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 MCQ Questions with Answers.

PSEB 6th Class Maths Chapter 3 Playing with Numbers MCQ Questions

Multiple Choice Questions

Question 1.
Which number is a factor of every, number?
(a) 0
(b) 1
(c) 2
(d) 3.
Answer:
(b) 1

Question 2.
How many even numbers are prime?
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(a) 1

Question 3.
The smallest composite number is:
(a) 1
(b) 2
(c) 3
(d) 4.
Answer:
(d) 4.

Question 4.
Which of the following number is a perfect number?
(a) 8
(b) 6
(c) 12
(d) 18.
Answer:
(b) 6

Question 5.
Which of the following is not a multiple of 7?
(a) 35
(b) 48
(c) 56
(d) 91.
Answer:
(b) 48

Question 6.
Which of the following is not a factor of 36?
(a) 12
(b) 6
(c) 9
(d) 8.
Answer:
(d) 8.

Question 7.
The number of prime numbers upto 25 are:
(a) 9
(b) 10
(c) 8
(d) 12.
Answer:
(a) 9

Question 8.
Which mathematician gave the method to find prime and composite numbers?
(a) Aryabhatta
(b) Ramayan
(c) Eratosthenes
(d) Goldbach.
Answer:
(c) Eratosthenes

Question 9.
The statement “Every even number greater than 4 can be expressed as the sum of two odd prime numbers” is given by:
(a) Goldbach
(b) Eratosthenes
(c) Aryabhatta
(d) Ramanujan.
Answer:
(a) Goldbach

Question 10.
Which of the following is a prime number?
(a) 221
(b) 195
(c) 97
(d) 111.
Answer:
(c) 97

Question 11.
Which of the following number is divisible by 4?
(a) 52369
(b) 25746
(c) 21564
(d) 83426.
Answer:
(c) 21564

Question 12.
Which of the following is not true?
(a) If a number is factor of two numbers then it is also factor of their sum.
(b) If a number is factor of two numbers then it is also factor of their difference.
(c) 15 and 24 are co-prime to each other.
(d) 1 is neither prime nor composite.
Answer:
(c) 15 and 24 are co-prime to each other.

Question 13.
Which of the following pair is co-prime?
(a) (12, 25)
(b) (18, 27)
(c) (25, 35)
(d) (21, 56).
Answer:
(a) (12, 25)

Question 14.
Which of the following number is divisible by 8?
(a) 123568
(b) 412580
(c) 258124
(d) 453230.
Answer:
(a) 123568

Question 15.
Prime factorisation of 84:
(a) 2 × 2 × 3 × 2 × 7
(b) 7 × 2 × 3 × 3
(c) 2 × 3 × 7 × 2
(d) 3 × 2 × 3 × 2 × 7.
Answer:
(c) 2 × 3 × 7 × 2

Question 16.
H.C.F. of 25 and 45 is:
(a) 15
(b) 5
(c) 225
(d) 135.
Answer:
(b) 5

Question 17.
If L.C.M. of two numbers is 36 then which of the following can not be their H.C.F.?
(a) 9
(b) 12
(c) 8
(d) 18.
Answer:
(c) 8

Question 18.
The L.C.M. of two co-prime numbers is 143. If one number is 11 then find other number.
(a) 132
(b) 154
(c) 18
(d) 13.
Answer:
(d) 13.

Question 19.
Find the greatest number which divides 145 and 235 leaving the remainder 1 in each case.
(a) 24
(b) 18
(c) 19
(d) 17.
Answer:
(b) 18

Question 20.
The greatest 4 digit number which is divisible by 12,15 and 20.
(a) 9990
(b) 9000
(c) 9960
(d) 9999.
Answer:
(c) 9960

Question 21.
Which of the following is a prime number?
(a) 23
(b) 51
(c) 39
(d) 26.
Answer:
(a) 23

Question 22.
Which of die following is a prime number?
(a) 32
(b) 30
(c) 31.
(d) 33.
Answer:
(c) 31.

Question 23.
Which of the following is a composite number?
(a) 12
(b) 19
(c) 29
(d) 31.
Answer:
(a) 12

Question 24.
Which of the following is an even number?
(a) 13
(b) 15
(c) 16
(d) 19.
Answer:
(c) 16

Question 25.
Which of the following is an odd number?
(a) 12
(b) 13
(c) 14
(d) 20.
Answer:
(b) 13

Fill in the blanks:

Question 1.
…………… is an even prime number?
Answer:
2

Question 2.
…………… is the greatest prime number between 1 and 10.
Answer:
7

Question 3.
……………. is neither prime nor composite number.
Answer:
1

Question 4.
A number which has only two factors is called a …………….. number.
Answer:
prime number

Question 5.
A number which has more than two factors is called a ……………… number.
Answer:
composite number

Write True/False:

Question 1.
The sum of three odd number is even. (True/False)
Answer:
False

Question 2.
All prime numbers are odd. (True/False)
Answer:
False

Question 3.
All even numbers are composite numbers. (True/False)
Answer:
False

Question 4.
1 neither prime nor composite. (True/False)
Answer:
True

Question 5.
If a number is factor of two numbers then it is also factor of their sum. (True/False)
Answer:
True

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