Class 6

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 4 Integers Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 4 Integers Ex 4.1

1. Write two examples from day-to-day life in which we can use positive and negative integers.
Solution:
1. If positive represents above sea level, then negative represents below sea level.
2. If positive represents a deposit, negative represents a withdrawal.

2. Write the opposite of the following:

Question (a)
A profit of ₹ 500
Solution:
A loss of ₹ 500

Question (b)
A withdrawal of ₹ 70 from bank account
Solution:
Deposit of ₹ 70 in bank account

Question (c)
A deposit of ₹ 1000
Solution:
Withdrawal of ₹ 1000

Question (d)
326 B.C
Solution:
326 AD

Question (e)
500 m below sea level
Solution:
500 m above sea level

Question (f)
25° above 0°C.
Solution:
25° below 0°C.

3. Represent the situations mentioned in integers.
Solution:
(a) + 500
(b) – 70
(c) + 1000
(d) – 326
(e) – 500 m
(f) + 25.

4. Represent the following situations in integers.

Question (a)
A deposit of ₹ 500.
Solution:
+ 500

Question (b)
An Aeroplane is flying at a height two thousand metre above the sea level.
Solution:
+ 2000

Question (c)
A withdrawal of ₹ 700 from Bank Account.
Solution:
– 700

Question (d)
A diver dives to a depth of 6 feet below ground level.
Solution:
– 6.

5. Represent the following numbers on number line.

Question (i)
(a) – 5
(b) + 6
(c) o
(d) + 1
(e) – 9
(f) – 4
(g) + 8
(h) + 3.
Solution:

6. Integers are represented on a horizontal number line as shown where A represents – 2. With reference to the number line, answer the following questions:

(a) Which point represent – 3?
(b) Locate the point which represents the opposite of B and name it P.
(c) Write integers for the points C and E.
(d) Which point marked on the number line has the least value?
Solution:

(a) Point B represents – 3.
(b) Point P represents + 3.
(c) Point C represents -7 and Point E represents + 4.
(d) Point C has the least value – 7.

7. In each of the following pairs, which number is to the right of other on the number line?

Question (i)
(a) 2 9
(b) -3, -8
(c) 0, -5
(d) -11, 10
(e) -9, 9
(f) 2, – 200.
Solution:
(a) 9
(b) – 3
(c) 0
(d) 10
(e) 9
(f) 2

8. Write all the integers between the given pairs (write them in increasing order)

Question (a)
0 and -6
Solution:
-5, -4, -3, -2, -1

Question (b)
-6 and +6
Solution:
-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

Question (c)
-9 and -17
Solution:
-16, -15, -14, -13, -12, -11, -10

Question (d)
-19 and -5.
Solution:
-18, -17, -16, -15, -14, -13, -12, -11, -10, -9, -8, -7, -6.

9.

Question (a)
Write five negative integers greater than ‘-15’.
Solution:
Five negative integers greater than ‘-15’ are:
-14, -13, -12, -11, -10

Question (b)
Write five integers smaller than ‘-20’.
Solution:
Five integers smaller than ‘-20’ are:
-21, -22, -23, -24, -25

Question (c)
Write five integers greater than 0.
Solution:
Five integers greater than 0 are:
1,2, 3, 4, 5

Question (d)
Write five integers smaller than 0.
Solution:
Five integers smaller than 0 are:
-1, -2, -3, -4, -5.

10. Encircle the greater integer in each given pair.

(a) -5, -7
(b) 0,-3
(e) 5, 7
(d) -9, 0
(e) -9, -11
(f) -4, 4
(g) -10, -100
(h) 10, 100.
Solution:
(a) -5
(b) 0
(c) 7
(d) 0
(e) -9
(f) 4
(g) -10
(h) 100.

11. Arrange the following integers in ascending order:

Question (a)
0, -7, -9, 5, -3, 2, -4
Solution:
Ascending order of given integers is:
-9, -7, -4, -3, 0, 2, 5

Question (b)
8, -3, 7, 0, -9, -6.
Solution:
Ascending order of given integers is:
-9, -6, -3, 0, 7, 8.

12. Arrange the following integers in descending order:

Question (a)
-9, 3, 4, -6, 8, -3
Solution:
8, 4, 3, -3 -6, -9

Question (b)
4, 8,-3,-2, 5, 0.
Solution:
8, 5, 4, 0, -2, -3.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 3 Playing with Numbers Ex 3.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.5

1. Find LCM of following numbers by prime factorization method:

Question (i)
45, 60
Solution:

∴ 45 = 3 × 3 × 5
60 = 2 × 2 × 3 × 5
We find that in these prime factorizations, 2 occurs maximum two times, 3 occurs maximum two times and 5 occurs maximum once
∴ L.C.M. of 45 and 60
= 2 × 2 × 3 × 3 × 5 = 180

Question (ii)
52, 56
Solution:

We find that in these prime fatorisation, 2 occurs maximum 3 times, 13 and 7 occurs maximum once.
∴ L.C.M. of 52 and 56
= 2 × 2 × 2 × 13 × 7 = 728

Question (iii)
96, 360
Solution:

∴ 96 = 2 × 2 × 2 × 2 × 2 × 3
360 = 2 × 2 × 2 × 3 × 3 × 5
We find that in these prime factorisation, 2 occurs maximum 5 times, 3 occurs maximum 2 times and 5 occurs maximum once.
∴ L.C.M. of 96 and 360
= 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440

Question (iv)
36, 96, 180
Solution:

∴ 36 = 2 × 2 × 3 × 3
96 = 2 × 2 × 2 × 2 × 2 × 3
and 180 = 2 × 2 × 3 × 3 × 5
We find that in these factorisation, 2 occurs maximum 5 times, 3 occurs maximum 2 times and 5 occurs maximum once.
∴ L.C.M. of 36, 96 and 182
= 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440

Question (v)
18, 42, 72.
Solution:

∴ 18 = 2 × 3 × 3
42 = 2 × 3 × 7
72 = 2 × 2 × 2 × 3 × 3
We find that in these factorization 2 occurs maximum 3 times, 3 occurs maximum 2 times and 7 occurs maximum once.
∴ L.C.M. of 18, 42 and 72
= 2 × 2 × 2 × 3 × 3 × 7 = 504

2. Find LCM of the following by common division method:

Question (i)
24, 64
Solution:

∴ L.C.M. of 24, 64
= 2 × 2 × 2 × 3 × 8 = 192

Question (ii)
42, 63
Solution:

∴ L.C.M. of 42 and 63
= 3 × 7 × 2 × 3 = 126

Question (iii)
108, 135, 162
Solution:

∴ L.C.M. of 108, 135 and 162
= 2 × 3 × 3 × 3 × 2 × 5 × 3 = 1620

Question (iv)
16, 18, 48
Solution:

∴ L.C.M. of 16, 18 and 48
= 2 × 2 × 2 × 2 × 3 × 3 = 144

Question (v)
48, 72, 108
Solution:

∴ L.C.M. of 48, 72 and 108
= 2 × 2 × 2 × 3 × 3 × 2 × 3 = 144

3. Find the smallest number which is divisible by 6, 8 and 10.
Solution:
We know that the smallest number divisible by 6, 8 and 10 is their L.C.M.
So, we calculate L.C.M. 6, 8 and 10

∴ L.C.M. = 2 × 3 × 4 × 5 = 120
Hence, required number =120

4. Find the least number when divided by 10,12 and 15 leaves remainder 7 in each case.
Solution:
We know that the least number divisible by 10, 12 and 15 is their L.C.M.

So, the required number will be 7 more than their L.C.M.
We calculate their L.C.M.
L.C.M of 10, 12 and 15 = 2 × 3 × 5 × 2 = 60
Hence, Required number = 60 + 7 = 67

5. Find the greatest 4-digit number exactly divisible by 12, 18 and 30.
Solution:
First find the L.C.M. of 12, 18, 30

∴ L.C.M. of 12, 18, 30
= 2 × 3 × 2 × 3 × 5 = 180
Now the greatest 4 digit number = 9999
We find that when 9999 is divided by 180, the remainder is 99.
Hence, the greatest number of 4 digits which is exactly divisible by 12, 18, 30
= 9999 – 99 = 9900

6. Find the sandiest 4-digit number exactly divisible by 15, 24 and 36.
Solution:
First find L.C.M. of 15, 24, 36

L.C.M. of 15, 24, 36
= 2 × 2 × 3 × 5 × 2 × 3 = 360 Now, 4 digit smallest number is 1000 We find that when 1000 is divided by 360, the remainder is 280.
∴ Smallest 4 digits number, which is exactly divisible by 15, 24 and 36
= 1000 + (360 – 280) = 1000 + 80 = 1080.
Hence, required number = 1080

7. Four bells toll at intervals of 4, 7, 12 and 14 seconds. The bells toll together at 5 a.m. When will they again toll together?
Solution:
The bells will toll together at a time which is multiple of four intervals 4, 7, 12 and 14 seconds
So, first we find L.C.M. of 4, 7, 12 and 14

∴ L.C.M. = 2 × 2 × 7 × 3 = 84
Thus the bells will toll together after 84 seconds or 1 minute 24 seconds.
First they toll together at 5 a.m., then they will toll together after 1 minutes 24 seconds i.e. 5 : 01 : 24 a.m.

8. Three boys step off together from the same spot their steps measures 56 cm, 70 cm and 63 cm respectively. At what distance from the starting point will they again step together?
Solution:
The distance covered by each one of them has to be same as well as minimum walk So, the minimum distance each should their steps will be L.C.M. of the distances L.C.M. of the measure of their steps.

∴ L.C.M. = 2 × 7 × 4 × 5 × 9 = 2520cm
Hence, the will again step to gether after a distance of 2520 cm.

9. Can two numbers have 15 as their HCF and 65 as their LCM. Give reasons in support of your answer.
Solution:
We know that H.C.F. of given numbers is a factor of their L.C.M.
But 15 is not a factor of 65
So, there can not be two numbers with H.C.F. 15 and L.C.M. 65.

10. Can two numbers have 12 as their HCF and 72 as their LCM. Give reasons in support of your answer.
Solution:
We know that H.C.F. of given numbers is a factor of their L.C.M.
Here, 12 divides 72 exactly. So 12 is a factor of 72
Hence, there can be two numbers with H.C.F. 12 and L.C.M 72.

11. The HCF and LCM of two numbers are 13 and 182 respectively. If one of the numbers is 26. Find other numbers.
Solution:
H.C.F. = 13 and L.C.M. = 182,
1st number = 25
Now, 1st number × 2nd number = H.C.F. × L.C.M.
26 × 2nd number = 13 × 182
∴ 2nd number = \(\frac {13×182}{26}\)
= 91

12. The LCM of two co-prime numbers is 195. If one number is 15 then find the other number.
Solution:
L.C.M. of two co-prime numbers = 195
H.C.F. of two co-prime numbers = 1
One number = 15
1st number × 2nd number = H.C.F. × L.C.M.
15 × 2nd number= 1 × 195
∴ 2nd number = \(\frac {1×195}{15}\)
= 13

13. The H.C.F. of two numbers is 6 and product of two numbers is 216. Find their L.C.M.
Solution:
H.C.F. of two numbers = 6
Product of two numbers = 216
We know that
H.C.F. × L.C.M. = Product of two numbers
∴ 6 × L.C.M. = 216
∴ L.C.M. = \(\frac {216}{6}\) = 36

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 3 Playing with Numbers Ex 3.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.4

1. Find H.C.F. of the following numbers by prime factorisation:

Question (i)
30, 42
Solution:
First we write the prime factorization of the given numbers

We find that 2 occurs two times and 3 occurs two times as common factors.
∴ HCF of 30 and 42 = 2 × 3 = 6

Question (ii)
135,225
Solution:
First we write the prime factorization of the given number

We find that 3 occurs two times and 5 occurs once as common factors
∴ HCF of 135 and 225 = 3 x 3 x 5 = 45

Question (iii)
180,192
Solution:
First we write the prime factorisation of the given numbers

We find that 2 occurs twice and 3 occurs once as common factors
HCF of 180 and 192
= 2 × 2 × 3 = 12

Question (iv)
49,91,175
Solution:
First we write the prime factorization of the given numbers

We find that 7 occurs once as a common factor.
∴ HCF of 49, 91 and 175 = 7

Question (v)
144, 252, 630.
Solution:
First we write the prime factorisation of the given numbers

We find that 2 occurs once and 3 occurs twice as common factors.
∴ HCF of 144, 252 and 630
= 2 × 3 × 3 = 18

2. Find H.C.F. of the following numbers using division method:

Question (i)
170, 238
Solution:
Given numbers are 170 and 238

Hence, H.C.F. of 170 and 238 = 34

Question (ii)
54, 144
Solution:
Given numbers are 54 and 144

Hence, H.C.F. of 54 and 144 = 18

Question (iii)
72, 88
Solution:
Given numbers are 72 and 88.

Hence, H.C.F. of 72 and 88 = 8

Question (iv)
96, 240, 336
Solution:
Given numbers are 96, 240 and 336 Consider any two numbers say 96 and 240

∴ H.C.F. of 96 and 240 = 48
Now, we find H.C.F. of 48 and 336
∴ H.C.F. of 48 and 336 = 48
Hence, H.C.F. of 96, 240 and 336 = 48

Question (v)
120, 156, 192.
Solution:
Given numbers are 120, 156 and 192 Consider any two numbers say 120 and 156

∴ H.C.F. of 12 and 192 = 12
Hence, H.C.F. of 120, 156 and 192 = 12

3. What is the H.C.F. of two prime numbers?
Solution:
H.C.F. of two prime numbers = 1.

4. What is the H.C.F. of two consecutive even numbers?
Solution:
The H.C.F. of two consecutive even numbers = 2.

5. What is the H.C.F. of two consecutive natural numbers?
Solution:
H.C.F. of two consecutive natural numbers = 1.

6. What is the H.C.F. of two consecutive odd numbers?
Solution:
H.C.F. of two conseutive odd numbers = 1.

7. Find the greatest number which divides 245 and 1029, leaving a remainder 5 in each case.
Solution:
Given that, required number when divides 245 and 1029, the remainder is 5 in each case.
⇒ 245 – 5 = 240 and 1029 – 5 = 1024 are completely divisible by the required number.
⇒ Required number is the highest common factor of 240 and 1024. Since it is given that required number is the greatest number.
∴ Required number is the H.C.F. 240 and 1024.

Hence, required number (H.C.F.) of 240 and 1024 = 16

8. Find the greatest number that can divide 782 and 460 leaving remainder 2 and 5 respectively.
Solution:
Required greatest number = H.C.F. of (782 – 2) and (460 – 5)
= H.C.F. of 780 and 455 = 65

Hence required greatest number = 65

9. Find the greatest number that will divide 398,437 and 540 leaving remainders 7,12 and 13 respectively.
Solution:
Required greatest number = H.C.F. of (398 – 7), (437 – 12) and (540 – 13)
= H.C.F. of 391, 425 and 527

∴ 391 = 17 × 23
425 = 5 × 5 × 17
and 527 = 17 × 31
∴ H.C.F. = 17
Hence, required greatest number = 17

10. Two different containers contain 529 litres and 667 litres of milk respectively. Find the maximum capacity of container which can measure the milk of both containers in exact number of times.
Solution:
We have to find, maximum capacity of a container which measure both conainers in exact number of times.
⇒ We required the maximum number which divides 529 and 667
⇒ Required number = H.C.F. of 529 and 667 = 23

Hence required capacity of container = 23 litres

11. There are 136 apples, 170 mangoes and 255 oranges. These are to be packed in boxes containing the same number of fruits. Find the greatest number of fruits possible in each box.
Solution:
We have to find the greatest number of fruits in each box ,
So, we required greatest numbers which divides 136, 170 and 255
∴ Required greatest number of fruits possible in each box
= H.C.F. of 136, 170 and 255
Now take any two numbers, say 136 and 170
H.C.F. of 136 and 170 = 34
Now find H.C.F. of 34 and 255
∴ H.C.F. of 34 and 255 = 17
H.C.F. of 136, 170 and 255 = 17

∴ Hence the greatest number of fruits possible in each box = 17

12. Three pieces of timber 54 m, 36 m and 24 m long, have to be divided into planks of the same length. What is the greatest possible length of each plank?
Solution:
We have to find the greatest possible length of each plank.
So, we required the maximum number which divides 54 m, 36 m and 24 m.
∴ Required length of each plank = H.C.F. of 54 m, 36 m and 24 m
Now, take any two numbers, say 54 and 36
H.C.F. of 54 and 36 = 18
Now find the H.C.F. of 18 and 24
H.C.F. 18 and 24 = 6
H.C.F. 54, 36 and 24 = 6

Hence, the greatest length of each plank = 6m

13. A room Measures 4.8 m and 5.04 m. Find the size of the largest square tile that can be used to tile the floor without cutting any tile.
Solution:
We have to find the size of largest square tile that can be used to the floor without cutting any tile.
∴ Required size of tile = H.C.F. of 4.8 and 5.04 m
= H.C.F. of 480 cm and 504 cm [1 m – 100 cm]
∴ H.C.F. of 480 cm and 504 cm = 24 cm

Hence size of each square tile = 24 cm

14. Reduce each of the following fractions to lowest forms:

Question (i)
\(\frac {85}{102}\)
Solution:
In order to reduce given fraction to the lowest terms,
We divide numerator and denominator by their H.C.F.
Now we find H.C.F. of 85 and 102 Clearly H.C.F. of 85 and 102 = 17

Question (ii)
\(\frac {52}{130}\)
Solution:
We find H.C.F. of 52 and 130
Clearly H.C.F. of 52 and 130 = 26

Question (iii)
\(\frac {289}{391}\)
Solution:
We find H.C.F. of 289 and 391
Clearly, H.C.F. of 289 and 391 = 17

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 3 Playing with Numbers Ex 3.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.1

1. Write down all the factors of each of the following:

Question (i)
18
Solution:
18 = 1 × 18
18 = 2 × 9
18 = 3 × 6
So, 1, 2, 3, 6, 9 and 18 are factors of 18

Question (ii)
24
Solution:
24 = 1 × 24
24 = 2 × 12
24 = 3 × 8
24 = 4 × 6
So, 1, 2, 3, 4, 6, 8, 12 and 24 are factors of 24

Question (iii)
45
Solution:
45 = 1 × 45
45 = 3 × 15
45 = 5 × 9
So, 1, 3, 5, 9, 15 and 45 are factors of 45

Question (iv)
60
Solution:
60 = 1 × 60
60 = 2 × 30
60 = 3 × 20
60 = 4 × 15
60 = 5 × 12
60 = 6 × 10
So, the factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60

Question (v)
65.
Solution:
65 = 1 × 65
65 = 5 × 13
So, 1, 5, 13 and 65 are the factors of 65

2. Write down the first six multiples of each of the following:

Question (i)
6
Solution:
First six multiples of 6 are:
6, 12, 18, 24, 30 and 36

Question (ii)
9
Solution:
First six multiples of 9 are:
9, 18, 27, 36, 45 and 54

Question (iii)
11
Solution:
First six multiples of 11 are:
11, 22, 33, 44, 55 and 66

Question (iv)
15
Solution:
First six multiples of 15 are:
15, 30, 45, 60, 75 and 90

Question (v)
24.
Solution:
First six multiples of 24 are:
24, 48, 72, 96, 120 and 144

3. List all the numbers less than 100 that are multiples of:

Question (i)
17
Solution:
Multiples of 17 less than 100 are:
17, 34, 51, 68 and 85

Question (ii)
12
Solution:
Multiples of 12 less than 100 are:
12, 24, 36,48, 60, 72, 84 and 96

Question (iii)
21.
Solution:
Multiples of 21 less than 100 are:
21, 42, 63 and 84

4. Which of the following are prime numbers?

Question (i)
39
Solution:
Given number = 39
We find that 39 is divisible by 3.
∴ It has more than two factors.
∴ So, 39 is not a prime number

Question (ii)
129
Solution:
Given number =129
It is divisible by 1 and itself So, it has exactly two factors.
∴ 129 is a prime number

Question (iii)
177
Solution:
Given number = 177
We find that 177 is divisble by 3
∴ It has more than two factors.
So, 177 is not a prime number

Question (iv)
203
Solution:
Given number = 203
It is divisible by 1 and itself
So, 203 is a prime number

Question (v)
237
Solution:
Given number = 237
We find that 237 is divisible by 3
∴ It has more than two factors.
So, 237 is not a prime number

Question (vi)
361.
Solution:
Given number = 361
We find that 361 is divisible by 19
∴ It has more than two factors.
So, 361 is not a prime number

5. Express each of the following as sum of two odd prime numbers:

Question (i)
16
Solution:
16 = 3 + 13
= 5 + 11

Question (ii)
28
Solution:
28 = 11+ 17

Question (iii)
40.
Solution:
40 = 3 + 37
= 11 + 29
= 17 + 23

6. Write all the prime numbers between the given numbers:

Question (i)
1 to 25
Solution:
Prime numbers between 1 to 25 are:
2, 3, 5, 7, 11, 13, 17, 19, 23

Question (ii)
85 to 105
Solution:
Prime numbers between 85 to 105 are:
89, 97, 101, 103

Question (iii)
120 to 140.
Solution:
Prime numbers between 120 to 140 are:
127, 129, 131, 137, 139

7. Is 36 a perfect number?
Solution:
Factors of 36 are:
2, 3, 4, 6, 9, 12, 18, 36
Sum of all the factors of 36
= 2 + 3 + 4 + 6 + 9 + 12+18 + 36
= 90
= 2 × 45
But sum of all factors of a number = 2 × Number
Thus, 36 is not a perfect number

8. Find the missing factors:

Question (i)
(i) 5 × …. = 30
(ii) …. × 6 = 48
(iii) 7 × …. = 63
(iv) …. × 8 = 104
(v) …. × 7 = 105.
Solution:
(i) 5 × 6 =30
(ii) 8 × 6 = 48
(iii) 7 × 9 = 63
(iv) 13 × 8 = 104
(v) 15 × 7 = 105.

9. List all 2-digit prime numbers, in which both the digits are prime numbers.
Solution:
All 2-digit numbers, in which both the digits are prime numbers are:
23, 37, 53, 73

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 2 Whole Numbers MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 2 Whole Numbers MCQ Questions

Multiple Choice Questions

Question 1.
The smallest whole number is:
(a) 0
(b) 1
(c) 2
(d) 3.
Answer:
(a) 0

Question 2.
The smallest natural number is:
(a) 0
(b) 1
(c) 2
(d) 3.
Answer:
(b) 1

Question 3.
The successor of 38899 is:
(a) 39000
(b) 38900
(c) 39900
(d) 38800.
Answer:
(b) 38900

Question 4.
The predecessor of 24100 is:
(a) 24999
(b) 24009
(c) 24199
(d) 24099.
Answer:
(d) 24099.

Question 5.
The statement 4 + 3 = 3 + 4 represents:
(a) Closure
(b) Associative
(c) Commutative property
(d) Identity.
Answer:
(c) Commutative property

Question 6.
Which of the following is the additive identity?
(a) 0
(b) 1
(c) 2
(d) 3.
Answer:
(a) 0

Question 7.
The multiplicative identity is ………………. .
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 8.
15 × 32 + 15 × 68 = …………….. .
(a) 1400
(b) 1600
(c) 1700
(d) 1500
Answer:
(d) 1500

Question 9.
The largest 4 digit number divisible by 13 is:
(a) 9997
(b) 9999
(c) 9995
(d) 9991.
Answer:
(a) 9997

Question 10.
The successor of 3 digit largest number is:
(a) 100
(b) 998
(c) 1001
(d) 1000
Answer:
(d) 1000

Question 11.
Which of the following is shown on the given number line?

(a) 2 + 5
(b) 5 + 2
(c) 7 – 2
(d) 7 – 5.
Answer:
(d) 7 – 5

Question 12.
The whole number which comes just before 10001 is:
(a) 10000
(b) 10002
(c) 9999
(d) 9998.
Answer:
(a) 10000

Question 13.
The smallest natural number is:
(a) 1
(b) 0
(c) 9
(d) 10
Answer:
(a) 1

Question 14.
Which is the smallest whole number?
(a) 1
(b) 0
(c) -1
(d) 9
Answer:
(b) 0

Question 15.
Which is the successor of 100199?
(a) 100198
(b) 100197
(c) 100200
(d) 100201.
Answer:
(c) 100200

Question 16.
Which is the predecessor of 10000?
(a) 10001
(b) 9999
(c) 10002
(d) 9998.
Answer:
(b) 9999

Question 17.
How many whole numbers are there between 32 and 53?
(a) 21
(b) 22
(c) 19
(d) 20.
Answer:
(d) 20

Fill in the blanks:

1. 25 …………… 205
2. 10001 …………. 9999
3. 15 × 0 = …………….
4. 0 ÷ 25 = ………….
5. 1 ÷ 1 = ……………

Answer:

1. <
2. >
3. 0
4. 0
5. 1

Write True or False:

Question 1.
Zero is smallest natural number. (True/False)
Answer:
False

Question 2.
All natural numbers are whole numbers. (True/False)
Answer:
True

Question 3.
All whole numbers are, natural numbers. (True/False)
Answer:
False

Question 4.
The naitural number 1 has no predecessor. (True/False)
Answer:
True

Question 5.
500 is the predecessor of 490. (True/False)
Answer:
False

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 2 Whole Numbers Ex 2.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.3

1. If the product of two whole numbers is zero. Can we say that one or both of them will be zero? Justify through examples.
Solution:
One of them can be Zero i.e. 0 × 5 = 0
Both of them can be Zero i.e. 0 × 0 = 0.

2. If the product of two whole numbers is 1. Can we say that one or both of them will be 1? Justify through examples.
Solution:
Both of them will be 1.
Example: 1 × 1 = 1.

3. Observe the pattern in the following and fill in the blanks:

Solution:

4. Observe the pattern and fill in the blanks:

Solution:

5. Represent numbers from 24 to 30 according to rectangular, square or triangular pattern.
Solution:
Numbers from 24 to 30 are 24, 25, 26, 27, 28, 29, 30.

6. Study the following pattern:

Question (i)

Hence find the sum of
(a) First 12 odd numbers
(b) First 50 odd numbers.
Solution:
(a) Sum of first 12 odd numbers = 12 × 12 = 144
(b) Sum of first 50 odd numbers = 50 × 50 = 2500

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 3 Playing with Numbers Ex 3.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.3

1. Find prime factors of the following numbers by factor tree method:

Question (i)
96
Solution:

∴ Prime factorisation of 96 = 2 × 2 × 2 × 2 × 2 × 3

Question (ii)
120
Solution:

∴ Prime factorisation of 120 = 2 × 2 × 2 × 3 × 5

Question (iii)
180.
Solution:

∴ Prime factorisation of 180 = 2 × 2 × 3 × 3 × 5

2. Complete each factor tree:

Question (i)

Solution:

Question (ii)

Solution:

Question (iii)

Solution:

3. Find the prime factors of the following numbers by division method:

Question (i)
420
Solution:

∴ 420 = 2 × 2 × 3 × 5 × 7

Question (ii)
980
Solution:

∴ 980 = 2 × 2 × 5 × 7 × 7

Question (iii)
225
Solution:

∴ 225 = 3 × 3 × 5 × 5

Question (iv)
150
Solution:

∴ 150 = 2 × 3 × 5 × 5

Question (v)
324
Solution:

∴ 324 = 2 × 2 × 3 × 3 × 3 × 3

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 3 Playing with Numbers Ex 3.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.2

1. Find the common factors of the followings:

Question (i)
16 and 24
Solution:
The factors of 16
= 1, 2, 4, 8, 16
The factors of 24
= 1, 2, 3, 4, 6, 8, 12, 24
Common factors of 16 and 24
= 1, 2, 4, 8

Question (ii)
25 and 40
Solution:
The factors of 25
= 1, 5, 25
The factors of 40
= 1, 2, 4, 5, 8, 10, 20, 40
Common factors of 25 and 40
= 1, 5

Question (iii)
24 and 36
Solution:
The factors of 24
= 1, 2, 3, 4, 6, 8, 12, 24
The factors of 36
= 1, 2, 3, 4, 6, 12, 18, 36
Common factors of 24 and 36
= 1, 2, 3, 4, 6, 12

Question (iv)
14, 35 and 42
Solution:
The factors of 14
= 1, 2, 7, 14
The factors of 35
= 1, 5, 7, 35
The factors of 42
= 1,2,3, 6, 7, 21, 42
Common factors of 14, 35 and 42
= 1, 7

Question (v)
15, 24 and 35.
Solution:
The factors of 15
= 1, 3, 5, 15
The factors of 24
= 1, 2, 3, 4, 6, 8, 12, 24
The factors of 35
= 1, 5, 7, 35
Common factors of 15, 24 and 35.
= 1

2. Find first three common multiples of the followings:

Question (i)
3 and 5
Solution:
The multiples of 3
= 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45
The multiples of 5
= 5, 10, 15, 20, 25, 30, 35, 40,45
First three common multiples of 3 and 5
= 15, 30 and 45

Question (ii)
6 and 8
Solution:
The multiples of 6
= 6, 12, 18, 24, 30, 36, 42, 48 54, 60, 66, 72
The multiples of 8
= 8, 16, 24, 32, 40, 48, 56, 64, 72
First three common multiples of 6 and 8
= 24, 48 and 72

Question (iii)
2, 3 and 4.
Solution:
The multiples of 2
= 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36
The multiples of 3
= 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36
The multiples of 4
= 4, 8, 12, 16, 20, 24, 28, 32, 36
First three common multiples of 2, 3 and 4
= 12, 24 and 36

3. Which of the following numbers are divisible by 2 or 4?

Question (i)
52314
Solution:
52314 is divisible by 2 as it is even number.
52314 is not divisible by 4 because the last two digits i.e. 14 which is not divisible by 4

Question (ii)
678913
Solution:
678913 is not divisible by 2. As it is an odd number.
678913 is not divisible by 4 because the last two digits i.e. 13 is not divisible by 4.

Question (iii)
4056784
Solution:
4056784 is divisible by 2. As it is an even number.
4056784 is also divisible by 4 because the last two digits i.e. 84 which is divisible by 4.

Question (iv)
21536
Solution:
21536 is divisible by 2. As it is an even number.
21536 is divisible by 4. As number formed by their last two digits is divisible by 4.

Question (v)
412318.
Solution:
412318 is divisible by 2. As it is an even number.
412318. is not divisible by 4. As number formed by their last two digits is not divisible by 4.

4. Which of the following numbers are divisible by 3 or 9?

Question (i)
654312
Solution:
654312 is divisible by 3.
As sum of its digits = 6 + 5 + 4 + 3 + 1 + 2 = 21, which is divisible by 3.
654312 is not divisible by 9.
As sum of its digits = 21, which is not divisible by 9.

Question (ii)
516735
Solution:
516735 is divisible by 3.
As sum of its digits = 5 + 1 + 6 + 7 + 3 + 5 = 27, which is divisible by 3.
516735 is also divisible by 9.
As sum of its digits = 27, which is divisible by 3.

Question (iii)
423152
Solution:
423152 is divisible by 3.
As sum of its digits = 4 + 2 + 3 + 1 + 5 + 2=17, which is not divisible by 3.
423152 is also not divisible by 9.
As sum of its digits = 17, which is not divisible by 9.

Question (iv)
704355
Solution:
704355 is divisible by 3.
As sum of its digits = 7 + 0 + 4 + 3 + 5 + 5 = 24, which is divisible by 3.
704355 is not divisible by 9.
As sum of its digits = 24, which is not divisible by 9.

Question (v)
215478.
Solution:
215478 is divisible by 3.
As sum of its digits = 2 + 1 + 5 + 4 + 7 + 8 = 27, which is divisible by 3.
215478 is divisible by 9.
As sum of its digits = 27, which is divisible by 9.

5. Which of the following numbers are divisible by 5 or 10?

Question (i)
456803
Solution:
456803 is not divisible by 5
As its last digit is not 0 or 5.
456803 is not divisible by 10
As its last digit is not 0.

Question (ii)
654130
Solution:
654130 is divisible by both 5 and 10
As its last digit is 0.

Question (iii)
256785
Solution:
256785 is divisible by 5
As its last digit is 5.
256785 is not divisible by 10
As its last digit is not 0.

Question (iv)
412508
Solution:
412508 is not divisible by 5
As its last digit is not 0 or 5.
412508 is not divisible by 10
As its last digit is not 0.

Question (v)
872565.
Solution:
872565 is divisible by 5
As its last digit is 5.
872565 is not divisible by 10
As its last digit is not 0.

6. Which of the following numbers are divisible by 8?

Question (i)
457432
Solution:
457432 is divisible by 8, because its last three digits are 432, which is divisible by 8.

Question (ii)
5134214
Solution:
5134214 is not divisible by 8, because its last three digits are 214, which is not divisible by 8.

Question (iii)
7232000
Solution:
7232000 is divisible by 8, because its last three digits are 000, which is divisible by 8.

Question (iv)
5124328
Solution:
5124328 is divisible by 8, because its last three digits are 328, which is divisible by 8.

Question (v)
642516.
Solution:
642516 is not divisible by 8, because its last three digits are 516, which is not divisible by 8.

7. Which of the following numbers are divisible by 6?

Question (i)
425424
Solution:
425424 is divisible by 2 because, it has 4 in its units place.
Sum of digits = 4 + 2 + 5+4 + 2 + 4 = 21
Sum of digits of 425424 is divisible by 3.
∴ 425424 is divisible by 2 as well as 3
Hence, 425424 is divisible by 6.

Question (ii)
617415
Solution:
617415 is not divisible by 2 because, it has 5 in its units place.
∴ 617415 is not divisible by 6.

Question (iii)
3415026
Solution:
3415026 is divisible by 2 because, it has 6 in its units place.
Sum of digits = 3 + 4 + 1 + 5 + 0 + 2 + 6 = 21
Sum of digits of 3415026 is divisible by 3
So, 3415026 is divisible by 3
∴ 3415026 is divisible by 2 as well as 3
Hence, 3415026 is divisible by 6.

Question (iv)
4065842
Solution:
4065842 is divisible by 2 because, it has 2 in its units place.
Sum of digits = 4 + 0 + 6 + 5 + 8 + 4 + 2 = 29
Sum of digits of 4065842 is not divisible by 3.
So, 4065842 is not divisible by 3.
∴ 4065842 is divisible by 2 but not by 3.
Hence, 4065842 is not divisible by 6.

Question (v)
725436.
Solution:
725436 is divisible by 2 because, it has 6 in its units place.
Sum of digits = 7 + 2 + 5 + 4 + 3 + 6 = 27
Sum of digits of 725436 is divisible by 3.
So, 725436 is divisible by 3.
∴ 725436 is divisible by 2 as well as 3
Hence, 725436 is divisible by 6.

8. Which of the following numbers are divisible by 11?

Question (i)
4281970
Solution:
4281970 is divisible by 11.

Since sum of its digits in odd places = 4 + 8 + 9 + 0 = 21 and
sum of its digits in even places = 2 + 1 + 7 = 10
Their difference = 21 – 10=11, which is odd places digits divisible by 11.

Question (ii)
8049536
Solution:
8049536 is divisible by 11.

Since sum of its digits in odd places = 8 + 4 + 5 + 6 = 23
and sum of its digits in even places = 0 + 9 + 3 = 12
Difference = 23 – 12 = 11, which is divisible by 11.

Question (iii)
1234321
Solution:
1234321 is divisible by 11.

Since sum of its digits in odd places = 1 + 3 + 3 + 1 = 8
and sum of its digits in even places = 2 + 4 + 2 = 8
Difference = 8 – 8 = 0.

Question (iv)
6450828
Solution:
6450828 is not divisible by 11.

Since sum of its digits in odd places = 6 + 5 + 8 + 8 = 27
and sum of its digits in even places = 4 + 0 + 2 = 6
Difference = 27 – 6 = 21, which is not divisible by 11.

Question (v)
5648346.
Solution:
5648346 is divisible by 11.

Since sum of its digits in odd places = 5 + 4 + 3 + 6 = 18 and
sum of its digits in even places = 6 + 8 + 4 = 18.
Difference = 18 – 18 = 0.

9. State True or False:

Question (i)
If a number is divisible by 24, then it is also divisible by 3 and 8.
Solution:
True

Question (ii)
60 and 90 both are divisible by 10 then their sum is not divisible by 10.
Solution:
False

Question (iii)
If a number is divisible by 8 then it is also divisible by 16.
Solution:
False

Question (iv)
If a number is divisible by 15 then it is also divisible by 3.
Solution:
True

Question (v)
144 and 72 are divisible by 12 then their difference is also divisible by 12.
Solution:
True

10. If a number is divisible by 5 and 9 then by which other numbers will that number be always divisible?
Solution:
If a number is divisible by 5 and 9. Then the number is also divisible by their product i.e. 5 × 9 = 45.

11. Which of the following pairs are co-prime?

Question (i)
25, 35
Solution:
Two numbers are said to be co-prime if they do not have a common factor other than 1.
Given numbers are 25 and 35 Factors of 25 = 1, 5, 25
Factors of 35 = 1, 5, 7, 35
Since 25 and 35 have 1 and 5 two common factors
∴ 25 and 33 are not co-prime.

Question (ii)
16,21
Solution:
Given numbers are 16 and 21
Factors of 16 = 1, 2, 4, 8, 16
Factors of 21 = 3, 7, 21
There is only 1 common factors 16 and 21 are co-prime
∴ 16 and 21 are co-prime.

Question (iii)
24, 41
Solution:
Given numbers are 24 and 41
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Factors of 41 = 1, 41
There is only one (1) common factors.
∴ 24 and 41 are co-prime.

Question (iv)
48,33
Solution:
Given numbers are 48 and 33
Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Factors of 33 = 1, 3, 11
There are two common factors 1 and 3.
∴ 48 and 33 are not co-prime.

Question (v)
20, 57.
Solution:
Given numbers are 20 and 57
Factors of 20 = 1, 2, 4, 5, 10, 20
Factors of 57 = 1, 3, 19, 57
There is only only one (1) common factors.
∴ 20 and 57 are co-prime.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 2 Whole Numbers Ex 2.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2

1. Find the sum by suitable arrangement of terms:

Question (a)
837 + 208 + 363
Solution:
837 + 208 + 363
= (837 + 363) + 208
= 1200 + 208
= 1408

Question (b)
1962 + 453 + 1538 + 647.
Solution:
1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647)
= 3500 + 1100
= 4600

2. Find the product by suitable arrangement of terms:

Question (a)
2 × 1497 × 50
Solution:
= (2 × 50) × 1497
= 100 × 1497
= 149700

Question (b)
4 × 263 × 25
Solution:
= (4 × 25) × 263
= 100 × 263
= 26300

Question (c)
8 × 163 × 125
Solution:
= (8 × 125) × 163
= 1000 × 163
= 163000

Question (d)
963 × 16 × 25
Solution:
= 963 × (16 × 25)
= 963 × 400
= 385200

Question (e)
5 × 171 × 60
Solution:
= (5 × 60) × 171
= 300 × 171
= 51300

Question (f)
125 × 40 × 8 × 25
Solution:
= (125 × 40) × (8 × 25)
= 5000 × 200
= 1000000

Question (g)
30921 × 25 × 40 × 2
Solution:
= 30921 × (25 × 40) × 2
= 30921 × 1000 × 2
= 61842000

Question (h)
4 × 2 × 1932 × 125
Solution:
4 × 2 × 1932 × 125
= 1932 × (4 × 2 × 125)
= 1932 × 1000
= 1932000

Question (i)
5462 × 25 × 4 × 2.
Solution:
= 5462 × 2 × 25 × 4
= 10924 × 100
= 1092400

3. Find the value of each of the following using distributive property:

Question (a)
(649 × 8) + (649 × 2)
Solution:
(649 × 8) + (649 × 2)
= 649 × (8 + 2)
= 649 × 10
= 6490

Question (b)
(6524 × 69) + (6524 × 31)
Solution:
(6524 × 69) + (6524 × 31)
= 6524 × (69 + 31)
= 6524 × 100
= 652400

Question (c)
(2986 × 35) + (2986 × 65)
Solution:
(2986 × 35) + (2986 × 65)
= 2986 × (35 + 65)
= 2986 × 100
= 298600

Question (d)
(6001 × 172) – (6001 × 72).
Solution:
(6001 × 172) – (6001 × 72)
= 6001 × (172 – 72)
= 6001 × 100
= 600100

4. Find the value of the following:

Question (a)
493 × 8 + 493 × 2
Solution:
(a) 493 × 8 + 493 × 2
= 493 × (8 + 2) = 493 × 10
= 4930

Question (b)
24579 × 93 + 7 × 24579
Solution:
24579 × 93 + 7 × 24579
= 24579 × (93 + 7)
= 24579 × 100
= 2457900

Question (c)
3845 × 5 × 782 + 769 × 25 × 218
Solution:
3845 × 5 × 782 + 769 × 25 × 218
= 769 × 5 × 5 × 782 + 769 × 5 × 5 × 218
= 769 × 5 × 5 × (782 + 218)
= 769 × 25 × 1000
= 19225 × 1000
= 19225000

Question (d)
3297 × 999 + 3297.
Solution:
3297 × 999 + 3297
= (3297) × (999 + 1)
= 3297 × 1000
= 3297000

5. Find the product, using suitable properties:

Question (a)
738 × 103
Solution:
= 738 × (100 + 3)
= (738 × 100) + (738 × 3)
[Using a × (b + c) = (a × b) + (a × c)]
= 73800 + 2214
= 76014

Question (b)
854 × 102
Solution:
= 854 × (100 + 2)
= (854 × 100) + (854 × 2)
[Using a × (b + c) = (a × b) + (a × c)]
= 85400 + 1708
= 87108

Question (c)
258 × 1008
Solution:
= 258 × (1000 + 8)
= (258 × 1000) + (258 x 8)
[Using a × (b + c) = (a × b) + (a × c)]
= 258000 + 2064
= 260064

Question (d)
736 × 93
Solution:
= 736 × (100 – 7)
= 736 × 100 = 736 × 7
[Using a × (b – c) = (a × b) – (a × c)]
= 73600 – 5152
= 68448

Question (e)
816 × 745
Solution:
= (800 + 16) × 745
= 800 × 745 + 16 × 745
[Using a × (b + c) = (a × b) + (a × c)]
= 596000 + 11920
= 607920

Question (f)
2032 × 613
Solution:
= 2032 × (600 + 13)
= 2032 × 600 + 2032 × 13
[Using a × (b + c) = (a × b) + (a × c)]
= 1219200 + 26416
= 1245616

6. A taxi driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs ₹ 78 per litre, how much he spend in all on petrol?
Solution:
Petrol filled on Monday = 40 Litres
Petrol filled on Tuesday = 50 Litres
Total Petrol filled = 40 + 50 = 90 Litres
Cost per litre = ₹ 78
Total Cost = 90 × ₹ 78
= ₹ 7020

7. A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs ₹ 35 per litre, how much money is due to the vendor per day?
Solution:
Milk supplied in the morning = 32 Litres
Milk supplied in the evening = 68 Litres
Total milk supplied = 32 + 68
= 100 Litres
Cost Per litre = ₹ 35
∴ Total cost of milk per day = 100 × ₹ 35
= ₹ 3500

8. We know that 0 × 0 = 0. Is there any other whole number which when multiplied by itself gives the product equal to the number itself? Find out the number.
Solution:
Yes, there is a whole number 1.
Here 1 × 1 = 1.

9. Fill in the blanks:

Question (i)
(a) 15 × 0 = ………….. .
(b) 15 + 0 = ………….. .
(c) 15 – 0 = ………….. .
(d) 15 ÷ 0 = ………….. .
(e) 0 × 15 = ………….. .
(f) 0 + 15 = ………….. .
(g) 0 ÷ 15 = ………….. .
(h) 15 × 1 = ………….. .
(i) 15 ÷ 1 = ………….. .
(j) 1 ÷ 1 = ………….. .
Solution:
(a) 0,
(b) 15,
(c) 15,
(d) Not defined,
(e) 0,
(f) 15,
(g) 0,
(h) 15,
(i) 15,
(j) 1.

Question 10.
The product of two Whole numbers is zero. What do you conclude. Explain with example.
Solution:
We conclude that one number must be zero such 25 × 0 = 0.

Question 11.
Match the following:

1. 537 × 106 = 537 ×100 + 537 × 6	(a)  Commutativity under multiplication
2. 4 × 47 × 25 = 4 × 25 × 47	(b) Commutativity under addition
3. 70 + 1923 + 30 = 70 + 30 + 1923	(c)  Distributivity of multiplication over addition.

Solution:

1. 537 × 106 = 537 × 100 + 537 × 6	(c)  Distributivity of multiplication over addition.
2. 4 × 47 × 25 = 4 × 25 × 47	(a)  Commutativity under multiplication
3. 70 + 1923 + 30 = 70 + 30 + 1923	(b) Commutativity under addition

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 2 Whole Numbers Ex 2.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.1

1. Answer the following questions:

Question (a)
Write the smallest whole number.
Solution:
The smallest Whole number = 0

Question (b)
Write the smallest natural number.
Solution:
The smallest natural number = 1

Question (c)
Write the successor of 0 in whole numbers.
Solution:
Successor of 0 = 0 + 1 = 1

Question (d)
Write the predecessor of 0 in whole numbers.
Solution:
Predecessor of 0 is whole number is not possible.

Question (e)
Write the Largest whole number.
Solution:
Largest whole number is not possible.

2. Which of the following statements are True (T) and which are False (F)?

Question (a)
Zero is the smallest natural number.
Solution:
False

Question (b)
Zero is the smallest whole number.
Solution:
True

Question (c)
Every whole number is a natural number.
Solution:
False

Question (d)
Every natural number is a whole number.
Solution:
True

Question (e)
1 is the smallest whole number.
Solution:
False

Question (f)
The natural number 1 has no predecessor in natural numbers.
Solution:
True

Question (g)
The whole number 1 has no predecessor in whole numbers.
Solution:
False

Question (h)
Successor of the largest two-digit number is smallest three-digit number.
Solution:
True

Question (i)
The successor of a two-digit number is always a two-digit number.
Solution:
False

Question (j)
300 is the predecessor of 299.
Solution:
False

Question (k)
500 is the successor of 499.
Solution:
True

Question (l)
The predecessor of a two-digit number is never a single-digit number.
Solution:
False

3. Write the successor of each of following:

Question (a)
100909
Solution:
Successor of 100909
= 100909 + 1
= 100910

Question (b)
4630999
Solution:
Successor of 4630999
= 4630999 + 1
= 4631000

Question (c)
830001
Solution:
Successor of 830001
= 830001 + 1
= 830002

Question (d)
99999.
Solution:
Successor of 99999
= 99999 + 1
= 100000

4. Write the predecessor of each of following:

Question (a)
1000
Solution:
Predecessor of 1000 = 1000 – 1
= 999

Question (b)
208090
Solution:
Predecessor of 208090 = 208090 – 1
= 208089

Question (c)
7654321
Solution:
Predecessor of 7654321 = 7654321 – 1
= 7654320

Question (d)
12576.
Solution:
Predecessor of 12576 = 12576 – 1
= 12575

5. Represent the following numbers on the number line: 2, 0, 3, 5, 7, 11, 15.
Solution:
Draw a line. Mark a point on it. Label it ‘O’. Mark a second point to the right of 0. Label it 1. The distance between these points labelled as 0 and 1 is called unit distance. On this line, mark a point to the right of 1 and at unit distance from 1 and label it 2. In this way go on labeling points at unit distance as 3, 4, 5, …………… on the line.

6. How many whole numbers are there between 22 and 43?
Solution:
Whole numbers between 22 and 43 are 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42
∴ There are 20 whole numbers between 22 and 43.
Or [(43 – 22) – 1 = 21 – 1 = 20].

7. Draw a number line to represent each of following on it.

Question (a)
3 + 2
Solution:
We draw a number line and move 3 steps from 0 to the right and mark this point as A.
Now, starting from A we move 2 steps towards right and move at B.

OA = 4, AB = 2, OB = 5
Hence, OB = 3 + 2 = 5.

Question (b)
4 + 5
Solution:
We draw a number line.
Starting from point 0 (i.e. zero), we move 4 steps to the right and mark this point as A.
Now, starting from A we move 5 steps towards right and arrive at B.

OA = 4, AB = 5, OB = 9
Hence, OB = 4 + 5 = 9.

Question (c)
6 + 2
Solution:
We draw a number line.
Starting from point 0 (i.e. zero), we move 6 steps to the right and mark this point as A.
Now, starting from A we move 2 steps towards right and arrive at B.

OA = 6, AB = 2, OB = 8
Hence, OB = 6 + 2 = 8.

Question (d)
8 – 3
Solution:
We draw a number line.
Starting from point 0 (i.e. zero), we move 8 steps to the right and arrive at A.
Now, starting from A we move 3 steps to the left of A and arrive at B.

OA = 8, AB = 3, OB = 5
Hence, OB = 8 – 3 = 5.

Question (e)
7 – 4
Solution:
We draw a number line.
Starting from point 0 (i.e. zero), we move 7 steps to the right and arrive at A.
Now, starting from A we move 4 steps to the left of A and arrive at B.

OA = 7, AB = 4, OB = 3
Hence, OB = 7 – 4 = 3.

Question (f)
7 – 2
Solution:
We draw a number line.
Starting from point 0 (i.e. zero), we move 7 steps to the right and arrive at A.
Now, starting from A, we move 2 steps to the left of A and arrive at B.

OA = 7, AB = 2, OB = 5
Hence, OB = 7 – 2 = 5.

Question (g)
3 × 3
Solution:
We draw a number line.
Starting from 0 we move 3 units to the right of 0 to arrive at A.
We make two more such same moves starting from A (total 3 moves of 3 units each) to reach finally at C which represents 9.

Hence, 3 × 3 = 9.

Question (h)
2 × 5
Solution:
We draw a number line.
We start from 0 move 5 units at a time to right.
We make 2 such moves. We shall reach at 10.

So, 2 × 5 = 10.

Question (i)
3 × 5
Solution:
We draw a number line.
We start from 0, move 5 units at a time to right.
We make 3 such moves. We shall reach at 15.

So, 3 × 5 = 15

Question (j)
9 ÷ 3
We draw a number line.
Starting from 0, we move 9 units to the right of 0 to arrive at A.
Now, from A take moves of 3 units to the left of A till we reach at ‘O’. We observe that there are 3 moves.

So, 9 ÷ 3 = 3.

Question (k)
12 ÷ 4
We draw a number line.
Starting from 0, we move 12 units to the right of 0 to arrive at A.
Now, from A take moves of 4 units to the left of A till we reach at ‘O’. We observe that there are 3 moves.

So, 12 ÷ 4 = 3.

Question (l)
10 ÷ 2
Solution:
We draw a number line.
Starting from 0, we move 10 units to the right of 0 to arrive at A.
Now, from A take moves of 2 units to the left c A till we reach at ‘O’. We observe that there are 5 moves.

So, 10 ÷ 2 = 5.

8. Fill in the blanks with the appropriate symbol < or > :

Question (i)
(a) 25 ……………. 205
(b) 170 …………… 107
(c) 415 …………… 514
(d) 10001 ………….. 9999
(e) 2300014 ………….. 2300041
(f) 99999 …………… 888888.
Solution:
(a) 25 < 205 (b) 170 > 107
(c) 415 < 514 (d) 10001 > 9999
(c) 2300014 < 2300041
(f) 99999 < 888888.