Students often refer to Maths Mela Class 5 Solutions Chapter 6 The Dairy Farm Question Answer NCERT Solutions to verify their answers.
Class 5 Maths Chapter 6 The Dairy Farm Question Answer Solutions
The Dairy Farm Class 5 Maths Solutions
Class 5 Maths Chapter 6 Solutions
Let Us Think (Pages 70-71)
Question 1.
The given shapes stand for numbers between 1 and 24. The same shape denotes the same number across all problems. Find the numbers hiding in all the shapes.
Solution:
Question 2.
Place the digits 2, 5, and 3 appropriately to get a product close to 100. Share your reasoning in class.
Solution:
53 × 2 = 106, which is close to 100.
Question 3.
A dairy has packed butter milk pouches in the following manner. Find the number of pouches kept in each arrangement. One is done for you.
Solution:
(b) 15 × 4 = 60
(c) 10 × 6 = 60
(d) 12 × 5 = 60 (Another group that can be made: 60 × 1 = 60)
Question 4.
Which number am I?
I am a two-digit number. Find me with the help of the following clues.
(a) I am greater than 8.
(b) I am not a multiple of 4.
(c) I am a multiple of 9.
(d) I am an odd number.
(e) I am not a multiple of 11.
(f) I am less than 50.
(g) My ones digit is even
(h) My tens digit is odd.
Did you use all the clues to find the number? Which clues did not help you in finding the number?
Solution:
I am number 27. The clues (g) & (h) did not help me.
Question 5.
Make your own numbers.
Choose any two numbers and one operation from the grid. Try to make all the numbers between 0 and 20. For example, 2 can be formed as 4 – 2.
Could you make all the numbers?
Solution:
| Number | Combination(s) |
| 0 | 3 – 3 |
| 1 | 4 – 3 |
| 2 | 4 – 2 |
| 3 | 12 ÷ 4 |
| 4 | 100 ÷ 25 |
| 5 | 10 – 5 |
| 6 | 10 – 4 |
| 7 | 12 – 5 |
| 8 | 12 – 4 |
| 9 | 12 – 3 |
| 10 | 100 ÷ 10 |
| 11 | 36 – 25 |
| 12 | 10 + 2 |
| 13 | 10 + 3 |
| 14 | 12 + 2 |
| 15 | 12 + 3 |
| 16 | 12 + 4 |
| 17 | 12 + 5 |
| 18 | 36 ÷ 2 |
| 20 | 100 ÷ 5 |
Number 19 could not be made.
Yes, it is possible to make these numbers using two operations & three numbers.
19 → 25 – 10 + 4
Numbers which could be made in more than one way are: 0, 1, 2, 3, 7, 9 and 11.
Order of Numbers in Multiplication (Page 72)
Daljeet Kaur runs a milk processing unit. She has arranged the butter packets in the following ways. Find the number of butter packets in each case. What pattern do you notice (or observe)? Discuss in class.
The number of groups and the group size are interchanged in each case above, but the total number of butter packets remain the same.
Solution:
(a) 2 × 3 = 6; 3 × 2 = 6
(b) 5 × 8 = 40; 8 × 5 = 40
(c) 10 × 5 = 50; 5 × 10 = 50
(d) 8 × 20 = 160; 20 × 8 = 160
(e) 12 × 9 = 108; 9 × 12 = 108
Question
What is 9 × 0? 0 × 9?
Solution:
Zero (0)
Patterns in Multiplication by 10s and 100s (Pages 72-73)
Question 1.
Let us revise multiplication by 10s and 100s.
a) 4 × 10 = _____
b) 20 × 10 = _____
c) 10 × 40 = _____
d) 10 × 10 = 100
e) 20 × 50 = ______
f) 80 × 10 = ______
g) 3 × 100 = 100 × 3 = 300
h) 8 × 100 = _____ = ______
i) 10 × 100 = _____ = ______
Solution:
(a) 4 × 10 = 40
(b) 20 × 10 = 200
(c) 10 × 40 = 400
(d) 10 × 10 = 100
(e) 20 × 50 = 100
(f) 80 × 10 = 800
(g) 3 × 100 = 100 × 3 = 300
(h) 8 × 100 = 100 × 8 = 800
(i) 10 × 100 = 100 × 10 = 1000
Question 2.
Find answers to the following questions. Fill in the table below and describe the pattern. Discuss in class.
(a) 100 × 90 = _____,000
(b) 400 × 10 = _____
(c) 60 × 50 = _____
(d) 30 × 20 = 600
(e) 700 × 4 = _____, _____, 00
(f) 10 × 45 = _____
Solution:
(a) 100 × 90 = 9,000
(b) 400 × 10 = 4,000
(c) 60 × 50 = 3,000
(d) 30 × 20 = 600
(e) 700 × 4 = 2,800
(f) 10 × 45 = 450
Solution:
Doubling and Halving (Pages 75-76)
Butter packets are arranged in the following ways. Let us find some strategies to calculate the total number of packets.
c) Solve the following problems like the previous ones
This halving and doubling strategy works well when we have to multiply with numbers like 5 and 25. Discuss why?
Solution:
Multiplying a number by 5 and 25 can be tricky, but the halving and doubling strategy makes it much easier because they are easily convertible to powers of ten, simplifying the multiplication process. Instead of multiplying by 5, we can think of it as multiplying by 10 and then dividing by 2. In the same way, since 25 = 100/4, we can multiply a number by 100 and then divide the result by 4 for multiplying by 25.
(d) Find the product by halving and doubling either the multiplier or the multiplicand.
1. 5 × 18
2. 50 × 28
3. 15 × 22
4. 25 × 12
5. 12 × 45
6. 16 × 45
Solution:
(e) Given 5 examples of multiplication problems where halving and doubling will help in finding the product easily. Find the products as well.
Solution:
Do it yourself.
Nearest Multiple (Page 76)
(a) 4 × 19
Solution:
(b) 14 × 21
Solution:
(c) Give 5 examples of problems where you can use the nearest multiple to find the product easily. Find the products as well.
Solution:
1. 19 × 7 = 20 × 7 – 7 = 133
2. 48 × 6 = 50 × 6 – 6 – 6 = 288
3. 99 × 8 = 100 × 8 – 8 = 792
4. 7 × 11 = 7 × 10 + 7 = 77
5. 205 × 4 = 200 × 4 + 5 × 4 = 820
(d) Find the products of the following numbers by finding the nearest multiple.
1. 7 × 52
2. 12 × 28
3. 75 × 31
4. 99 × 15
5. 8 × 25
6. 22 × 42
Solution:
1. 7 × 52 = 7 × 50 + 7 × 2 = 364
2. 12 × 28 = 12 × 30 – 24 = 336
3. 75 × 31 = 75 × 30 + 75 = 2325
4. 99 × 15 = 100 × 15 – 15 = 1485
5. 8 × 25 = 10 × 25 – 50 = 200
6. 22 × 42 = 22 × 40 + 44 = 924
Let Us Solve (Page 77)
Use strategies flexibly to answer the following questions.
Discuss your thoughts in class.
1. A school has an auditorium with 35 rows, with 42 seats in each row. How many people can sit in this auditorium?
Solution:
Total no. of people = 35 × 42 = 1,470 people
2. Priya jogs 4 kilometres every day. How many kilometers will she jog in 31 days?
Solution:
Total distance = 4 × 31 = 124 km
3. A school has received 36 boxes of books with 48 books in each box. How many total books did the school receive in the boxes?
Solution:
Total books = 36 × 48 = 1,728 books
4. Priya uses 16 metres of cloth to make 4 kurtas. How much cloth would she need to make 8 kurtas?
Solution:
Total length of kurtas = 2 × 16 = 32 meters
5. Gollappa has 29 cows on his farm. Each cow produces 5 litres of milk per day. How many litres of milk do the cow produce in total, each day?
Solution:
Quantity of milk = 29 × 5 = 145 litres
6. Maska Cow Farm has 297 cows. Each cow requires 18 kg of fodder per day. How much total fodder is needed to feed 297 cows every day?
Solution:
Total fodder = 297 × 18 = 5,346 kgs
Let Us Multiply (Page 79)
(a) 32 × 8
Solution:
(b) 69 × 45
Solution:
Let Us Do (Pages 79-80)
Question 1.
Solve the following problems like Nida did.
(a) 78 × 4
Solution:
(b) 83 × 9
Solution:
(c) 67 × 28
Solution:
(d) 53 × 37
Solution:
Question 2.
Solve the following problems like Kanti.
a) 94 × 5
Solution:
(b), (c), (d) do yourself.
b) 49 × 6
Solution:
49 × 6 = 294
c) 37 × 53
Solution:
37 × 53 = 1961
d) 28 × 79
Solution:
28 × 79 = 2212
Question 3.
Solve the following problems like John.
a) 86 × 3
Solution:
86 × 3 = 258
b) 72 × 7
Solution:
72 × 7 = 504
c) 94 × 36
Solution:
(a), (b) & (d) do yourself.
d) 66 × 22
Solution:
66 × 22 = 1452
Question 4.
Solve the following problems:
(a) A movie theater has 8 rows of seats, and each row has 12 seats. If half the seats are filled, how many people are watching the movie? If 3 more rows get filled, how many total people will be there?
Solution:
Number of rows = 8
Number of seats in each row = 12
Total capacity of theater = 12 × 8 = 96
Half of seats are filled = \(\frac{96}{2}\) = 48
If three more rows are filled = 12 × 3 = 36
Total people now = 48 + 36 = 84
(b) In a test match between India and West Indies, the Indian team hit twenty-four 4s and eighteen 6s across the two innings. How many runs were scored in 4s and 6s each? 234 runs were made by running between the wickets. If 23 runs were extras, how many runs were scored by Indian team in the two innings?
Solution:
Runs scored by Indian team
Runs from 4s = 24 × 4 = 96 runs
Runs from 6s = 18 × 6 = 108 runs
Runs between wickets = 234 runs
Extra runs = 23 runs
Total runs scored = 96 + 108 + 234 + 23 = 461
(c) Anjali buys 15 bulbs and 12 tube lights from Sudha Electricals. Each bulb costs ₹ 25 and each tube light costs ₹ 34. How much money should Anjali give to the shopkeeper?
Solution:
Cost of bulb and tubelights
(a) Bulb = 15 × ₹ 25 = ₹ 375
(b) Tubelight = 12 × ₹ 34 = ₹ 408
Total cost = ₹ 375 + ₹ 408 = ₹ 783
(d) A shopkeeper sold 28 bags of rice. Each bag costs ₹ 350. How much money did he earn by selling rice bags?
Solution:
Earnings from rice bags
= 28 × ₹ 350 = ₹ 9800
(e) A school library has 86 shelves and each shelf has 162 books. Find the number of books in the library.
Solution:
Total books in the library Shelves = 86
Books per shelf = 162
Total books = 86 × 162 = 13932 books
Let Us Solve (Pages 83-84)
Question 1.
Solve the following problems like Nida.
(a) 548 × 6
(b) 682 × 3
(c) 324 × 18
(d) 507 × 23
(e) 190 × 65
Solution:
Do yourself.
(a) 548 × 6 = 3288
(b) 682 × 3 = 2046
(c) 324 × 18 = 5832
(d) 507 × 23 = 11661
(e) 190 × 65 = 12350
Question 2.
Solve the following problems like John.
(a) 123 × 84
(b) 368 × 32
(c) 159 × 324
(d) 239 × 401
(e) 592 × 5
(f) 101 × 22
Solution:
Do yourself.
(a) 123 × 84 = 10,332
(b) 368 × 32 = 11,776
(c) 159 × 324 = 51,516
(d) 239 × 401 = 95,839
(e) 592 × 5 = 2960
(f) 101 × 22 = 2222
Question 3.
Let us solve a few questions like Mili’s father.
Now use Mili’s father’s method to solve the following questions.
(a) 807 × 5
(b) 143 × 28
(c) 309 × 9
(d) 450 × 38
(e) 584 × 23
(f) 302 × 13
(g) 604 × 54
(h) 112 × 23
(i) 237 × 19
Solution:
(c)-(i) – Do yourself.
(c) 309 × 9 = 2781
(d) 450 × 38 = 17,100
(e) 584 × 23 = 13,432
(f) 302 × 13 = 3926
(g) 604 × 54 = 32,616
(h) 112 × 23 = 2576
(i) 237 × 19 = 4503
Let Us Do (Pages 86-87)
Question 1.
Identify the problems that have the same answer as the one given at the top of each box. Do not calculate.
Solution:
Question 2.
Find easy ways of solving these problems.
(a) 16 × 25
Solution:
16 × 25 = = (10 + 6) × 25
= 10 × 25 + 6 × 25
= 250 + 150 = 400
(b) 12 × 125
Solution:
12 × 125 = 12 × (100 + 25)
= 1200 + 300
= 1500
(c) 24 × 250
Solution:
24 × 250 = 24 × (200 + 50)
= 4800 + 1200
= 6000
(d) 36 × 25
Solution:
36 × 25 = (30 + 6) × 25
=750 + 150
= 900
(e) 28 × 75
Solution:
28 × 75 = (20 + 8) × 75
= 1500 + 600
= 2100
(f) 300 × 15
Solution:
300 × 15 = 300 × (10 + 5)
= 3000 + 1500
= 4500
(g) 50 × 78
Solution:
50 × 78 = 50 × (70 + 8)
= 3500 + 400
= 3900
(h) 199 × 63
Solution:
199 × 63 = (200 – 1) × 63
= 12,600 – 63
= 12
(i) 128 × 35
Solution:
128 × 35 = 128 × (30 + 5)
= 3840 + 640
= 4480
Question 3.
Write 5 other examples for which you can find easy ways of getting products.
Solution:
Do Yourself
Question 4.
Find the answers to the following questions based on the given information.
(a) 17 × 23 = 391
(b) 17 × 24 = _______
(c) 17 × 22 = _______
(d) 16 × 23 = _______
(e) 8 × 9 = 72
(f) 18 × 9 = _______
(g) 28 × 9 = _______
(h) 108 × 9 = _______
(i) 18 × 23 = _______
Solution:
(a) 17 × 23 = 391
(b) 17 × 24 = 408
(c) 17 × 22 = 17 × (23 – 1)
= 17 × 23 – 17
= 391 – 17
= 374
(d) 16 × 23
= (17 – 1) × 23
= 17 × 23 – 23
= 391 – 23
= 368
(e) 8 × 9 = 72
(f) 18 × 9 = (10 + 8) × 9
= 90 + 72
= 162
(g) 28 × 9 = 28 × (10 – 1)
= 280 – 28
= 252
(h) 108 × 9 = 108 × (10 – 1)
= 1080 – 108
= 972
(i) 18 × 23 = (17 + 1) × 23
= 17 × 23 + 23
= 414
To find 17 × 24, how much is to be added to 17 × 23 — 17 or 23?
Solution:
This is because 17 × 24 can be rewritten as:
17 × (23 + 1) = (17 × 23) + (17 × 1) = (17 × 23) + 17.
So, we should add 17.
To find 18 × 23, how much is to be added to 17 × 23 — 17 or 23?
Solution:
18 × 23 = (17 + 1) × 23 = 17 × 23 + 23.
Hence, to find the product of 18 × 23, we should add 23 to the product of 17 × 23.
Let Us Think (Pages 87-88)
Question 1.
Find the possible values of the coloured boxes in each of the following problems. The same colour indicates the same number in a problem. Some problems can have more than one answer.
Solution:
Question 2.
Estimate the products on the left and match them to the numbers given on the right.
Solution:
The King’s Reward (Pages 88-89)
One day, a king decided to reward three of his most talented ministers. The king called them to his court and said, “You all have served my empire with great dedication. As a reward, I give you three choices of gold.
Choose wisely!
Choice 1: Take 5 gold coins and double the number of coins every day for 7 days.
Choice 2: Take 3 gold coins and triple the number of coins every day for 7 days.
Choice 3: Take 1 gold coin and multiply the number of coins by 5 every day for 7 days.
Minister 1 — I will take 5 gold coins and double the number of coins every day for 7 days.
Minister 2 — I will take 3 gold coins and triple the number of coins every day for 7 days.
Minister 3 — I will take 1 gold coin and multiply the number of coins by 5 every day for 7 days.
The King gave 5 coins to Minister 1, 3 coins to Minister 2 and 1 coin to Minister 3.
Which of the rewards would you have chosen?
After a week, the 3 ministers were surprised at the final amount of gold coins. Guess who received the most gold coins? Calculate how much gold coins each minister received.
Solution:
My Choice : Reward 3
Minister 1 – 5 × 128 = 640 gold coins.
Minister 2 – 3 × 2187 = 6561 gold coins.
Minister 3 – 1 × 78, 125 = 78,125 gold coins
Multiplication Patterns (Pages 89-90)
Question 1.
Notice how the multiplier, multiplicand, and products are changing in each of the following. What is the relationship of the new product with the original product? Solve a) completely, and then predict the answers for the rest.
a) 16 × 44 = 704
1) 8 × 88 = 704
2) 8 × 22 = 176
3) 16 × 22 = ______
4) 32 × 44 = ______
Solution:
16 × 44 = 704
1. 8 × 88 = 704
2. 8 × 22 = 176
3. 16 × 22 = 352
4. 32 × 44 = 1408
b) 12 × 32 = 384
1) 6 × 16 = ______
2) 24 × 16 = ______
3) 24 × 64 = ______
4) 12 × 16 = ______
Solution:
12 × 32 = 384
1. 6 × 16 = 96
2. 24 × 16 = 384
3. 24 × 64 = 1536
4. 12 × 16 = 192
Question 2.
Observe and complete the given patterns
Solution:
Smallest Product:- 11 × 14 = 154
Largest Product:- 75 × 35 = 2,625
Let Us Solve (Page 91)
Question 1.
Mala went to a book exhibition and bought 18 books. The shop was selling 3 books for ₹ 150. After buying the books, she still had ₹ 20 left. How much money did Mala have at the beginning?
Solution:
Price of each book = ₹ \(\frac{150}{3}\) = ₹ 50
Number of books Mala bought = 18
Money paid by Mala = 18 × ₹ 50 = ₹ 900
Money Mala have at the beginning = ₹ 900 + ₹ 20 = ₹ 920
Question 2.
A village sports club organises a women’s football tournament. The club earned money by selling match tickets and charging fees for team participation.
They sold 57 tickets for ₹ 115 each.
They had 3 teams joining the tournament, with each team paying a participation fee of ₹ 1,599.
The teams paid ₹ 1,750 in total to rent the football ground and ₹ 1,129 for food and water.
(a) How much money did the club collect in total from ticket sales and team participation fees?
(b) What were the total expenses on renting the ground and food and water?
Solution:
(a) Total collected = ticket revenue + team fees
= 57 × 115 + 3 × 1599
= 6555 + 4797
= ₹ 11,352
(b) Total expenses = rent + food/water = 1750 + 1129
= ₹ 2,879
Question 3.
Ananya is watching Republic Day celebrations on the city’s public ground. There are 12 rows of students sitting in front of her and 17 rows behind her. There are 18 students to her right and 22 students to her left.
(a) How many rows of students are there in total?
(b) How many students are there in Ananya’s row?
(c) What is the total number of students on the ground?
Solution:
(a) There are 30 rows of students in total.
(b) There are 41 students in Ananya’s row.
(c) There are 1230 students on the ground.
Question 4.
Multiply.
(a) 67 × 78
(b) 34 × 56
(c) 45 × 263
(d) 86 × 542
(e) 432 × 107
(f) 310 × 120
Solution:
(a) 67 × 78 = 5226
(b) 34 × 56 = 1904
(c) 45 × 263 – 11,835
(d) 86 × 542 = 46,612
(e) 432 × 107 = 46,224
(f) 310 × 120 = 37,200
Question 5.
If 67 × 67 = 4489, without multiplication find 67 × 68.
Solution:
67 × 68 = 67 x (67 + 1)
= 67 × 67 + 67 = 4489 + 67 = 4556
Question 6.
If 99 × 100 = 9900, without multiplication find 99 × 99.
Solution:
99 × 99 = 99 × (100 – 1)
= 99 × 100 – 99
= 9900 – 99
= 9801