NCERT Solutions for Class 5 Mathematics Math-Mela Chapter 4 We the Travellers

NCERT solutions for Class 5 Mathematics Chapter We the Travellers–II – complete answers & explanations

This blog provides NCERT solutions for Class 5 Mathematics Chapter We the Travellers–II in a clear and student-friendly way. This chapter focuses on addition, subtraction, number patterns, and logical thinking through real-life situations. It is important because it helps students understand numbers deeply and improve problem-solving skills. This blog includes complete NCERT solutions strictly based on the worksheet, ensuring accuracy and alignment with NCERT standards.

What this NCERT chapter covers?

1. Understanding addition and subtraction relationships
2. Solving problems using fuel arithmetic methods
3. Working with consecutive numbers and patterns
4. Learning shortcuts to find sums without direct addition
5. Practicing large number operations
6. Understanding even and odd numbers
7. Identifying palindrome numbers and patterns
8. Solving real-life word problems and logical reasoning questions

How to use these NCERT solutions?

1. First, students should try solving all questions on their own
2. Use these solutions to check answers and correct mistakes
3. Follow the exact worksheet order while practicing
4. Parents and teachers can guide students step-by-step
5. Use these solutions during revision for better understanding
6. Practice regularly to improve speed and accuracy

Important tips & tricks for students

1. Always write steps clearly in addition and subtraction
2. Be careful with large numbers and place values
3. Use shortcut methods only after understanding concepts
4. Double-check answers to avoid calculation mistakes
5. Practice patterns to improve logical thinking
6. Pay attention to instructions in activity-based questions

NCERT solutions – complete answer key

Making sums equal

(a) Both sums become 20
(b) Both sums become 41
(c) Both sums become 72
(d) Both sums become 322

Fuel arithmetic

1. 15 + 79 = 94
2. 46 + 99 = (46 + 100) − 1 = 145
3. 38 + 35 = 73
4. 5 + 89 = 94
5. 76 + 28 = 104
6. 69 + 20 = 89

Relationship between addition and subtraction

(a) If 46 + 21 = 67 67 − 21 = 46 and 67 − 46 = 21
(b) If 198 − 98 = 100 100 + 98 = 198 and 198 − 100 = 98
(c) If 189 + 98 = 287 287 − 98 = 189 and 287 − 189 = 98
(d) If 872 − 672 = 200 200 + 672 = 872 and 872 − 200 = 672

(a) 242 – 164 = 78
1301 – 462 = 839
(b) 1301 – 839 = 462
242 – 78 = 164

More fuel arithmetic

1. 82 − 37 = 45 (Check: 37 + 45 = 82)
2. 57 − 11 = 46
3. 23 − 19 = 4
4. 49 − 21 = 28
5. 56 − 18 = 38
6. 93 − 35 = 58
7. 84 − 23 = 61
8. 70 − 43 = 27
9. 65 − 47 = 18

Sums of consecutive numbers

Sum of 2 consecutive numbers is always odd because one number is even and the other is odd. Their sum becomes odd.

Sum of 3 consecutive numbers is always a multiple of 3 because they include one number divisible by 3.

Sum of 4 consecutive numbers is always even because two even and two odd numbers together give an even sum.

The difference between successive sums remains constant.

For 2 consecutive numbers, difference = 2.
For 3 consecutive numbers, difference = 3.
For 4 consecutive numbers, difference = 4.

This happens because each new sum increases by adding the next number and removing the first

(a) 5 consecutive numbers → 5
(b) 6 consecutive numbers → 6

Find the sum without adding directly

(a) 67 + 68 + 69 The middle number is 68. Sum = 68 × 3 = 204
(b) 24 + 25 + 26 + 27 Middle pair = (25 + 26)/2 = 25.5 → Sum = 25.5 × 4 = 102
(c) 48 + 49 + 50 + 51 + 52 Middle number = 50 → Sum = 50 × 5 = 250
(d) 237 + 238 + 239 + 240 + 241 + 242 Middle pair = (239 + 240)/2 = 239.5 → Sum = 239.5 × 6 = 1437

1,855 km + 1,862 km = 3,717 km
2. 21,880 km + 38,900 km = 60,780 km
3. 267 + 54 = 321

₹21,880 + ₹38,900 = ₹60,780

Let us solve (addition)

(a) 605
(b) 13,579
(c) 135
(d) 59,010
(e) 9,667
(f) 19,239

Total distance travelled

2. 590 + 1,055 + 670 + 1,600 = 3,915 km

3. (a) 10,000 → 5,205 + 4,840
(b) 15,000 → 7,095 + 8,455
(c) 13,000 → 6,220 + 7,095
(d) 16,000 → 8,455 + 7,095

Let us solve (subtraction)

(a) 2,356
(b) 3,544
(c) 5,000
(d) 111
(e) 77,000
(f) 74

Mary’s train journey

Initial money = 12,540
Less: Spent = 3,275
Ticket = 2,645
Souvenirs = 1,275 = 7,295
Remaining = 5,345
Add: Gift received = 4,900
Final money left = 10,245

Remaining distance = 1,617 km
4922 km
459

School council problem

(a) estimated total expense ≈ 40,000 + 10,000 + 20,000 = 70,000. Money available = 70,500. So, estimate shows enough money.
(b) Exact calculation Total expense = 39,785 + 9,545 + 19,548 = 68,878. Money left = 70,500 − 68,878 = 1,622.

(a) 6,525 kg
(b) 1,725 kg

Quick sums and differences

32 + 68 = 100
59 + 41 = 100

877 + 123 = 1,000 and 1000 − 877 = 123
666 + 334 = 1,000 and 1000 − 666 = 3343.
4,103 + 5,897 = 10,000 and 10000 − 4103 = 5897
5,555 + 4,445 = 10,000 and 10000 − 5555 = 4445

Will this method work if the units digit is 0?

Yes, the method still works.
When the units digit is 0, we can still find the missing number by subtracting from the nearest base (10, 100, 1000, etc.).
For example, 820 + ___ = 1000 → 1000 − 820 = 180. The method works because it depends on place value, not just the units digit.

Solve the following

820
240
600

Subtract 9 or 99 method explanation

To subtract 9, subtract 10 and then add 1.
Example: 67 − 9 = (67 − 10) + 1 = 58.

To subtract 99, subtract 100 and then add 1.
Example: 187 − 99 = (187 − 100) + 1 = 88. This method makes calculation faster.

(a) 67 − 9 = (67 − 10) + 1 = 58
(b) 87 − 9 = (87 − 10) + 1 = 78
(c) 144 − 9 = (144 − 10) + 1 = 135
(d) 187 − 99 = (187 − 100) + 1 = 88
(e) 247 − 99 = (247 − 100) + 1 = 148
(f) 736 − 99 = (736 − 100) + 1 = 637

(a) 32 − 23 = 9 and 32 − 23 = 9
(b) 56 − 47 = 9 and 56 − 47 = 9
(c) 877 − 778 = 99 and 778 − 877 = 9
(d) 666 − 567 = 99 and 567 − 666 = 99

Let us think and solve

Palindrome numbers read the same forwards and backwards. Example: 121, 131. They are formed by keeping digits symmetric.

Let Us Think (Pattern) Add 2 to 18 and 23 Adding 2 does not change the arrangement pattern. Only numbers increase, but the structure remains same.

Palindrome numbers between 100 and 200:
101, 111, 121, 131, 141, 151, 161, 171, 181, 191

Palindrome numbers between 900 and 1,200:
909, 919, 929, 939, 949, 959, 969, 979, 989, 999, 1001, 1111

Palindrome numbers between 25,000 and 27,000:
25052, 25152, 25252, 25352, 25452, 25552, 25652, 25752, 25852, 25952
26062, 26162, 26262, 26362, 26462, 26562, 26662, 26762, 26862, 26962

Grid activity

Increasing Order

9 8 7
6 5 4
3 2 1

Decreasing Order

1 2 3
4 5 6
7 8 9

Columns decreasing, rows increasing

7 8 9
4 5 6
1 2 3

9 8 1
7 5 2
6 4 3

Mixed: rows → dec, dec, inc | columns → dec, dec, inc

Even and odd numbers

Even numbers:
498
724
100
846
222

Observation explanation
Even + Even = Even because both numbers are divisible by 2.
Odd + Odd = Even because two odd numbers together make a multiple of 2.
Odd + Even = Odd because only one number contributes a factor of 2.

Add 2 to 18 → arrangement pattern remains same
Add 2 to 23 → arrangement pattern remains same

(a) Pair Sum
12 + 6 18
8 + 4 12
10 + 2 12
14 + 6 20
16 + 8 24

Observation The sum of two even numbers is always even.

(b) Pair Sum
13 + 9 22
7 + 5 12
11 + 3 14
15 + 1 16
17 + 9 26

Observation The sum of two odd numbers is always even.

Observation The sum of one odd and one even number is alwaysodd.

Let us think

₹5 × 5 = 25
₹2 × 9 = 18
₹1 × 8 = 8
Total = ₹41

Torch after 23rd press → ON

ON for odd number of presses
OFF for even number of presses

(a) Mount Everest
(b) 8,848 − 5,642 = 3,206 m
(c) 8,586 − 5,642 = 2,944 m
(d) She was born in 1993

Chittoor Extra certificates = 18,225 − 18,104 = 121
JJaunpur Shortage = 19,265 − 19,043 = 222 short
Raigad Extra = 20,863 − 19,974 = 889 extra

Let us do

Add

(a) 2,009 + 7,388 9,397
(b) 26,444 + 71,111 97,555
(c) 777 + 888 1,665
(d) 1,234 + 1,234 2,468
(e) 56 + 56,789 56,845
(f) 777 + 77,777 78,554
(g) 5,922 + 9,221 15,143
(h) 4,321 + 8,765 13,086
(i) 50,050 + 55,000 1,05,050

Subtract

(a) 458 – 226 232
(b) 7,777 – 4,449 3,328
(c) 65,447 – 47,299 18,148
(d) 1,234 – 123 1,111
(e) 12,345 – 1,234 11,111
(f) 56,789 – 56 56,733
(g) 87,326 – 11,111 76,215
(h) 878 – 52 826
(i) 749 – 222 527

Total money saved 92,375
Cost of cow + goat + machine 26,000 + 17,000 + 19,873 = 62,873
Money left 92,375 – 62,873 = 29,502
Final Answer Yes, he has enough money. ₹29,502 left.

Order quantity 85,300
Produced per day 54,000
More needed 85,300 – 54,000 = 31,300
Final Answer 31,300 more nuts and bolts needed.

Virat's runs 27,599
Difference 6,758
Sachin's runs 27,599 + 6,758 = 34,357
Final Answer Sachin Tendulkar scored 34,357 runs.

Why NCERT solutions help students?

NCERT solutions help students understand concepts clearly and prepare effectively for exams. They ensure answers are accurate and follow NCERT guidelines. These solutions build confidence and improve problem-solving skills.

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NCERT Solutions for Class 5 Mathematics Math-Mela Chapter 4 We the Travellers – II