NCERT Solutions for Class 5 Maths Mela Chapter 4 We the Travellers – II

Page 42

Making Sums Equal

In each of the following, there are two groups of numbers. Look carefully at the numbers in each group and their sums. Interchange pairs of numbers between the two groups to make their sums equal. Try to do this using the least number of moves. You could write each number on a small piece of paper.

Answer:

Page 43

Let Us Solve

Add the following numbers. Wherever possible, find easier ways to add the pairs of numbers.
1. 15 + 79
2. 46 + 99
3. 38 + 35
4. 5 + 89
5. 76 + 28
6. 69 + 20
Answer:

Page 44

Relationship Between Addition and Subtraction

1. Find the relationship between the numbers in the given statements and fill in the blanks appropriately.
(a) If 46 + 21 = 67, then,
67 – 21 = _______.
67 – 46 = _______.
See SolutionIf 46 + 21 = 67, then,
67 – 21 = “46”.
67 – 46 = “21”.

(b) If 198 – 98 = 100, then,
100 + _______ = 198.
198 – _______ = 98.
See SolutionIf 198 – 98 = 100, then,
100 + “98” = 198.
198 – “100” = 98.

(c) If 189 + 98 = 287, then,
287 – 98 = _______.
287 – 189 = _______.
See SolutionIf 189 + 98 = 287, then,
287 – 98 = “189”.
287 – 189 = “98”.

(d) If 872 – 672 = 200, then,
200 + _______ = 872.
872 – _______ = 672.
See SolutionIf 872 – 672 = 200, then,
200 + “672” = 872.
872 – “200” = 672.

2. In each of the following, write the subtraction and addition sentences that follow from the given sentence.
(a) If 78 + 164 = 242, then,
See Solution242 – 78 = 164
242 – 164 = 78

(b) If 462 + 839 = 1301, then,
See Solution1301 – 462 = 839
1301 – 839 = 462

(c) If 921 – 137 = 784, then,
See Solution921 – 784 = 137
137 + 784 = 921

(d) If 824 – 234 = 590, then,
See Solution824 – 590 = 234
590 + 234 = 824

Page 45

Let Us Solve

1. What is the difference between 82 and 37?

2. 57 – 11 = –––––––––––
3. 23 – 19 = –––––––––––
4. 49 – 21 = –––––––––––
5. 56 – 18 = –––––––––––
6. 93 – 35 = –––––––––––
7. 84 – 23 = –––––––––––
8. 70 – 43 = –––––––––––
9. 65 – 47 = –––––––––––
Answer:

Page 45

Sums of Consecutive Numbers

1. In each of the boxes above, state whether the sums are even or odd. Explain why this is happening.
See SolutionIn box 1, sums are odd.
Explanation: Sum of a odd number and an even number is always odd number.
In box 2, sums are even.
Explanation: Sum of two odd numbers and even number is always even.
In box 3, sums are even.
Explanation: Sum of two odd numbers and even numbers is always even.

2. What is the difference between two successive sums in each box? Is it the same throughout?
See SolutionIn box 1, the difference is 2 and same throughout.
In box 2, the difference is 3 and same throughout.
In box 3, the difference is 4 and same throughout.

3. What will be the difference between two successive sums for —
(a) 5 consecutive numbers (b) 6 consecutive numbers
See Solution(a) Sum of 5 consecutive numbers.
1 + 2 + 3 + 4 + 5 = 15
2 + 3 + 4 + 5 + 6 = 20
3 + 4 + 5 + 6 + 7 = 25
4 + 5 + 6 + 7 + 8 = 30
The difference is 5 and same throughout.
(b) Sum of 5 consecutive numbers.
1 + 2 + 3 + 4 + 5 + 6 = 21
2 + 3 + 4 + 5 + 6 + 7 = 27
3 + 4 + 5 + 6 + 7 + 8 = 33
4 + 5 + 6 + 7 + 8 + 9 = 39
The difference is 6 and same throughout.

Page 48

Let Us Solve

1. 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
Answer:

2. 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.

See SolutionSolving it step by step:
Delhi → Mumbai = 1,600 km
Mumbai → Goa = 590 km
Goa → Hyderabad = 670 km
Hyderabad → Puri = 1,055 km
Now adding them carefully:
1,600 + 590 + 670 + 1,055 = 3,915 km
So, Nazrana and her friends travelled a total of 3,915 km on their great Indian road trip.

3. 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
See SolutionListing all pair sums:
► 5,205 + 6,220 = 11,425
► 5,205 + 7,095 = 12,300
► 5,205 + 8,455 = 13,660
► 5,205 + 4,840 = 10,045
► 6,220 + 7,095 = 13,315
► 6,220 + 8,455 = 14,675
► 6,220 + 4,840 = 11,060
► 7,095 + 8,455 = 15,550
► 7,095 + 4,840 = 11,935
► 8,455 + 4,840 = 13,295
Now, comparing with the given numbers:
(a) 10,000
Closest is 5,205 + 4,840 = 10,045 (just +45 away).
(b) 15,000
Closest is 7,095 + 8,455 = 15,550 (difference 550).
(c) 13,000
Options: 13,295 (difference 295), 13,315 (difference 315), 12,300 (difference 700).
So, the closest is 8,455 + 4,840 = 13,295 (difference 295).
(d) 16,000
Closest is 7,095 + 8,455 = 15,550 (difference 450).

Page 50

Let Us Solve

1. 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
Answer:

2. 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?

See SolutionInitial Amount at start: ₹12,540
Spent on food (−3,275): ₹12,540 − ₹3,275 = ₹9,265
Uncle’s gift (+4,900): ₹9,265 + ₹4,900 = ₹14,165
Delhi ticket (−2,645): ₹14,165 − ₹2,645 = ₹11,520
Souvenirs (−1,275): ₹11,520 − ₹1,275 = ₹10,245
Mary is left with ₹10,245 at the end of her Delhi trip.

3. Members of a school council have raised ₹70,500. They plan to setup 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.
See SolutionEstimation
Maths Lab ≈ ₹40,000
Books ≈ ₹10,000
Sports ≈ ₹20,000
Total estimate ≈ ₹70,000
Since the council has raised ₹70,500, the estimate shows that they have just about enough money.

(b) Check your estimate with calculations.
See SolutionExact Calculation
Maths Lab = ₹39,785
Books = ₹9,545
Sports = ₹19,548
Total amount = 39,785 + 9,545 + 19,548 = 68,878
Money raised = ₹70,500
Difference = 70,500 − 68,878 = ₹1,622
Yes, the school council has enough money. After making all the purchases, they will still have ₹1,622 left.

4. 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?
See SolutionTotal weight loaded
Cement = 3,675 kg
Steel = 2,850 kg
Total = 3,675 + 2,850 = 6,525 kg
So, the truck is carrying 6,525 kg.

(b) How much more weight can the truck carry before reaching its maximum capacity?
See SolutionRemaining capacity
Maximum = 8,250 kg
Loaded = 6,525 kg
Left weight = 8,250 − 6,525 = 1,725 kg
So, the truck can still carry 1,725 kg more before it reaches full capacity.

Page 53

Let Us Think and Solve

1. Nitin likes numbers that read the same when read from left to right or from right to left. Such numbers are called palindrome numbers. The
numbers 22, 363, 404, and 8,558 are some examples.
List all palindrome numbers between 100 and 200.
List all palindrome numbers between 900 and 1,200.
List all palindrome numbers between 25,000 and 27,000.
See SolutionBetween 100 and 200
A 3-digit palindrome looks like aba.
So possible numbers: 101, 111, 121, 131, 141, 151, 161, 171, 181, 191
Between 900 and 1,200
Palindromes here are 3-digit and 4-digit.
3-digit: 909, 919, 929, 939, 949, 959, 969, 979, 989, 999
4-digit: 1001, 1111
So, total list = 909, 919, 929, 939, 949, 959, 969, 979, 989, 999, 1001, 1111
3. Between 25,000 and 27,000
Here, 5-digit palindromes look like ab c ba.
Starting with 25…52 → 25052, 25152, 25252, 25352, 25452, 25552, 25652, 25752, 25852, 25952
Starting with 26…62 → 26062, 26162, 26262, 26362, 26462, 26562, 26662, 26762, 26862, 26962
Total palindromes in this range = 20.

2. 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.

Answer:

This time, fill the grid such that each row and column has numbers in decreasing (dec) order.

Answer:

Now, fill the grids below with numbers (1–9) based on the inc increasing) and dec (decreasing) conditions, as indicated below.

Answer:

Page 54

Even and Odd Numbers

1. 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
Answer:

2. Observe the given arrangement.

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?
See SolutionPaired arrangement for 18
Pairs:1 + 17, 2 + 16, 3 + 15, 4 + 14, 5 + 13, 6 + 12, 7 + 11, 8 + 10, 9 + 9.
Now, add 2 to 18 → total becomes 20.
New pairs: 1 + 19, 2 + 18, 3 + 17, 4 + 16, 5 + 15, …, 10 + 10.
Changes: The partner of each number increases by 2.
Unchanged: The number of pairs and the symmetry (first goes up, second goes down).
Paired arrangement for 23
Pairs: 1 + 22, 2 + 21, 3 + 20, 4 + 19, …, 11 + 12.
Now, add 2 to 23 → total becomes 25.
New pairs: 1 + 24, 2 + 23, 3 + 22, …, 12 + 13.
Changes: Again, the second number in each pair increases by 2.
Unchanged: The pattern of pairing (smallest with largest, next-smallest with next-largest) remains exactly the same.
So, adding 2 to the target number increases each partner number by 2, but the structure and balance of the arrangement remain unchanged.

3. 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.
(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.
See SolutionGiven pair: 12 and 6
Both are even numbers.
Their sum = 12 + 6 = 18, which is also even.
Now 5 more pairs of even numbers:
• 2 + 4 = 6 (even)
• 8 + 10 = 18 (even)
• 14 + 20 = 34 (even)
• 26 + 44 = 70 (even)
• 100 + 222 = 322 (even)
Given pair: 13 and 9
We notice that the sum of two even numbers is always even.
Reason:
An even number can be written as 2n (n is a whole number).
Another even number = 2m.
Sum = 2n + 2m = 2( n + m), which is clearly divisible by 2.
Hence, the result is always even.

(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.
See SolutionBoth are odd numbers.
Their sum = 13 + 9 = 22, which is even.
Now 5 more pairs of odd numbers:
• 1 + 3 = 4 (even)
• 5 + 7 = 12 (even)
• 11 + 15 = 26 (even)
• 21 + 23 = 44 (even)
• 101 + 111 = 212 (even)
We notice that the sum of two odd numbers is always even.
Reason:
An odd number can be written as (2n + 1).
Another odd number = (2m + 1).
Sum = (2n + 1) + (2m + 1) = 2(n + m + 1), which is divisible by 2.
Hence, the result is always even.

(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.
See SolutionGiven pair: 7 (odd) and 12 (even)
7 + 12 = 19, which is odd.
Now 5 more pairs (odd + even):
• 3 + 8 = 11 (odd)
• 5 + 14 = 19 (odd)
• 9 + 20 = 29 (odd)
• 15 + 32 = 47 (odd)
• 101 + 200 = 301 (odd)
We notice that the sum of an odd number and an even number is always odd.
Reason:
Odd number = 2n + 1
Even number = 2m
Sum = (2n + 1) + 2m = 2( n + m) + 1, which is not divisible by 2.
So the result is always odd.

Page 54

Let Us Think

1. 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?
See SolutionCoins and Quantity:
₹1 coins = 8 (total ₹8)
₹2 coins = 9 (total ₹18)
₹5 coins = 5 (total ₹25)
Total Amount = ₹51
Target = ₹38
Combination 1:
₹5 × 5 = ₹25
₹2 × 6 = ₹12
₹1 × 1 = ₹1
Total = 25 + 12 + 1 = ₹38
Combination 2:
₹5 × 4 = ₹20
₹2 × 9 = ₹18
₹1 × 0 = ₹0
Total = 20 + 18 = ₹38
Combination 3:
₹5 × 3 = ₹15
₹2 × 8 = ₹16
₹1 × 7 = ₹7
Total = 15 + 16 + 7 = ₹38
So, the possible ways Jincy can pay ₹38:
• 5 coins of ₹5, 6 coins of ₹2, 1 coin of ₹1
• 4 coins of ₹5, 9 coins of ₹2
• 3 coins of ₹5, 8 coins of ₹2, 7 coins of ₹1

2. Raghu is fond of his grandfather’s torch. He starts playing with it. He presses the switch once and the light turns ON. He presses it a second time and the light turns 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 press? How do you know?
For what number of presses will the torch be ON? For what number of presses of the switch will the torch be OFF?
See SolutionPattern of pressing
• 1st press → ON
• 2nd press → OFF
• 3rd press → ON
• 4th press → OFF
• 5th press → ON … and so on.
So the torch switches state every press.
Here, we observe that:
Odd presses = ON
Even presses = OFF
So, 23 is an odd number, hence torch will be ON.
General Rule:
For all odd numbers of presses (1, 3, 5, 7, …) → Torch will be ON
For all even numbers of presses (2, 4, 6, 8, …) → Torch will be OFF

3. Mountain climbing
Priyanka Mohite is the first Indian woman to climb five Himalayan peaks above 8,000 metres. In addition to that, she has also climbed mountain peaks in other parts of the world. Read the table below and answer the questions that follow.

Mountain Range – Height (in metres) – Climbed in the Year
Mount Kanchenjunga (India and Nepal border) – 8,586 – 2022
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?
See SolutionHighest peak she climbed, as per the table given here:
Mount Everest = 8,848 m
This is the tallest.

(b) What is the difference in height between the highest and lowest peaks she has climbed, as per the table.
See SolutionDifference between highest and lowest peaks
Highest = Everest = 8,848 m
Lowest = Mount Elbrus = 5,642 m
Difference = 8,848 − 5,642 = 3,206 m

(c) What is the difference between heights of Mount Elbrus and Mount Kanchenjunga?
See SolutionSee SolutionDifference between Mount Elbrus and Mount Kanchenjunga
Kanchenjunga = 8,586 m
Elbrus = 5,642 m
Difference = 8,586 − 5,642 = 2,944 m

(d) If Priyanka was 20 years old when she summited Mount Everest in 2013, in which year was she born?
See SolutionYear of birth:
She climbed Everest in 2013 at age 20.
Year of birth = 2013 − 20 = 1993.

Page 56

Math Metric Mela

A grand Math Metric Mela was held at the district level to celebrate young math whizzes. Every participating student was to receive a certificate of participation. The organisers got certificates printed for each district before the Mela. The number of certificates printed and the number of students who attended the competition in each district are as follows.

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.
See SolutionDistrict-wise comparison
► Chittoor, A.P.
Certificates printed = 18,225
Students attended = 18,104
Extra certificates = 18,225 − 18,104 = 121
So, these are sufficient (121 extra).
► Jaunpur, U.P.
Certificates printed = 19,043
Students attended = 19,265
Shortage = 19,265 − 19,043 = 222
These are not sufficient (short by 222).
► Raigad, Maharashtra
Certificates printed = 20,863
Students attended = 19,974
Extra certificates = 20,863 − 19,974 = 889
So, these are sufficient (889 extra).

Page 56

Let Us Do

1. Add.
(a) 2,009 + 7,388
(b) 26,444 + 71,111
(c) 777 + 888
(d) 1,234 + 1,234
(e) 56 + 56,789
(f ) 777 + 77,777
(f) 5,922 + 9,221
(g) 4,321 + 8,765
(h) 50,050 + 55,000
Answer:

2. 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
Answer:

3. 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?
See SolutionAmbrish’s savings = ₹92,375
His purchases:
Cow = ₹26,000
Goat = ₹17,000
Milking machine = ₹19,873
Total Amount = ₹26,000 + ₹17,000 + ₹19,873 = ₹62,873
Compare with his savings:
Savings = 92,375
Spent = 62,873
Leftover = 92,375 − 62,873 = ₹29,502
So, yes, Ambrish has enough money. After buying everything, he still has ₹29,502 left.

4. 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?
See SolutionFactory’s production in one day = 54,000 nuts and bolts
Order placed = 85,300 nuts and bolts
Difference (need more) = 85,300 − 54,000 = 31,300
So, the factory needs to produce 31,300 more nuts and bolts to complete the order.

5. Virat Kohli has scored 27,599 runs. He has 6,758 runs less than Sachin Tendulkar. How many runs has Sachin Tendulkar scored?
See SolutionVirat Kohli’s runs = 27,599
He has 6,758 runs less than Sachin
So Sachin’s runs = Virat’s runs + 6,758
= 27,599 + 6,758 = 34,357
So, Sachin Tendulkar has scored 34,357 runs.