Practicing with Maths Mela Class 5 Solutions Chapter 6 The Dairy Farm Question Answer NCERT Solutions improves a student’s confidence in the subject.
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 (NCERT Pg 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.
Answer:
Question 2.
Place the digits 2, 5, and 3 appropriately to get a product close to 100. Share your reasoning in class.
Answer:
Now, using the digits 2,5 and 3 appropriately to get a product 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.
Answer:
(ii) 15 × 4 = 60
(iii) 10 × 6 = 60
(iv) 12 × 5 = 60
Different group we make like 20 × 3 = 60
Question 4.
(i) 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.
Answer:
Let’s use the given clues to find two-digit number:
- Two-digit Number This means the number is between 10 and 99.
- I am greater than 8 All the two digit numbers are greater than 8.
- I am not a multiple of 4 All the possible two digit numbers except the multiple of 4 i.e. 12,16,20,24,…..
- I am a multiple of 9 All the two digit numbers which are multiples of 9 are 18,27,45,54,63,81,90 and 99.
- I am an odd number The odd numbers from the multiples of 9 are 27,45,63,81 and 99.
- I am less than 50 The number less than 50 from the remaining numbers are 27 and 45.
- My ones digit is even None of the remaining numbers have even ones digit.
- My tens digit is odd None of the remaining numbers have odd tens digit.
(ii) Did you use all the clues to find the number? Which clues did not help you in finding the number?
Answer:
There is no two digit number that can satisfy all the given clues simultaneously because clues (d) “I am an odd number” and (g) My ones digit is even” are contradictory because an odd number must have an odd digit in ones place Therefore, the clues (d) and (g) did not help in finding a specific number as they create a logical inconsistency.
Question 5.
Make your own numbers.
(i) 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?
Answer:
All numbers between 0 and 20 that are formed by two numbers and an operation
5 – 4 or 4 – 3 or 3 – 2 = 1
10 + 5 or 4 – 2 = 2
36 ÷ 12 or 5 – 2 or 12 ÷ 4 = 3
12 + 3 or 100 + 25 = 4
25 * 5 or 10 * 2 or 10 – 5 =5
10 – 4 or 3 × 2 = 6
4 + 3 or 10 – 3 = 7
4 × 2 or 12 – 4 = 8
36 + 4 or 12 – 3 = 9
100 + 10 or 5 × 2 = 10
36 – 25 = 11
10 + 2 or 36
3 or 4 × 3 = 12
25 – 12 or 10 + 3 = 13
10 + 4 or 12 + 2 = 14
10 + 5 or 5 × 3 or 12 + 3 = 15
12 + 4 = 16
12 + 5 = 17
36 + 2 = 18
No, we can not make all the numbers.
(ii) Which numbers could you not make? Is it possible to make these numbers using three numbers? You can use two operations, if needed.
Which numbers between 0-20 can you get in more than one way?
Answer:
The number 19 cannot be made by taking two numbers and one operation.
So, we use three numbers and two operations to make the number 19
25 – (10 – 4) = 19
So, many numbers can be obtained.
The numbers between 0-20 in which to get use more than one way are 1, 2, 3, 4, 5, 6, 7,8, 9, 10, 12, 13, 14 and 15.
Order of Numbers in Multiplication (NCERT Pg 72)
Question 6.
(i) Daljeet Kaur runs a milk processing unit. She has arranged the butter packets in the following ways. Fin
Answer:
(a) We have, 2 × 3 = 6 and 3 × 2 = 6
(b) We have, 5 × 8 = 40 and 8 × 5 = 40
(c) We have, 6 × 13 = 78 and 13 × 6 = 78
(d) We have, 10 × 5 = 50 and 5 × 10 = 50
(e) We have, 8 × 20 = 160 and 20 × 8 = 160
(f) We have, 12 × 9 = 108 and 9 × 12 = 108
This is the commutative property of multiplication i.e. a × b = b × a.
So, swapping the factors does not change the product.
(ii) Is this true for the product of any two numbers? Discuss in class.
Answer:
Yes, this is true for the product of any two numbers because the number of groups and the group size are interchanged, but the result remains the same.
(iii) What is 9 × 0 ? 0 × 9 ?
Answer:
We have, 9 × 0 = 0 and 0 × 9 = 0
Therefore, the product of any number and 0 is always 0.
Patterns in Multiplication by 10s and 100s (NCERT Pg 72-73)
Question 7.
Let us revise multiplication by 10 s and 100 s.
(i) 4 × 10= _______
(ii) 20 × 10= _______
(iii) 10 × 40= _______
(iv) 10 × 10=100
(v) 20 × 50= _______
(vi) 80 × 10= _______
(vii) 3 × 100=100 × 3 = 300
(viii) 8 × 100= _______ = _______
(ix) 10 × 100= _______ = _______
Answer:
(i) We have, 4 × 10 = 40
(ii) We have, 20 × 10 = 200
(iii) We have, 10 × 40 = 400
(v) We have, 20 × 50 = 1000
(vi) We have, 80 × 10 = 800
(viii) We have, 8 × 100 = 100 × 8 = 800
(xi) We have, 10 × 100 = 100 × 10 = 1000
Question 8.
(i) Find answers to the following questions. Fill in the table below and describe the pattern. Discuss in class.
Answer:
(ii) How should we write 450 in the table below?
Answer:
The following table as shown below
(iii)
Answer:
The following tables as shown below
Question 9.
Let us fill in the table and observe the patterns.
Answer:
The following table as shown below
Doubling and Halving (NCERT Pg 75-76)
Question 10.
Butter packets are arranged in the following ways. Let us find some strategies to calculate the total number of packets.
(i)
Answer:
We have,
(iii) 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?
Answer:
The halving and doubling strategy works well with numbers like 5 and 25 because they are easily convertible to power of 10 through multiplication.
(iv) Find the product by halving and doubling either the multiplier or the multiplicand.
(a) 5 × 18
(b) 50 × 28
(c) 15 × 22
(d) 25 × 12
(e) 12 × 45
(f) 16 × 45
Answer:
(v) Give 5 examples of multiplication problems where halving and doubling will help in finding the product easily. Find the products as well.
Answer:
There are following five examples of multiplication problems.
Nearest Multiple (NCERT Pg 76)
Question 11.
(iii) Give 5 examples of problems where you can use the nearest multiple to find the product easily. Find the products as well.
Answer:
Here are 5 examples of problems where we use the nearest multiple.
(a) 21 × 7 = 20 × 7 + 7 = 140 + 7 = 147
(b) 19 × 6 = 20 × 6 – 6 = 120 – 6 = 114
(c) 51 × 4 = 50 × 4 + 4 = 200 + 4 = 204
(d) 8 × 21 = 20 × 8 + 8 = 160 + 8 = 168
(e) 10 × 19 = 20 × 10 – 10 = 200 – 10 = 190
(iv) Find the products of the following numbers by finding the nearest multiple.
(a) 7 × 52
(b) 12 × 28
(c) 75 × 31
(d) 99 × 15
(d) 8 × 25
(f) 22 × 42
Answer:
(a) We have, 7 × 52 = 7 × (50+2)
= 7 × 50 + 7 × 2
= 350 + 14 = 364
(b) We have, 12 × 28 = 12 ×(30 – 2)
=12 × 30 – 12 × 2
= 360 – 24 = 336
(c) We have, 75 × 31=75 × 30+75
= 2250 + 75 = 2325
(d) We have, 99 × 15 = 100 × 15 – 15
= 1500 – 15 = 1485
(e) We have, 8 × 25 = (10-2) × 25
= 25 × 10 – 2 × 25
= 250-50 = 200
(f) We have, 22 × 42 = (20 + 2) × 42
= 20 × 42 + 2 × 42
= 840 + 84 = 924
Let Us Solve (NCERT Pg 77)
Question 12.
Use strategies flexibly to answer the following questions. Discuss your thoughts in class.
(i) A school has an auditorium with 35 rows, with 42 seats in each row. How many people can sit in this auditorium?
(ii) Priya jogs 4 kilometres every day. How many kilometers will she jog in 31 days?
(iii) 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?
(iv) Priya uses 16 metres of cloth to make 4 kurtas. How much cloth would she need to make 8 kurtas?
(v) 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?
(vi) 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?
Answer:
(i) Given, the number of rows in an auditorium = 35
and the number of seats in one row = 42
So, the number of people who can sit in an auditorium = 35 × 42
Therefore, 1470 people can sit in the auditorium.
(ii) Given, the distance covered by Priya per day = 4 km
So, the total distance covered by Priya in 31 days = 31 × 4 = 30 × 4 + 4
= 120 + 4 = 124 km
Therefore, Priya will jog 124 kilometres in 31 days.
(iii) Given, the total number of boxes = 36 and the number of books in one box = 48 So, the total number of books in 36 boxes
= 36 × 48 = 36 × (50-2)
= 36 × 50-36 × 2 = 1800 – 72 = 1728
Therefore, the school receive 1728 books in the boxes.
(iv) Given, total length of cloth used for making 4 kurtas = 16 m
∴ Length of cloth used for making 1 kurta
= \(\frac{16}{4}\) = 4 m
So, the total length of cloth used for making 8 kurtas = 8 × 4 = 32 m
(v) Given, the number of cows in farm = 29 and total amount of milk per cow = 5
So, the total amount of milk of 29 cows
= 29 × 5 = 30 × 5 – 5
= 150 – 5 = 145 L
Therefore, the cow produce 145 L of milk each day.
(vi) Given, the number of cows = 297
and total fodder per day required by a cow
= 18 kg
So, total fodder per day required for
297 cows = 297 × 18 = (300-3) × 18
= 300 × 18-3 × 18
= 5400 – 54
= 5346 kg
Therefore, 297 cows need 5346 kg food every day.
Waste and Composting (NCERT Pg 77-78)
Question 13.
A family of 4 produces around 35 kg of kitchen waste in a month. How much waste will the family produce in a year?
Quantity of kitchen waste in 1 month is 35 kg.
Quantity of kitchen waste in 12 months is 12 × 35 kg.
Kanti and John tried to solve it in the following ways.
(i) The family produces _______ kg of waste in a year.
(ii) How are these solutions same or different? Discuss in class.
Answer:
(i) The family produces 420 kg of waste in a year
(ii) Here, Kanti breaks down both 35 and 12 into their expanded forms like
35 = 30+5 and 12 = 10+2
Then, kanti multiples each part of 35 by each part of 12 separately and adds the resulting products,
John also breaks down 35 and 12 into their expanded forms,
i.e. 35 = 30 + 5
and 12 = 10 + 2
Then, John first multiplies 35 by 10 and then by 2 and adds the results,
i.e. (30 + 5) × 10 = 300 + 50 = 350
and (30 + 5) × 2 = 60 + 10 = 70
And then John adds these two numbers, i.e. 350 + 70 = 420
So, these methods are same.
Let Us Multiply (NCERT Pg 79)
Question 14.
(i) 32 × 8
Answer:
(i) 32 × 8
(ii) 69 × 45
Let Us Do (NCERT Pg 79-80)
Question 15.
Solve the following problems like Nida did.
(i) 78 × 4
Answer:
(ii) 83 × 9
Answer:
(iii) 67 × 28
Answer:
(iv) 53 × 37
Answer:
Question 16.
Solve the following problems like Kanti.
(i) 94 × 5
(ii) 49 × 6
(iii) 37 × 53
(iv) 28 × 79
Answer:
Question 17.
Solve the following problems like John.
(i) 86 × 3
(ii) 72 × 7
(iii) 94 × 36
(iv) 66 × 22
Answer:
(i) We have, 86 × 3
Question 18.
Solve the following problems:
(i) 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?
Answer:
Given, the total number of rows in movie theater = 8
and number of seats in each row = 12
If half the seats are filled,
∴ Number of people watching movie
Now, 3 more rows get filled
∴ Number of people in 3 rows = 12 × 3 = 36
Therefore, the total number of people who are watching the movie = 48 + 36
= 84 people.
(ii) 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 4 s and 6 s 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?
Answer:
Given, the Indian team hit twenty-four 4 s and eighteen 6 s across the two innings.
Now, total runs from 4 s = 24 × 4 = 96
and total runs from 6 s = 18 × 6 = 108 s
If 234 runs were made by running between the wickets and 23 extra runs were extra.
Therefore, total runs scored by the Indian team in the two innings
= 96 + 108 + 234 + 23
= 461
(iii) 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?
Answer:
Given, the total number of bulbs bought by Anjali = 15
and the total number of tube lights bought by Anjali =12
If each bulb costs ₹ 25 and each tube lights costs ₹ 34.
Therefore, the total money given to shopkeeper by Anjali
= 15 × 25 + 12 × 34
= 375 + 408
= ₹ 783
(iv) A shopkeeper sold 28 bags of rice. Each bag costs ₹ 350. How much money did he earn by selling rice bags?
Answer:
Given, number of bags sold = 28
and cost of each bag = ₹ 350
Therefore, total earnings
= 28 × 350
= ₹ 9800
Hence, the shopkeeper earned ₹ 9800 by selling rice bags.
(v) A school library has 86 shelves and each shelf has 162 books. Find the number of books in the library.
Answer:
Given, number of shelves in school library = 86
and number of books per shelf =162
So, the total number of books in the library
= 86 × 162
= 13932
Let Us (NCERT Pg 83-84)
Question 19.
Solve the following problems like Nida.
(i) 548 × 6
(ii) 682 × 3
(iii) 324 × 18
(iv) 507 × 23
(v) 190 × 65
Answer:
(i) We have, 548 × 60
(ii) We have, 682 × 3
(iii) We have, 324 × 18
(iv) We have, 507 × 23
(v) We have, 190 × 65
Question 20.
Solve the following problems like John.
(i) 123 × 84
(ii) 368 × 32
(iii) 159 × 324
(iv) 239 × 401
(v) 592 × 5
(vi) 101 × 22
Answer:
(i) We have, 123 × 84
(ii) We have,368 × 32
(iii) We have, 159 × 324
(iv) We have, 239× 401
(v) We have, 592 × 5
(vi) We have, 101 × 22
Question 21.
(i) Let us solve a few questions like Mili’s father.
Answer:
(ii) 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
Answer:
(a) We have, 807 × 5
(b) We have, 143 × 28
(c) We have, 309 × 9
(d) We have, 450 × 38
(e) We have, 584 × 23
(f) We have, 302 × 13
(g) We have, 604 × 54
(h) We have, 112 × 23
(i) We have, 237 × 19
Check, Check! (NCERT Pg 85)
Question 22.
Check if the following children’s solutions are correct. If correct, explain why the solution is correct. If it is incorrect, then identify the error and correct the solution.
Answer:
(i) Here, 46 × 59 = 2714
So, Asma’s solution is correct because she used Kanti’s method to break down the numbers into easiser form i.e.
(ii) Here, Pankaj’s used John’s method for solution i.e.
So, Pankaj’s solution is incorrect.
(iii) Here, Lado used John’s method for solution i.e.
So, Lado’s solution is correct.
(iv) Here, Kira used John’s method for solution i.e.
So, Kira’s solution is incorrect.
(v) Here, Asher’s used Mili’s father method for solution i.e.
Let Us Do (NCERT Pg 86-87)
Question 23.
Identify the problems that have the same answer as the one given at the top of each box. Do not calculate.
Answer:
(i) We have,
(ii) We have, 26 × 11 = 26 × (10 + 1)
= 26 × 10 and 26 × 1
or
26 × 11 = (20 + 6) × 11
= 20 × 11 and 6 × 11
(iii) We have,
or 18 × 4 = 20 × 4 – 8
(iv) We have, 55 × 9 = (50 + 5) × 9
= 50 × 9 and 5 × 9
or 55 × 9 = 55 × 10 – 55
(v) We have, 101 × 42 = (100 + 1) × 42
= 100 × 42 + 1 × 42
= 100 × 42 and 42
(vi) We have, 247 × 8 = 250 × 8 – 24
(vii) We have, 1001 × 5 = (1000 + 1) × 5
= 1000 × 5 + 1 × 5
= 1000 × 5 and 5
(viii) We have, 1999 × 2 = (2000 – 1) × 2
= 2000 × 2 – 2
Question 24.
Find easy ways of solving these problems.
(i) 16 × 25
(ii) 12 × 125
(iii) 24 × 250
(iv) 36 × 25
(v) 28 × 75
(vi) 300 × 15
(vii) 50 × 78
(viii) 199 × 63
(ix) 128 × 35
Answer:
Question 25.
Write 5 other examples for which you can find easy ways of getting products.
Answer:
Here are some examples for easy ways of getting products.
(i) Here,
(ii) Here, 46 × 32
(iii) Here, 42 × 36
(iv) Here,
31 × 6 = (30 + 1) × 6 = 30 × 6 + 1 × 6
= 180 + 6 = 186
(v) Here, 11 × 31
Question 26.
Find the answers to the following questions based on the given information. _______
(i) 17 × 23 = 391
(ii) 17 × 24 = _______
(iii) 17 × 22 = _______
(iv) 16 × 23 = _______
(v) 8 × 9 = 72
(vi) 18 × 9 = _______
(vii) 28 × 9 = _______
(viii) 108 × 9 = _______
(ix) 18 × 23 = _______
Answer:
(i) We have, 17 × 23 = 391
(ii) To find, 17 × 24, we will add 17 in 17 × 23 i.e. 17 × 24 = 17 × 23 + 17 = 391 + 17 = 408
(iii) To find 17 × 22, we will subtract 17 from
17 × 23 i.e.
17 × 22 = 17 × 23-17
= 391-17 = 374
(iv) To find 16 × 23, we will subtract 23 from
17 × 23 i.e.
16 × 23 = 17 × 23-23
= 391-23 = 368
(vi) We have, 8 × 9 = 72
To find 18 × 9, we will add 90 in 8 × 9
i.e.
18 × 9 =(10 + 8) × 9=10 × 9 + 8 × 9
= 90 + 72 = 162
(vii) To find 28 × 9, we will add 180 in 8 × 9
i.e.
28 × 9 = (20+8) × 9 = 20 × 9 + 8 × 9
= 180 + 72 = 252
(viii) To find 108 × 9, we will add 900 in 8 × 9
i.e.
108 × 9 = (100 + 8) × 9 = 100 × 9 + 8 × 9
= 900 + 72 = 972
(ix) To find 18 × 23, we will add 23 from
17 × 23 i.e.
18 × 23 = 17 × 23 + 23
= 391 + 23 = 414
Let Us Think (NCERT Pg 87-88)
Question 27.
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.
Answer:
Question 28.
Estimate the products on the left and match them to the numbers given on the right.
Answer:
(i) (d) We have, 25 × 31
Estimated value of 25 × 31 = 30 × 25 = 750
(ii) (a) We have, 132 × 19
Estimated value of 132 × 19 = 130 × 20 = 2600
(iii) (e) We have, 101 × 11
Estimated value of 101 × 11 = 100 × 10 = 1000
(iv) (b) We have, 248 × 49
Estimated value of 248 × 49 = 250 × 50 = 12,500
(v) (c) We have, 12 × 25
Estimated value of 12 × 25 = 30 × 10 = 300
The King’s Reward (NCERT Pg 88-89)
Question 29.
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 cons each minister reaceived.
Answer:
For Minister 1
Minister 1 starts with 5 gold coins and doubles the amount every day for 7 days.
Therefore, for day 1,5 × 2 = 10
for day 2,10 × 2 = 20
for day 3,20 × 2 = 40
for day 4,40 × 2 = 80
for day 5,80 × 2 = 160
for day 6,160 × 2 = 320
for day 7,320 × 2 = 640
So, the number of coins after 7 days = 640
For Minister 2
Minister 2 starts with 3 gold coins and triple the amount every day for 7 days
Therefore, for day 1, 3 × 3 = 9
for day 2,9 × 3 = 27
for day 3,27 × 3 = 81
for day 4,81 × 3 = 243
for day 5,243 × 3 = 729
for day 6,729 × 3 = 2187
for day 7,2187 × 3 = 6561
So, the number of coins after 7 days = 6,561
For Minister 3
Minister 3 starts with 1 gold coin and multiply the number of coins by 5 every day for 7 days.
Therefore, for day 1,1 × 5 = 5
for day 2,5 × 5 = 25
for day 3,25 × 5 = 125
for day 4,125 × 5 = 625
for day 5,625 × 5 = 3125
for day 6,3125 × 5 = 15625
for day 7,15625 × 5 = 78125
So, the number of cons after 7 days = 78125 Based on the calculations, choice 3 which minister 3 chose, yields the most gold coins. Therefore, we would have chosen choice 3.
Multiplication Patterns (NCERJ Pg 89-90)
Question 30.
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 (i) completely, and then predict the answers for the rest.
(i) 16 × 44 = 704
(a) 8 × 88 = 704
(b) 8 × 22 =176
(c) 16 × 22 =
(d) 32 × 44 =
(ii) 12 × 32 = 384
(a) 6 × 16 = _______
(b) 24 × 16 = _______
(c) 24 × 64 = _______
(d) 12 × 16 = _______
Answer:
Here, the first number is halved and the second number is doubled.
The relationship between the new product and the original product is the half, double, one fourth or four times the original product.
Question 31.
(i) Observe and complete the given patterns.
(ii) Here are some numbers. Remember number pairs from Grade 4? Any two adjacent numbers in a row or a column are number pairs. Can you identify the pair whose product is the smallest and another pair whose product is the largest? Do you need to find every product or can you find this by looking at the numbers?
Answer:
(ii) Do yourself
Let Us Solve (NCERT Pg 91)
Question 32.
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?
Answer:
Given, number of books bought by Mala = 18
∵ The shop was selling 3 books for ₹ 150.
∴ The cost of 1 book = 150 + 3 = ₹ 50
So, the cost of 18 books = 50 × 18 = ₹ 900
Since, Mala still had 20 rupees left.
Hence, the money she had at the beginning
= 900 + 20 = ₹ 920
Question 33.
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.
(i) How much money did the club collect in total from ticket sales and team participation fees?
(ii) What were the total expenses on renting the ground and food and water?
Answer:
(i) Given, the number of tickets sold = 57 and price for each ticket = ₹ 115 Now, earnings from ticket sales
= 57 × 115 = ₹ 6,555
Also, given number of teams = 3 and participation fee per team = ₹ 1,599
∴ Earnings from participation fees
= 1,599 × 3 = ₹ 4,797
Total money collected = 4,797 + 6,555
= ₹ 11,352
(ii) Since, the teams paid ₹ 1,750 in total to rent the football ground and ₹ 1,129 for food and water.
So, total expenses =1,750 + 1,129 = ₹ 2,879
Question 34.
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?
Answer:
(a) Given, number of rows in front of Ananya = 12 and number of rows behind Ananya = 17 Also, there is one Ananya’s row.
∴ The total number of rows in ground.
= 12 + 17 + 1 = 30
(b) Since, there are 18 students to Ananya’s right and 22 students to Ananya’s left. Also, 1 is Ananya herself.
∴ The total number of students in Ananya’s row = 18 + 22 + 1 = 41
(c) Number of students in Ananya’s row =41 and total number of rows = 30
∴ Total number of students in ground
= 41 × 30 = 1230
Question 35.
Multiply.
(i) 67 × 78
(ii) 34 × 56
(iii) 45 × 263
(iv) 86 × 542
(v) 432 × 107
(vi) 310 × 120
Answer:
(i) We have, 67 × 78 = 5226
(ii) We have, 34 × 56 = 1904
(iii) We have, 45 × 263 = 11,835
(iv) We have, 86 × 542 = 46,612
(v) We have, 432 × 107 = 46,224
(vi) We have, 310 × 120 = 37200
Question 36.
If 67 × 67 = 4489, without multiplication, find 67 × 68.
Answer:
Given, 67 × 67 = 4489
We have, 67 × 68 = 67 × (67+1)
= 67 × 67 + 67
= 4489 + 67
= 4556
Question 37.
If 99 × 100 = 9900, without multiplication, find 99 × 99.
Answer:
Given, 99 × 100 = 9900
We have, 99 × 99 = 99 × (100-1)
= 99 × 100-99
= 9900 – 99
= 9801