer.
g l a
a
The words how much longer mean we must find the difference, so we subtract the shorter road
from the longer one.

co m
m .
m as e
.co l
Difference = 15,150 km − 10,228 km
a g
a s em
a gl Ones: 0 is smaller than 8. Take 1 ten from the 5 tens. Now the tens are 4 and the ones are 10.
1.

m
10 − 8 = 2.

a se
com l
2. Tens: 4 − 2 = 2.
. a g
e m
3. Hundreds: 1 is smaller than 2. Take 1 thousand from the 5 thousands. Now the thousands

g l as11 − 2 = 9.
are 4 and the hundreds are 11.
4. Thousands: 4 − 0 = 4. a
m
5. Ten-thousands: 1 − 1 = 0.

. co
se m
com l a
TTH TH H T O

.1 a g
e m
as
agl
4 (was 5) 11 (was 1) 4 (was 5) 10 (was 0)

.c
m
−1 0 2 2 8

m a s e
. co agl
0 4 9 2 2

e m
g l as
a
The Talon–Sagres road is 4,922 km longer.

com
m .
m as e
.co


a g l Page 29 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Check it yourself: 10,228 + 4,922 = 15,150 ✓

Let Us Solve — Pages 50–51
Subtracting Large Numbers

LET US SOLVE

Q1 Subtract the following. Try not to write TTh, Th, H, T, and O at the top. Align the
digits carefully. (a) 4,578 – 2,222 (b) 15,324 – 11,780 (c) 5,423 – 423 (d) 123 – 12 (e)
77,777 – 777 (f) 826 – 752

SUBTRACTION ANSWER CHECK BY ADDING BACK

(a) 4,578 − 2,222 2,356 2,356 + 2,222 = 4,578 ✓

(b) 15,324 − 11,780 3,544 3,544 + 11,780 = 15,324 ✓

(c) 5,423 − 423 5,000 5,000 + 423 = 5,423 ✓

(d) 123 − 12 111 111 + 12 = 123 ✓

(e) 77,777 − 777 77,000 77,000 + 777 = 77,777 ✓

(f) 826 − 752 74 74 + 752 = 826 ✓

(a) 4,578 − 2,222 = 2,356

1. Ones: 8 − 2 = 6. Tens: 7 − 2 = 5. Hundreds: 5 − 2 = 3. Thousands: 4 − 2 = 2.
2. No regrouping was needed anywhere.

(b) 15,324 − 11,780 = 3,544

1. Ones: 4 − 0 = 4.
2. Tens: 2 is smaller than 8. Take 1 hundred from the 3 hundreds. Tens become 12, hundreds
become 2. 12 − 8 = 4.
3. Hundreds: 2 is smaller than 7. Take 1 thousand from the 5 thousands. Hundreds become 12,
thousands become 4. 12 − 7 = 5.
4. Thousands: 4 − 1 = 3.
5. Ten-thousands: 1 − 1 = 0.

(c) 5,423 − 423 = 5,000

1. Ones: 3 − 3 = 0. Tens: 2 − 2 = 0. Hundreds: 4 − 4 = 0. Thousands: 5 stays.

Page 30 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

2. Everything below the thousands cancels out, leaving 5,000.

(d) 123 − 12 = 111

1. Line up from the right: 2 under 3, 1 under 2.
2. Ones: 3 − 2 = 1. Tens: 2 − 1 = 1. Hundreds: 1 stays.

(e) 77,777 − 777 = 77,000

1. Ones: 7 − 7 = 0. Tens: 7 − 7 = 0. Hundreds: 7 − 7 = 0.
2. Thousands: 7 stays. Ten-thousands: 7 stays.

(f) 826 − 752 = 74

1. Ones: 6 − 2 = 4.
2. Tens: 2 is smaller than 5. Take 1 hundred from the 8 hundreds. Tens become 12, hundreds
become 7. 12 − 5 = 7.
3. Hundreds: 7 − 7 = 0, so nothing is written there.

Tip: In (d) the danger is lining 12 up under 123 from the left. Always start from the
ones end.

Q2 Mary's train journey to Delhi. Mary is on a train journey. She starts from Kolkata
with ₹12,540. She spends ₹3,275 on food and other expenses during her trip to
Varanasi. In Varanasi, her uncle gives her a gift worth ₹4,900. She then travels to
Delhi, spending ₹2,645 on the train ticket. She spends ₹1,275 on souvenirs in Delhi.
How much money is Mary left with at the end of the Delhi trip?

Mary is left with ₹10,245.
Follow her purse stop by stop. Money spent is taken away; the gift is added.

STAGE WHAT HAPPENS MONEY IN HAND

Leaves Kolkata starting amount ₹12,540

Reaches Varanasi spends ₹3,275 ₹9,265

In Varanasi gift of ₹4,900 added ₹14,165

Ticket to Delhi spends ₹2,645 ₹11,520

In Delhi spends ₹1,275 ₹10,245

1. 12,540 − 3,275 = 9,265

Page 31 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

2. 9,265 + 4,900 = 14,165
3. 14,165 − 2,645 = 11,520
4. 11,520 − 1,275 = 10,245

Money left with Mary = ₹10,245

Why we work stop by stop: Each stop changes only one thing. Doing them one at a
time keeps the numbers small and lets you see, at every station, exactly how much
Mary is carrying.

Check it yourself, a second way: Money coming in = 12,540 + 4,900 = 17,440.
Money going out = 3,275 + 2,645 + 1,275 = 7,195. Left over = 17,440 − 7,195 =
₹10,245 ✓

Q3 Members of a school council have raised ₹70,500. They plan to set up a Maths Lab
with some games and models worth ₹39,785, buy library books worth ₹9,545, and
purchase sports equipment worth ₹19,548. (a) Estimate whether the school council
has raised enough money to make the purchases. Share your thoughts in the class.
(b) Check your estimate with calculations.

(a) Estimate — yes, but only just.

1. Round each cost to the nearest thousand: ₹39,785 → about ₹40,000; ₹9,545 → about
₹10,000; ₹19,548 → about ₹20,000.
2. Add the rounded costs: 40,000 + 10,000 + 20,000 = ₹70,000.
3. They have ₹70,500. That is a little more than ₹70,000.
4. So it looks like they can just manage it — but the gap is small, so we must check properly.

Sample answer for the class discussion: "Each of the three costs is a little under the round
number I used — 39,785 is below 40,000, 9,545 is below 10,000 and 19,548 is below 20,000. So
the real total will be a bit less than ₹70,000, and ₹70,500 should be enough."
(b) The exact calculation — yes, and ₹1,622 is left over.

1. 39,785 + 9,545 = 49,330
2. 49,330 + 19,548 = 68,878 (total cost)
3. 70,500 − 68,878 = 1,622

Page 32 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Total cost = ₹68,878

Money raised = ₹70,500

Money left over = ₹1,622

Check it yourself: 68,878 + 1,622 = 70,500 ✓ The estimate said "just enough", and
the exact sum agrees.

Q4 A truck can carry 8,250 kg of goods. A factory loads 3,675 kg of cement and 2,850 kg
of steel on it. (a) What is the total weight loaded onto the truck? (b) How much more
weight can the truck carry before reaching its maximum capacity?

(a) 6,525 kg is already loaded. (b) 1,725 kg more can still be loaded.
(a) Total weight loaded
Cement and steel are both sitting on the truck, so we add them.

1. Ones: 5 + 0 = 5.
2. Tens: 7 + 5 = 12. Write 2, carry 1.
3. Hundreds: 6 + 8 + 1 = 15. Write 5, carry 1.
4. Thousands: 3 + 2 + 1 = 6.

3,675 kg + 2,850 kg = 6,525 kg

(b) Weight the truck can still take
The truck can hold 8,250 kg in all. Take away what is already on it.

1. Ones: 0 is smaller than 5. Take 1 ten from the 5 tens. Ones become 10, tens become 4. 10 − 5
= 5.
2. Tens: 4 − 2 = 2.
3. Hundreds: 2 is smaller than 5. Take 1 thousand from the 8 thousands. Hundreds become 12,
thousands become 7. 12 − 5 = 7.
4. Thousands: 7 − 6 = 1.

8,250 kg − 6,525 kg = 1,725 kg

Page 33 of 60

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Class 5 Maths Chapter 4 We the Travellers—II
a g l AglaSem · NCERT Solutions

co m
m.
Check it yourself: 6,525 + 1,725 = 8,250 ✓ — exactly the truck's full capacity.

m as e
.co a g l
a s em
agl Sums and Differences — Pages 51–52
Quick
Quick Sums and Differences

co m
MATH TALK
em . ag
g l as
Q1
a
Sukanta likes the numbers 10, 100, 1,000, and 10,000. He wants to figure out what

m
number he should add to a given number such that the sum is 100 or 1,000. Help
co
m.
him fill in the blanks with an appropriate number. 32 + ______ = 100

as e
ANSWER .c
om a g l
a s em
agl
32 + 68 = 100
Piku's way, in three steps.

m a s
.co agl
1. First aim for 99 instead of 100. Take each digit of 32 away from 9: 9 − 3 = 6 and 9 − 2 = 7. So
32 + 67 = 99.
a s em
2. 100 is 1 more than 99, so addg1lto the 67: 67 + 1 = 68.
a
3. So 32 + 68 = 100.

co m
m .
as e
om l
32

+m6.c7 a g
ase
agl ———

se m
com a
99 → 32 + 68 = 100
. a g l
m
ase
agl
Why aiming for 99 first is easier: Every digit of 99 is a 9, and no digit can be bigger
than 9. So you never have to borrow — you just take each digit away from 9 in your
head.
co m
m .
o m l a se
m ag 1; tens 3 + 6 + 1 = 10. That
.cCheck it yourself: 32 + 68 → ones 2 + 8 = 10, write 0 carry
l a se
g
is 100 ✓
a c
m .
m a s e
e m . co agl
g l as
a

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m .
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.co


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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Q2 Do you think this method will always work?

Yes — for any whole number less than 100 when you are making 100, and in the same way for
1,000 and 10,000.

Why it always works: 100 is always 99 + 1. Reaching 99 needs no borrowing,
because 9 is the biggest a digit can be, so every digit of your number is 9 or less.
Once you are at 99 you are just one short of 100, so adding 1 finishes the job. The
same reasoning works for 1,000 (which is 999 + 1) and for 10,000 (which is 9,999 + 1).

The rule to remember

1. To make 100, subtract each digit from 9 to reach 99, then add 1.
2. To make 1,000, subtract each digit from 9 to reach 999, then add 1.
3. To make 10,000, subtract each digit from 9 to reach 9,999, then add 1.

Try it on 46: 9 − 4 = 5, 9 − 6 = 3, so 46 + 53 = 99

Then 53 + 1 = 54, so 46 + 54 = 100 ✓

Q3 59 + ______ = 100. Try this method for the number 59.

59 + 41 = 100

1. Aim for 99 first. 9 − 5 = 4 and 9 − 9 = 0. So 59 + 40 = 99.
2. Add 1 to reach 100: 40 + 1 = 41.
3. So the missing number is 41.

59

+40

———

99 → 59 + 41 = 100

Page 35 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Check it yourself: 59 + 41 → ones 9 + 1 = 10, write 0 carry 1; tens 5 + 4 + 1 = 10. That
is 100 ✓

Q4 Now, use this method to solve the following. 877 + ______ = 1,000 and 666 + ______ =
1,000. 4,103 + ______ = 10,000 and 5,555 + ______ = 10,000

QUESTION STEP 1 — REACH ALL 9S STEP 2 — ADD 1 ANSWER

877 + ___ = 1,000 999 − 877 = 122 122 + 1 = 123 123

666 + ___ = 1,000 999 − 666 = 333 333 + 1 = 334 334

4,103 + ___ = 10,000 9,999 − 4,103 = 5,896 5,896 + 1 = 5,897 5,897

5,555 + ___ = 10,000 9,999 − 5,555 = 4,444 4,444 + 1 = 4,445 4,445

877 + 123 = 1,000

1. Take each digit of 877 away from 9: 9 − 8 = 1, 9 − 7 = 2, 9 − 7 = 2. That gives 122, and 877 +
122 = 999.
2. Add 1 more to reach 1,000: 122 + 1 = 123.

666 + 334 = 1,000

1. 9 − 6 = 3 three times, so 666 + 333 = 999.
2. 333 + 1 = 334.

4,103 + 5,897 = 10,000

1. Use 9,999 now, because 10,000 has one more digit. 9 − 4 = 5, 9 − 1 = 8, 9 − 0 = 9, 9 − 3 = 6.
That gives 5,896.
2. 5,896 + 1 = 5,897.

5,555 + 4,445 = 10,000

1. 9 − 5 = 4 four times, so 5,555 + 4,444 = 9,999.
2. 4,444 + 1 = 4,445.

Check it yourself: 5,555 + 4,445 → ones 5 + 5 = 10, and each column after it
becomes 5 + 4 + 1 = 10. You end with 10,000 ✓

Page 36 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Q5 Will this method work if the units digit is 0? What do you think? What other
methods can you use to find the missing number to fill in the blanks? Share your
thoughts in the class. (a) 180 + ______ = 1,000 (b) 760 + ______ = 1,000 (c) 400 + ______ =
1,000

Yes, it still works. A units digit of 0 changes nothing, because 9 − 0 = 9 and there is still no
borrowing.

QUESTION REACH 999 ADD 1 ANSWER

(a) 180 + ___ = 1,000 999 − 180 = 819 819 + 1 = 820 820

(b) 760 + ___ = 1,000 999 − 760 = 239 239 + 1 = 240 240

(c) 400 + ___ = 1,000 999 − 400 = 599 599 + 1 = 600 600

Other methods you can use

1. Count on in easy jumps. For 180: go up to the next hundred, 180 + 20 = 200. Then 200 + 800
= 1,000. Add the jumps: 20 + 800 = 820.
2. Use the zeros. 760 ends in 0, so ask instead: 76 tens + how many tens = 100 tens? 100 − 76 =
24 tens, which is 240.
3. Straight subtraction. 1,000 − 400 = 600. This is easy when the number is a round hundred.

Sample answer for the class discussion: "The 99s trick still works when the last digit is 0,
because 9 − 0 is just 9. But when a number ends in one or two zeros, counting on in jumps is
even faster — for 400 I can see straight away that 600 more makes 1,000."

Check it yourself: 180 + 820 = 1,000 ✓ 760 + 240 = 1,000 ✓ 400 + 600 = 1,000 ✓

Q6 Namita likes the number 9. She wants to subtract 9 or 99 from any number. Find a
way to quickly subtract 9 or 99 from any number. (a) 67 – 9 (b) 83 – 9 (c) 144 – 9 (d)
187 – 99 (e) 247 – 99 (f) 763 – 99

The trick: to take away 9, take away 10 and then put 1 back. To take away 99, take away 100
and then put 1 back.

Page 37 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

QUESTION WORKING ANSWER

(a) 67 − 9 67 − 10 = 57, then 57 + 1 58

(b) 83 − 9 83 − 10 = 73, then 73 + 1 74

(c) 144 − 9 144 − 10 = 134, then 134 + 1 135

(d) 187 − 99 187 − 100 = 87, then 87 + 1 88

(e) 247 − 99 247 − 100 = 147, then 147 + 1 148

(f) 763 − 99 763 − 100 = 663, then 663 + 1 664

Why you put 1 back: 9 is one less than 10. If you take 10 away you have taken one
rupee too much, so you hand one rupee back. In the same way 99 is one less than
100, so after taking 100 away you put 1 back.

Taking away 10 or 100 is easy because only one digit changes — the tens digit for 10, the
hundreds digit for 100.

67 − 9

Step 1: 67 − 10 = 57

Step 2: 57 + 1 = 58

Check: 58 + 9 = 67 ✓

Check them all: 58 + 9 = 67 ✓, 74 + 9 = 83 ✓, 135 + 9 = 144 ✓, 88 + 99 = 187 ✓, 148 +
99 = 247 ✓, 664 + 99 = 763 ✓

Q7 Now, use the above solutions to find answers to the following problems. Do not
calculate again. Namita wonders if she can get 9 or 99 as the answer to any
subtraction problem. Find a way to get the desired answer. (a) 32 – ______ = 9 (b) 56 –
______ = 9 (c) 877 – ______ = 99 (d) 666 – ______ = 99

Yes, she can — from any number. The number to take away is simply the starting number
minus 9 (or minus 99), and that is exactly the quick trick from the last question.

Page 38 of 60

as e
Class 5 Maths Chapter 4 We the Travellers—II
a g l AglaSem · NCERT Solutions

co m
m.
QUESTION WORKING WITH THE TRICK ANSWER

m l a se23
.co ag
(a) 32 − ___ = 9 32 − 10 = 22, then 22 + 1 = 23

se m
g l a
a
(b) 56 − ___ = 9 56 − 10 = 46, then 46 + 1 = 47 47

(c) 877 − ___ = 99 877 − 100 = 777, then 777 + 1 = 778 778

co m
(d) 666 − ___ = 99
e m .
666 − 100 = 566, then 566 + 1 = 567 567 ag
g l as
How to see it a
1. The three numbers in a subtraction always make one family: whole − part = other part.
co m
2. Here 9 is one part and the starting number is the whole. So the missing part is starting
em.
c o m
number − 9.
g l as
3. Andm . away 9" is the quick trick — take away 10 and put 1 back.
a
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a gla
(a) 32 − 23 = 9 ✓
m a s
(b) 56 − 47 = 9 ✓
m .co agl
l a se
g
(c) 877 − 778 = 99 ✓

(d) 666 − 567 = 99 ✓
a

com
m .
m as e
. c o
Try This: Pick any number, say 415. Take away 9 the quick way: 415 − 10 = 405, then
a g l
s e+m1 = 406. So 415 − 406 = 9. It works every time.
a
agl
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Let Us Think and Solve — Page.c53 om g l a
em a
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Let Us Think and Solve

LET US THINK AND SOLVE

c o m
.
mto right or from right to
s e
Nitin likes numbers that read the same when read from left

comleft. lanumbers 22, 363, 404, and
Q1

. a g
Such numbers are called palindrome numbers. The

as em 8,558 are some examples. List all palindrome numbers between 100 and 200. List all
agl
palindrome numbers between 900 and 1,200. List all palindrome numbers between

.c
m
25,000 and 27,000.

m a s e
e m . co agl
as

a g l
A palindrome number reads the same forwards and backwards, so its first digit must match its
last digit, and its second digit must match its second-last digit.

co m
m .
m ase
.co


a g l Page 39 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Between 100 and 200 — ten numbers

1. Every number here starts with 1, so it must also end with 1. The shape is 1 _ 1.
2. The middle digit can be anything from 0 to 9.

101, 111, 121, 131, 141, 151, 161, 171, 181, 191

Between 900 and 1,200 — twelve numbers

1. First take the 3-digit ones, from 900 to 999. They start with 9, so they end with 9. The shape is
9 _ 9.
2. That gives 909, 919, 929, 939, 949, 959, 969, 979, 989, 999 — ten numbers.
3. Now the 4-digit ones, from 1,000 to 1,200. They start with 1, so they end with 1. The shape is
1 b b 1.
4. 1001 works and 1111 works. The next one, 1221, is bigger than 1,200, so it is out.

909, 919, 929, 939, 949, 959, 969, 979, 989, 999, 1001, 1111

Between 25,000 and 27,000 — twenty numbers

1. All these numbers start with 2, so they end with 2. The shape is 2 b c b 2.
2. The second digit b can only be 5 or 6, because 25 000–25 999 and 26 000–26 999 lie inside
the range, while 27 072 and above are outside it.
3. The middle digit c can be any of 0 to 9. So there are 2 × 10 = 20 numbers.

25052, 25152, 25252, 25352, 25452, 25552, 25652, 25752, 25852, 25952,
26062, 26162, 26262, 26362, 26462, 26562, 26662, 26762, 26862, 26962

Check it yourself: Read 25452 backwards — 2, 5, 4, 5, 2. The same. Read 26762
backwards — 2, 6, 7, 6, 2. The same again.

Page 40 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Q2 In a 3×3 grid, arrange the numbers 1 to 9 such that each row and each column has
numbers in an increasing (inc) order. Each number should be used only once. This
time, fill the grid such that each row and column has numbers in decreasing (dec)
order.

inc dec

inc dec

inc dec

inc inc inc dec dec dec
Two empty 3×3 grids: in the first, every row and every column must increase; in the
second, every row and every column must decrease.

First grid — every row and every column increasing

1 2 3 inc →

4 5 6 inc →

7 8 9 inc →

inc ↓ inc ↓ inc ↓

Read across: 1 < 2 < 3, 4 < 5 < 6, 7 < 8 < 9. Read down: 1 < 4 < 7, 2 < 5 < 8, 3 < 6 < 9. Every row and
column climbs.
Second grid — every row and every column decreasing

9 8 7 dec →

6 5 4 dec →

3 2 1 dec →

dec ↓ dec ↓ dec ↓

Page 41 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Read across: 9 > 8 > 7, 6 > 5 > 4, 3 > 2 > 1. Read down: 9 > 6 > 3, 8 > 5 > 2, 7 > 4 > 1. Every row and
column falls.

How to build them: Write 1 to 9 in order, three at a time, filling the rows from the
top. That automatically makes the rows climb, and it also makes each column climb
because every number in a row is 3 more than the one above it. For the decreasing
grid, do the same but count down from 9.

There is more than one answer. For example 1, 2, 3 / 4, 5, 7 / 6, 8, 9 also works.
There are 42 different ways to fill each of these two grids. Any filling where every row
climbs and every column climbs is correct.

Q3 Now, fill the grids below with numbers (1–9) based on the inc (increasing) and dec
(decreasing) conditions, as indicated below. Grid 1: rows dec, dec, dec and columns
inc, inc, inc. Grid 2: rows dec, dec, inc and columns dec, dec, inc. Grid 3: rows inc, inc,
dec and columns inc, dec, dec.

Grid 1 — all rows going down, all columns going up

3 2 1 dec →

6 5 4 dec →

9 8 7 dec →

inc ↓ inc ↓ inc ↓

1. Rows: 3 > 2 > 1 ✓, 6 > 5 > 4 ✓, 9 > 8 > 7 ✓
2. Columns: 3 < 6 < 9 ✓, 2 < 5 < 8 ✓, 1 < 4 < 7 ✓

Grid 2 — rows: dec, dec, inc; columns: dec, dec, inc

8 6 1 dec →

7 5 2 dec →

3 4 9 inc →

dec ↓ dec ↓ inc ↓

1. Rows: 8 > 6 > 1 ✓, 7 > 5 > 2 ✓, 3 < 4 < 9 ✓

Page 42 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

2. Columns: 8 > 7 > 3 ✓, 6 > 5 > 4 ✓, 1 < 2 < 9 ✓

Grid 3 — rows: inc, inc, dec; columns: inc, dec, dec

1 6 8 inc →

2 5 7 inc →

9 4 3 dec →

inc ↓ dec ↓ dec ↓

1. Rows: 1 < 6 < 8 ✓, 2 < 5 < 7 ✓, 9 > 4 > 3 ✓
2. Columns: 1 < 2 < 9 ✓, 6 > 5 > 4 ✓, 8 > 7 > 3 ✓

A way to find your own answer: Start with the corners. A cell that must be the
biggest in both its row and its column is a good home for 9; a cell that must be
smallest in both is a good home for 1. In Grid 2 the bottom-right cell has to be the
biggest in its row and in its column, so 9 belongs there. Then fill in the rest and
adjust.

There are many correct answers. Grid 1 has 42 solutions, and Grids 2 and 3 have
60 each. Check yours by reading every row across and every column down.

Even and Odd Numbers — Page 54
Even and Odd Numbers

Q1 Circle the numbers that are even. (a) 297 (b) 498 (c) 724 (d) 100 (e) 199 (f) 789 (g) 49
(h) 6,893 (i) 846 (j) 111 (k) 222 (l) 1,023

The even numbers are (b) 498, (c) 724, (d) 100, (i) 846 and (k) 222. Circle those five.
How to tell in one look: only the last digit matters. If the number ends in 0, 2, 4, 6 or 8 it is
even. If it ends in 1, 3, 5, 7 or 9 it is odd.

Page 43 of 60

as e
Class 5 Maths Chapter 4 We the Travellers—II
a g l AglaSem · NCERT Solutions

co m
m.
NUMBER LAST DIGIT EVEN OR ODD

m as e
(a) 297
.co
7 odd
a g l
se m
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(b) 498 8 even

(c) 724 4 even

co m
(d) 100 0
e m . even ag
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(e) 199 9
a odd

(f) 789 9 odd
co m
e m.
com l as
(g) 49 9 odd

. a g
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(h) 6,893 3 odd

agl(i) 846 6 even

m a s
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(j) 111 1 odd

se m
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(k) 222 2 even

(l) 1,023 3 odd

co m
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o m l a se
Why only the last digit matters: Tens, hundreds and thousands can always be
.c into two equal groups — 10 is 5 and 5, 100 is 50 anda g50. So they never leave
m
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shared

agl even or odd.
se m
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co m
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a c
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m a s e
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g l as
a

co m
m .
m as e
.co


a g l Page 44 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Q2 Observe the given arrangement.

Paired arrangement for 18 Paired arrangement for 23

The two paired arrangements printed with the question.

Add 2 to 18. What changes or does not change in the arrangement? Add 2 to 23.
What changes or does not change in the arrangement?

In both cases the arrangement gets one more full pair. Whether anything is left over does not
change.

18 — nine complete pairs, nothing left over

20 — one more pair joins; still nothing left over

The two green counters are the pair that 2 adds on.

Adding 2 to an even number simply adds one more complete pair, so it stays even.

Add 2 to 18

1. 18 is arranged as 9 complete pairs, with nothing left alone. That is why 18 is even.
2. The 2 new counters make one more pair on the end.
3. 20 is now 10 complete pairs. Nothing is left over.

Page 45 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

4. What changes: the number of pairs, from 9 to 10. What does not change: there is still no
counter standing alone — 20 is even, just as 18 was.

Add 2 to 23

1. 23 is arranged as 11 complete pairs and 1 counter left over. That is why 23 is odd.
2. The 2 new counters make one more pair.
3. 25 is now 12 complete pairs and still 1 counter left over.
4. What changes: the number of pairs, from 11 to 12. What does not change: the single
lonely counter is still there — 25 is odd, just as 23 was.

The big idea: Adding 2 always adds exactly one whole pair. It can never rescue a
lonely counter and it can never create one. So an even number stays even and an
odd number stays odd.

Q3 What do you notice about the sums in each of the following cases? Do you think it
will be true for all pairs of such numbers? Explain your observations. You may use
the paired arrangement to explain your thinking.

Paired arrangement for 18 Paired arrangement for 23

The two paired arrangements printed with the question.

(a) 12 and 6 are a pair of even numbers. Choose 5 such pairs of even numbers. Add
the numbers in each of the pairs. (b) 13 and 9 are a pair of odd numbers. Choose 5
such pairs of odd numbers. Add the numbers in each of the pairs. (c) 7 and 12 are a
pair of odd and even numbers. Choose 5 such pairs of odd and even numbers. Add
the numbers in each of the pairs.

(a) even + even is always even. (b) odd + odd is always even. (c) odd + even is always odd.
Yes, these are true for every such pair, without exception.
(a) Five pairs of even numbers

Page 46 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

PAIR SUM EVEN OR ODD

12 and 6 18 even

8 and 4 12 even

20 and 10 30 even

14 and 16 30 even

2 and 18 20 even

Each even number is made only of complete pairs. Put two such sets together and you still have
only complete pairs — so the total is even.
(b) Five pairs of odd numbers

PAIR SUM EVEN OR ODD

13 and 9 22 even

7 and 5 12 even

11 and 3 14 even

15 and 21 36 even

1 and 9 10 even

Each odd number leaves one counter alone. Put two odd numbers together and the two lonely
counters find each other and make a pair. Nothing is left over, so the sum is even.

5 (odd) + 3 (odd): the two lonely counters join up

5+3=8
4 full pairs

The two pink counters are the ones left over. Together
they make a pair.

Two odd numbers each leave one counter alone; the two lonely counters pair up, so the sum is even.

(c) Five pairs of one odd and one even number

Page 47 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

PAIR SUM EVEN OR ODD

7 and 12 19 odd

3 and 8 11 odd

9 and 6 15 odd

15 and 4 19 odd

11 and 20 31 odd

The even number brings only complete pairs. The odd number brings complete pairs and one
lonely counter. There is no second lonely counter for it to join, so it is still alone at the end — the
sum is odd.

Check it yourself: Try any pair you like — 100 and 46 give 146 (even), 101 and 47
give 148 (even), 101 and 46 give 147 (odd). The rule never breaks.

Let Us Think — Pages 54–55
Let Us Think

LET US THINK

Q1 Jincy opened her piggy bank. She found 8 coins of ₹1, 9 coins of ₹2 and 5 coins of ₹5.
She wants to buy stickers worth ₹38. What possible combination of coins can she
use to pay the exact amount?

There are eleven different ways. Here are three easy ones to start with.

5 coins of ₹5 + 6 coins of ₹2 + 1 coin of ₹1 = 25 + 12 + 1 = ₹38
4 coins of ₹5 + 9 coins of ₹2 = 20 + 18 = ₹38
3 coins of ₹5 + 9 coins of ₹2 + 5 coins of ₹1 = 15 + 18 + 5 = ₹38

How to search in an orderly way

1. First see how much money she has in all: 8 × 1 = ₹8, 9 × 2 = ₹18, 5 × 5 = ₹25. Total ₹51. That is
more than ₹38, so it can be done.
2. Start with the biggest coins. Decide how many ₹5 coins to use — she has at most 5.
3. Whatever is left must be made from ₹2 and ₹1 coins, using at most 9 twos and at most 8
ones.
4. Write down every case that fits.

Page 48 of 60

as e
Class 5 Maths Chapter 4 We the Travellers—II
a g l AglaSem · NCERT Solutions

co m
m.
₹5 COINS ₹2 COINS ₹1 COINS TOTAL

m l a s₹e38
.co ag
5 3 7

se m
g l a ₹38
a
5 4 5

5 5 3 ₹38

co m1
5 6
e m . ₹38 ag
g l as
4 5
a 8 ₹38

4 6 6
co
₹38
m
e m₹.38
com l as
4 7 4

. a g
m
ase
4 8 2 ₹38

agl4 9 0 ₹38

m a s
.co agl
3 8 7 ₹38

se m
g l a ₹38
a
3 9 5

c o m
Why she cannot use 2 or fewer ₹5 coins: With only 2 five-rupee coins she has ₹10
m .
s e
from them. The most she can add is 9 × 2 + 8 × 1 = 18 + 8 = ₹26. Together that is only
₹36, o
.
m is short of ₹38.
c which a gla
m
ase
agl Check one row: 4 fives + 7 twos + 4 ones = 20 + 14 + 4 = ₹38 ✓ And she owns

se m
com a
enough of each coin: 4 ≤ 5, 7 ≤ 9, 4 ≤ 8.
. a g l
m
gl ase
Q2
a
Raghu is fond of his grandfather's torch. He starts playing with it. He presses the

co
switch once and the light turns ON. He presses it a second time and the light turns
m
m .
e How do you know?
OFF. He presses the switch a third time and the light turns ON. He keeps doing this
several times. Will the torch be ON or OFF after the 23rd spress?
o m l a
m .c For what number of presses will the torch be ON? For ag what number of presses of
l a se
g
the switch will the torch be OFF?
a c
m .
m a s e
co agl

m .
e
After the 23rd press the torch is ON, because 23 is an odd number.

g l as
a
The torch is ON after an odd number of presses and OFF after an even number of presses.

com
m .
m ase
.co


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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

PRESS NUMBER 1 2 3 4 5 6

Torch ON OFF ON OFF ON OFF

Odd or even odd even odd even odd even

How to know without counting all 23 presses

1. The torch starts OFF. One press turns it ON.
2. Presses come in twos — one to switch it on, the next to switch it off. Each such pair leaves
the torch back where it began, OFF.
3. 23 presses make 11 such pairs, and 11 × 2 = 22 presses. After those 22 the torch is OFF.
4. One press is left over — the 23rd. It turns the torch ON.

Why odd and even is the whole answer: This is exactly the paired arrangement
from page 54. An even number of presses pairs off completely and cancels out. An
odd number always has one press left alone, and that lonely press is the one that
leaves the light shining.

Check it yourself: Press 7 times in the air, counting aloud — ON, OFF, ON, OFF, ON,
OFF, ON. The 7th press, an odd one, leaves it ON ✓

Page 50 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Q3 Mountain climbing. Priyanka Mohite is the first Indian woman to climb five
Himalayan peaks above 8,000 metres. Read the table and answer:

MOUNTAIN RANGE HEIGHT (IN CLIMBED IN THE
METRES) YEAR

Mount Kanchenjunga (India and Nepal 8,586 2022
border)

Mount Everest (Nepal–China border) 8,848 2013

Mount Makalu (China–Nepal border) 8,485 2019

Mount Lhotse (Tibet–Nepal border) 8,516 2018

Mount Kilimanjaro (Africa) 5,895 2016

Mount Elbrus (Russia) 5,642 2017

Mount Annapurna I (Nepal) 8,091 2021

(a) Which is the highest peak she climbed? (b) What is the difference in height
between the highest and lowest peaks she has climbed, as per the table. (c) What is
the difference between heights of Mount Elbrus and Mount Kanchenjunga? (d) If
Priyanka was 20 years old when she summited Mount Everest in 2013, in which year
was she born?

First, put the seven peaks in order of height so the table is easy to read.

PEAK HEIGHT (M) YEAR CLIMBED

Mount Everest (Nepal–China border) 8,848 2013

Mount Kanchenjunga (India and Nepal border) 8,586 2022

Mount Lhotse (Tibet–Nepal border) 8,516 2018

Mount Makalu (China–Nepal border) 8,485 2019

Mount Annapurna I (Nepal) 8,091 2021

Mount Kilimanjaro (Africa) 5,895 2016

Mount Elbrus (Russia) 5,642 2017

Page 51 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

(a) Mount Everest, 8,848 m. Compare the thousands digits first — all seven begin with 8 or 5,
so the 5s are out. Among the 8-thousands, look at the hundreds: 8, 5, 5, 4, 0. The biggest is 8, so
8,848 is the tallest.
(b) The difference is 3,206 m.

1. Highest peak: Mount Everest, 8,848 m. Lowest peak in the table: Mount Elbrus, 5,642 m.
2. Ones: 8 − 2 = 6.
3. Tens: 4 − 4 = 0.
4. Hundreds: 8 − 6 = 2.
5. Thousands: 8 − 5 = 3.

8,848 m − 5,642 m = 3,206 m

(c) The difference is 2,944 m.

1. Mount Kanchenjunga is 8,586 m and Mount Elbrus is 5,642 m.
2. Ones: 6 − 2 = 4. No regrouping needed here.
3. Tens: 8 − 4 = 4.
4. Hundreds: 5 is smaller than 6. Take 1 thousand from the 8 thousands. Hundreds become 15,
thousands become 7. 15 − 6 = 9.
5. Thousands: 7 − 5 = 2.

8,586 m − 5,642 m = 2,944 m

(d) She was born in 1993.

1. She climbed Everest in 2013, aged 20.
2. Go back 20 years from 2013: 2013 − 20 = 1993.

Year of birth = 1993

Check them: 5,642 + 3,206 = 8,848 ✓ 5,642 + 2,944 = 8,586 ✓ 1993 + 20 = 2013 ✓

Math Metric Mela — Page 56

Page 52 of 60

Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Math Metric Mela

Q1 For each district, find out if the number of certificates were sufficient? If
insufficient, calculate how many certificates fell short. If extra, calculate how many
certificates were in excess. Chittoor, A.P. — printed 18,225, attended 18,104. Jaunpur,
U.P. — printed 19,043, attended 19,265. Raigad, Maharashtra — printed 20,863,
attended 19,974.

Chittoor and Raigad had enough certificates. Jaunpur fell short by 222.

DISTRICT PRINTED ATTENDED ENOUGH? SHORT / EXTRA

Chittoor, A.P. 18,225 18,104 Yes 121 extra

Jaunpur, U.P. 19,043 19,265 No 222 short

Raigad, Maharashtra 20,863 19,974 Yes 889 extra

Step 1 — compare the two numbers for each district. If the printed number is bigger, there
are certificates to spare. If it is smaller, some students will not get one.
Chittoor — 121 certificates in excess

1. 18,225 printed is more than 18,104 attended, so there are enough.
2. Ones: 5 − 4 = 1. Tens: 2 − 0 = 2. Hundreds: 2 − 1 = 1. Thousands: 8 − 8 = 0. Ten-thousands: 1 −
1 = 0. No regrouping is needed anywhere.
3. 18,225 − 18,104 = 121.

Jaunpur — 222 certificates short

1. 19,043 printed is less than 19,265 attended, so there are not enough.
2. Take the smaller from the bigger to find the shortfall.
3. Ones: 5 − 3 = 2. Tens: 6 − 4 = 2. Hundreds: 2 − 0 = 2. Thousands: 9 − 9 = 0. Ten-thousands: 1 −
1 = 0.
4. 19,265 − 19,043 = 222.

Raigad — 889 certificates in excess

1. 20,863 printed is more than 19,974 attended, so there are enough.
2. Ones: 3 is smaller than 4. Take 1 ten from the 6 tens. Ones become 13, tens become 5. 13 − 4
= 9.
3. Tens: 5 is smaller than 7. Take 1 hundred from the 8 hundreds. Tens become 15, hundreds
become 7. 15 − 7 = 8.
4. Hundreds: 7 is smaller than 9. Take 1 thousand from the 0 thousands — but there are none,
so first take 1 ten-thousand from the 2 and turn it into 10 thousands. Now thousands are 10;
take 1 of them for the hundreds. Hundreds become 17, thousands become 9, ten-thousands

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Class 5 Maths Chapter 4 We the Travellers—II
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become 1. 17 − 9 = 8.
co m
5. Thousands: 9 − 9 = 0. Ten-thousands: 1 − 1 = 0.
em.
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.co a g
6. 20,863 − 19,974 = 889.

a s em
gl them all: 18,104 + 121 = 18,225 ✓ 19,043 + 222 = 19,265 ✓ 19,974 + 889 =
aCheck
20,863 ✓

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Let Us Do — Page 56 a
m
Math Metric Mela

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LET US DO

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a s em
l
Add. (a) 2,009 + 7,388 (b) 26,444 + 71,111 (c) 777 + 888 (d) 1,234 + 1,234 (e) 56 + 56,789
g
Q1

a (f) 777 + 77,777 (f) 5,922 + 9,221 (g) 4,321 + 8,765 (h) 50,050 + 55,000

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SUM ANSWER

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(a) 2,009 + 7,388 9,397

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co+ m as
(b) 26,444 + 71,111 97,555

. 1,665 a g l
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(c) 777 888

agl (d) 1,234 + 1,234 2,468

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.
(e) 56 + 56,789 56,845

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(f) 777 + 77,777 78,554

(f) 5,922 + 9,221 15,143

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(g) 4,321 + 8,765 13,086

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(h) 50,050 + 55,000 1,05,050

a s
agl (a) 2,009 + 7,388 = 9,397 — Ones: 9 + 8 = 17, write 7 carry 1. Tens: 0 + 8 + 1 = 9. Hundreds: 0 + 3 =
.c
3. Thousands: 2 + 7 = 9.
s e m
. c om a g
(b) 26,444 + 71,111 = 97,555 — no carrying at all. Ones: 4 + 1 = 5. Tens: 4 + 1 = 5. Hundreds: 4 + 1la
s e m
= 5. Thousands: 6 + 1 = 7. Ten-thousands: 2 + 7 = 9.
a
agl 7 + 8 = 15, write 5 carry 1. Tens: 7 + 8 + 1 = 16, write 6 carry 1.
(c) 777 + 888 = 1,665 — Ones:
Hundreds: 7 + 8 + 1 = 16, write 6 carry 1. Thousands: 1.

co m
m .
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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

(d) 1,234 + 1,234 = 2,468 — adding a number to itself is doubling. 4 + 4 = 8, 3 + 3 = 6, 2 + 2 = 4, 1
+ 1 = 2.
(e) 56 + 56,789 = 56,845 — line up from the right: 6 under 9, 5 under 8. Ones: 9 + 6 = 15, write 5
carry 1. Tens: 8 + 5 + 1 = 14, write 4 carry 1. Hundreds: 7 + 1 = 8. Thousands: 6. Ten-thousands: 5.
(f) 777 + 77,777 = 78,554 — Ones: 7 + 7 = 14, write 4 carry 1. Tens: 7 + 7 + 1 = 15, write 5 carry 1.
Hundreds: 7 + 7 + 1 = 15, write 5 carry 1. Thousands: 7 + 1 = 8. Ten-thousands: 7.
(f, second one) 5,922 + 9,221 = 15,143 — Ones: 2 + 1 = 3. Tens: 2 + 2 = 4. Hundreds: 9 + 2 = 11,
write 1 carry 1. Thousands: 5 + 9 + 1 = 15, write 5 carry 1. Ten-thousands: 1.
(g) 4,321 + 8,765 = 13,086 — Ones: 1 + 5 = 6. Tens: 2 + 6 = 8. Hundreds: 3 + 7 = 10, write 0 carry
1. Thousands: 4 + 8 + 1 = 13, write 3 carry 1. Ten-thousands: 1.
(h) 50,050 + 55,000 = 1,05,050 — Ones: 0. Tens: 5 + 0 = 5. Hundreds: 0. Thousands: 0 + 5 = 5. Ten-
thousands: 5 + 5 = 10, write 0 and carry 1 to the lakhs place.

One thing to notice in the book: The English book prints the letter (f) twice in this
list — once for 777 + 77,777 and again for 5,922 + 9,221. There are nine sums in all,
so read the last three as (g), (h) and (i). The Hindi book labels all nine correctly, from
(क) to (झ).

Q2 Subtract. (a) 458 – 226 (b) 7,777 – 4,449 (c) 65,447 – 47,299 (d) 1,234 – 123 (e) 12,345 –
1,234 (f) 56,789 – 56 (f) 87,326 – 11,111 (g) 878 – 52 (h) 749 – 222

SUBTRACTION ANSWER CHECK BY ADDING BACK

(a) 458 − 226 232 232 + 226 = 458 ✓

(b) 7,777 − 4,449 3,328 3,328 + 4,449 = 7,777 ✓

(c) 65,447 − 47,299 18,148 18,148 + 47,299 = 65,447 ✓

(d) 1,234 − 123 1,111 1,111 + 123 = 1,234 ✓

(e) 12,345 − 1,234 11,111 11,111 + 1,234 = 12,345 ✓

(f) 56,789 − 56 56,733 56,733 + 56 = 56,789 ✓

(f) 87,326 − 11,111 76,215 76,215 + 11,111 = 87,326 ✓

(g) 878 − 52 826 826 + 52 = 878 ✓

(h) 749 − 222 527 527 + 222 = 749 ✓

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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

(a) 458 − 226 = 232 — Ones: 8 − 6 = 2. Tens: 5 − 2 = 3. Hundreds: 4 − 2 = 2. No regrouping.
(b) 7,777 − 4,449 = 3,328

1. Ones: 7 is smaller than 9. Take 1 ten. Ones become 17, tens become 6. 17 − 9 = 8.
2. Tens: 6 − 4 = 2.
3. Hundreds: 7 − 4 = 3.
4. Thousands: 7 − 4 = 3.

(c) 65,447 − 47,299 = 18,148

1. Ones: 7 is smaller than 9. Take 1 ten. Ones become 17, tens become 3. 17 − 9 = 8.
2. Tens: 3 is smaller than 9. Take 1 hundred. Tens become 13, hundreds become 3. 13 − 9 = 4.
3. Hundreds: 3 − 2 = 1.
4. Thousands: 5 is smaller than 7. Take 1 ten-thousand. Thousands become 15, ten-thousands
become 5. 15 − 7 = 8.
5. Ten-thousands: 5 − 4 = 1.

(d) 1,234 − 123 = 1,111 — line up from the right. Ones: 4 − 3 = 1. Tens: 3 − 2 = 1. Hundreds: 2 − 1
= 1. Thousands: 1.
(e) 12,345 − 1,234 = 11,111 — Ones: 5 − 4 = 1. Tens: 4 − 3 = 1. Hundreds: 3 − 2 = 1. Thousands: 2
− 1 = 1. Ten-thousands: 1.
(f) 56,789 − 56 = 56,733 — the 56 lines up under the last two digits. Ones: 9 − 6 = 3. Tens: 8 − 5 =
3. Hundreds: 7. Thousands: 6. Ten-thousands: 5.
(f, second one) 87,326 − 11,111 = 76,215 — Ones: 6 − 1 = 5. Tens: 2 − 1 = 1. Hundreds: 3 − 1 = 2.
Thousands: 7 − 1 = 6. Ten-thousands: 8 − 1 = 7.
(g) 878 − 52 = 826 — Ones: 8 − 2 = 6. Tens: 7 − 5 = 2. Hundreds: 8.
(h) 749 − 222 = 527 — Ones: 9 − 2 = 7. Tens: 4 − 2 = 2. Hundreds: 7 − 2 = 5.

One thing to notice in the book: As in the addition list, the letter (f) is printed twice
here in the English book. There are nine subtractions in all. The Hindi book labels
them (क) to (झ) with no repeat.

Q3 Ambrish saved ₹92,375 over a year to buy cows and goats. He buys a cow for ₹26,000
and a goat for ₹17,000. He also buys a milking machine for ₹19,873. Does he have
enough money to buy these? How much more or less does he have than he needs?

Yes, he has enough. He is left with ₹29,502 after all three purchases.
Step 1 — find what everything costs together.

1. 26,000 + 17,000 = 43,000 (cow and goat)
2. 43,000 + 19,873 = 62,873 (add the milking machine)

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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

Total cost = ₹62,873

Step 2 — compare with his savings.

1. He saved ₹92,375. The cost is ₹62,873.
2. 92,375 is bigger than 62,873, so he can afford everything.

Step 3 — find how much is left.

1. Ones: 5 − 3 = 2.
2. Tens: 7 − 7 = 0.
3. Hundreds: 3 is smaller than 8. Take 1 thousand. Hundreds become 13, thousands become 1.
13 − 8 = 5.
4. Thousands: 1 is smaller than 2. Take 1 ten-thousand. Thousands become 11, ten-thousands
become 8. 11 − 2 = 9.
5. Ten-thousands: 8 − 6 = 2.

92,375 − 62,873 = ₹29,502 more than he needs

Check it yourself: 62,873 + 29,502 = 92,375 ✓

Q4 A factory produces 54,000 nuts and bolts in a day. An order is placed for 85,300 nuts
and bolts. How many more nuts and bolts does the factory need to produce to
complete the order?

The factory needs 31,300 more nuts and bolts.
The order asks for 85,300. One day's work gives 54,000. The words how many more mean we
subtract.

Still needed = 85,300 − 54,000

1. Ones: 0 − 0 = 0.
2. Tens: 0 − 0 = 0.
3. Hundreds: 3 − 0 = 3.
4. Thousands: 5 − 4 = 1.
5. Ten-thousands: 8 − 5 = 3.

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Class 5 Maths Chapter 4 We the Travellers—II AglaSem · NCERT Solutions

85,300 − 54,000 = 31,300 nuts and bolts

Check it yourself: 54,000 + 31,300 = 85,300 ✓ — exactly the size of the order.

Q5 Virat Kohli has scored 27,599 runs. He has 6,758 runs less than Sachin Tendulkar.
How many runs has Sachin Tendulkar scored?

Sachin Tendulkar has scored 34,357 runs.
Read the sentence carefully. Virat has 6,758 runs less than Sachin. So Sachin has more, and to
reach Sachin's score we must add 6,758 to Virat's.

Sachin's runs = 27,599 + 6,758

1. Ones: 9 + 8 = 17. Write 7, carry 1.
2. Tens: 9 + 5 + 1 = 15. Write 5, carry 1.
3. Hundreds: 5 + 7 + 1 = 13. Write 3, carry 1.
4. Thousands: 7 + 6 + 1 = 14. Write 4, carry 1.
5. Ten-thousands: 2 + 1 = 3.

Sachin Tendulkar's runs = 34,357 runs

Why we add and not subtract: "Virat has fewer runs than Sachin" tells us Sachin is
the bigger of the two. Subtracting would have made Sachin smaller, which cannot be
right. Whenever a question says less than, first decide which person or thing is the
bigger one.

Check it yourself: 34,357 − 6,758 = 27,599 ✓ — back to Virat's score.

Chapter at a glance
Adding with place value. Write ones under ones, tens under tens, hundreds under
hundreds. Add one column at a time, starting from the ones. If a column gives 10 or more,

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Class 5 Maths Chapter 4 We the Travellers—II
a g l AglaSem · NCERT Solutions

keep the ones digit there and carry the ten to the next column. Example: 28 + 75. Ones: 8 +
co m
se
5 = 13, so write 3 and carry 1. Tens: 2 + 7 + 1 = 10, so write 0 and carry 1. Hundreds: 1.
m.
o m l a
gto regroup — break 1
.c with place value. Same columns, but now you may have
Answer 103.

m a
e 10 ones, or 1 hundred into 10 tens. Example: 65 − 18. Ones: 5 is smaller than 8, so
sinto
Subtracting
l a
ag
ten
take 1 ten from 6 tens. Now the ones are 15 and the tens are 5. 15 − 8 = 7. Tens: 5 − 1 = 4.

m
Answer 47.

. co ag
m
Addition and subtraction are two sides of one fact. If 46 + 21 = 67, then 67 − 21 = 46 and

l a se
67 − 46 = 21. This is why you can always check a subtraction by adding: 65 − 18 = 47, and 18
ag started with means your answer is right.
+ 47 = 65. Getting the number you
Consecutive numbers are numbers that follow one another with no gaps — 5, 6, 7, 8. Their

. c om
sums make neat patterns. Add 3 of them and the answer is 3 times the middle number: 67 +

s e m 24 + 25 + 26 +
68 + 69 = 3 × 68 = 204. Add 4 of them and the answer is 2 times (first + last):

c m= 102.
27 = 2 ×o51
. a gla
a s em the run up by 1 and the sum grows by the count. 1 + 2 + 3 = 6 and 2 + 3 + 4 = 9 — a
Slide

a gl jump of 3, because there are 3 numbers and each one grew by 1. For 5 consecutive
numbers the jump is 5; for 6 it is 6.

m a s
coFor agl
Quick sums. To find what adds to 100, first find what adds to 99 — take each digit away
from 9, with no borrowing — then add .1.
e m 32: 99 − 32 = 67, and 67 + 1 = 68. So 32 + 68 =

g l as (use 999) and for 10,000 (use 9,999).
100. The same trick works for 1,000
a 9, subtract 10 and put 1 back: 67 − 9 = 67 − 10 + 1 = 58. To
Quick differences. To subtract

m
subtract 99, subtract 100 and put 1 back: 187 − 99 = 187 − 100 + 1 = 88.

. co
m
Even and odd. Line the counters up in pairs. An even number pairs off completely; an odd

o m l a se
number always leaves one counter alone. So even + even = even, odd + odd = even (the two

m .c counters join up), and odd + even = odd (one counter isagstill alone).
se Palindrome numbers read the same both ways — 22, 363, 404, 8,558. Between 100 and
lonely

a
agl 200 there are ten of them: 101, 111, 121, … 191.

se m
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