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Haryana State Board HBSE 6th Class Maths Solutions Chapter 3 Playing With Numbers Ex 3.7 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 3 Playing With Numbers Exercise 3.7

Question 1.
Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of fertiliser exact number of times.
Solution:
Maximum value of weight which can measure exact number of times will be the H.C.F. of 75 and 69.

∴ H.C.F. of 75 and 69 is 3:
Hence, maximum value of the required wt. = 3 kg

Question 2.
Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps ?
Solution:
The required minimum distance each should walk must be a least common multiple of the measures of their steps.
Thus, we find the L.C.M. of 63, 70, 77.

∴ L.C.M. of 63, 70, 77 = 7 × 2 × 5 × 3 × 3 × 11 = 6930.
Hence, the required minimum distance is 6930 cm.

Question 3.
The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.
Solution:
The longest tape which can measure the three dimensions of the room exactly will be the H.C.F. of 825, 675, 450.

∴ H.C.F. of 825, 675, 450 is 75.
Hence, the required longest tape will be 75 cm.

Question 4.
Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.
Solution:

∴ L.C.M. of 6, 8, 12= 2 × 2 × 3 × 2 = 24
Now, the smallest 3-digit number = 100

24 – 4 = 20
Hence, the required number
= 100 + 20 = 120

Question 5.
Determine the largest 3-digit number exactly divisible by 8,10 and 12.
Solution:

∴ L.C.M. of 8, 10, 12
=2 × 2 × 2 × 3 × 5 = 120
Now, largest 3-digit number = 999

Hence, the required number
= 999 – 39
= 960.

Question 6.
The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 A.M., at what time they will change simultaneously again?
Solution:
48 sec, 72 sec, 108 sec.

∴ L.C.M. of 48, 72, 108
= 2 × 2 × 2 × 3 × 3 × 2 × 3
= 432 sec. = 7 min. 12 sec.
They will change simultaneously after 7 min. 12 sec
i.e., at 7 hr. 7 min. 12 sec. A.M.

Question 7.
Three tankers contain 403 litres, .434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of three containers exact number of times.
Solution:
The maximum capacity of a container will be the
H.C.F. of 403, 434, 465.

∴ H.C.F. of 403, 434, 465 is 31.
Hence, maximum capacity of the required container = 31 litres. Ans.

Question 8.
Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
Solution:
We first find the L.C.M. of 6, 15 and 18 as follows :

∴ L.C.M. of 6, 15, 18 = 2 × 3 × 3 × 5 = 90
∴ 90 is the least number which when divided by the given numbers will leave remainder ‘0’ in each case. But we need the least number that leaves remainder 5 in each case.
Hence, the required number = 90 + 5 = 95.

Question 9.
Find the smallest four-digit number which is divisible by 18,24 and 32.
Solution:
We first find the L.C.M. of 18, 24, 32

∴ L.C.M. of 18, 24, 32 .
= 2 × 2 × 2 × 2 × 3 × 3 × 2
= 288
Now, the smallest 4-digit number = 1000

288 – 136 = 152
Hence, the required number
= 1000 + 1.52= 1152

Question 10.
Find the L.C.M. of the following numbers in which one number is always a multiple of 3 :
(a) 9 and 4
(b) 12 and 5
(c) 6 and. 5
(d) 15 and 4.
Observe a common property in the obtained L.C.M. Is L.C.M. the product of two numbers in each case ? Is L.C.M. always a multiple of 3 ?
Solution:

(a) L.C.M.of 9 and 4 = 2 × 2 × 3 × 3 = 36
= 9 × 4
(b) L.C.M. of 12 and 5 = 2 × 2 × 3 × 5
60 = 12 × 5
(c) LC.M. of 6 and 5 = 2 × 3 × 5 = 30 = 6 × 5
(d) L.C.M.of 15 and 4 = 2 × 2 × 3 × 5 = 60 = 15 × 4
Yes, L.C.M. is equal to the product of two numbers in each case.
Yes, L.C.M. is always a multiple of 3.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 3 Playing With Numbers Ex 3.6 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 3 Playing With Numbers Exercise 3.6

Question 1.
Find the H.C.F of the following numbers:
(a) 18, 48
(b) 30, 42
(c) 18, 60
(d) 27, 63
(e) 36, 84
(f) 34, 102
(g) 70, 105, 175
(h) 91, 112, 49
(i) 18, 54, 81
(j) 12, 45, 75.
Solution:
(a)

Question 2.
What is the H.C.F. of two consecutive :
(a) numbers ? .
(b) even numbers ?
(c) odd numbers ?
Solution:
(a) H.C.F. of two consecutive numbers is 1.
(b) H.C.F. of two consecutive even numbers is 2.
(c) H.C.F. of two consecutive odd numbers is 1.

Question 3.
H.C.F. of co-prime numbers 4 and 15 was found as follows : Factorisation : 4 = 2×2 and 15 = 3 × 5, since there is no common prime factor, so H.C.F. of 4 and 15 is 0. Is the answer correct ? If not, what is the correct H.C.F. ?
Solution:
No, the correct H.C.F. is 1.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 3 Playing With Numbers Ex 3.5 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 3 Playing With Numbers Exercise 3.5

Question 1.
Which of the following state¬ments are true ?
(a) If a number is divisible by 3, it must be divisible by 9.
(b) If a num ber is divisible is 9, it must be divisible by 3.
(c) A number is divisible by 18, if it is divisible by both 3 and 6.
(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.
(e) If two numbers are co-primes, at least one of them must be prime.
(f) All that numbers divisible by 4 must also be divisible by 8.
(g) All that numbers divisible by 8 must also be divisible by 4.
(h) The sum of two consecutive odd numbers is divisible by 4.
(i) If a number exactly divides two numbers separately, it must exactly divide their sum.
(j) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately.
Solution:
(a) F, (b) T, (c) T, (d) T, (e) F, (/) F, (g) T, (h) T, (i) T, (/) F.

Question 2.
Here are two different factor trees for 60. Write the missing numbers.

Solution:

Question 3.
Which factors are not included in the prime factorisation of a composite number ?
Solution:
Let us consider any composite number say 12.
12 = 1 x 12
= 2 x 6
= 3 x 4
and

∴ Factors of 12 are : 1, 2, 3, 4, 6, 12.
Prime factorisation of 12 = 2 x 2 x 3
We clearly see that composite factors 4, 6, 12 are not included in the prime factorisation of a composite number.
Hence, composite factors are not included in the prime factorisation of a composite number.

Question 4.
Write the greatest four-digit number and express it into the form of prime factorisation.
Solution:
Greatest 4-digit number is 9999.

∴ Prime factorisation of 9999
= 3 x 3 x 11 x 101

Question 5.
Write the smallest five-digit number and express it into the form of prime factorisation.
Solution:
Smallest 5-digit number is 10000

∴ Prime factorisation of 10000
= 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5.

Question 6.
Find all the prime factors of 1729 and arrange them in ascending order. Now state the relation, if any, between two consecutive prime factors.
Solution:

∴ Prime factors of 1729 are 7, 13, 19
Difference between two consecutive prime factors is 6.

Question 7.
The product of three consecutive numbers is always divisible by 6. Explain this statement with the help of some examples.
Solution:
(i) Let us take three consecutive numbers 5, 6, 7.
Their product = 5 x 6 x 7 = 210 which is divisible by 6.
(ii) Let us take three consecutive numbers 9, 10, 11.
Their product = 9 x 10 x 11 = 990 which is divisible by 6.
(iii) Let us take three consecutive numbers 23, 24, 25.
Their product = 23 x 24 x 25 = 13800 which is divisible by 6.
(iv) Let us take three consecutive numbers 2, 3, 4.
Their product = 2 x 3 x 4 = 24 which indivisible by 6.

Question 8.
In which of the following expressions, prime factorisation has been done :
(a) 24 = 2 x 3 x 4
(b) 56 = 1 x 7 x 2 x 2 x 2
(c) 70 = 2 x 5 x 7
(d) 54 = 2 x 3 x 9.
Solution:
(c) 70 = 2 x 5 x 7.
In this expression prime factorisation has been done.
(b) 56 = 1 x 7 x 2 x 2 x 2 In this expression prime factorisation has been done.

Question 9.
Write the prime factorisation of 15470.

∴ Prime factorisation of 15470
= 2 x 5 x 7 x 13 x 17.

Question 10.
Determine if 25110 is divisible by 45.
Solution:
∵ The unit’s digit of 25110 is 0,
∴ 25110 is divisible by 5.
Sum of the digits = 2 + 5+ l + l + 0 = 9 which is divisible by 9.
∴ 25110 is divisible by 9.
Now, 5 and 9 are co-prime numbers.
∴ 25110 is divisible by their product 5 x 9 = 45.

Question 11.
18 is divisible by both 2 and 3. It is also divisible by 2 x 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 x 6 = 24 ? If not, give an example to justify your answer.
Solution:
No, because 4 and 6 are not co-primes.
e.g. (i) 36 is divisible by both 4 and 6. But it is not divisible by 24.
(ii) 12 is divisible by both 4 and 6 but 12 is not divisible by 24.

Question 12.
I am the smallest number having four different prime factors. Can you find me ?
Solution:
The smallest four different prime factors are 2, 3, 5, 7.
Hence, the smallest number, having four different prime factors = 2 x 3 x 5 x 7 = 210.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 3 Playing With Numbers Ex 3.4 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 3 Playing With Numbers Exercise 3.4

Question 1.
Find the common factor of:
(a) 20 and 28
(b) 15 and 25
(c) 35 and 50
(d) 56 and 120.
Solution:
(a) 20 = 1 × 20
=2 × 10
=4 × 5

28 = 1 × 28
= 2 × 14
= 4 × 7
∴ Factors of 20 are : 1, 2, 4, 5, 10, 20.
∴ Factors of 28 are : 1, 2, 4, 7, 14, 28.

Hence, common factors of 20 and 28 are: 1, 2, 4.

(b) 15 = 1 × 15
= 3 × 5
25 = 1 × 25
= 5 × 5
∴ Factors of 15 are: 1, 3, 5, 15.
Factors of 25 are: 1, 5, 25.
Hence, common factors of 15 and 25 are: 1, 5 .

(c) 35 = 1 × 35
= 5 × 7

50 = 1 × 50
= 2 × 25
= 5 × 10
∴ Factors of 35 are: 1, 5, 7, 35.
Factors 50 are : 1, 2, 5, 10, 25, 50.
Hence, common factors of 35 and 50 are: 1, 5.

(d)
56 = 1 × 56
= 2 × 28
= 4 × 14
= 7 × 8

120 = 1 × 120
= 2 × 60
= 3 × 40
= 4 × 30
= 5 × 24

= 6 × 20
= 8 × 15
= 10 × 12
∴ Factors of 56 are : 1, 2, 4, 7, 8, 14, 28, 56.
Factors of 120 are : 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120.
Hence, common factors of 56 and 120 are : 1, 2, 4, 8.

Question 2.
Find the common factors of:
(a) 4, 8 and 12 (b) 5, 15 and 25.
Solution:
4 = 1 × 4
= 2 × 2

8 = 1 × 8
= 2 × 4

12,= 1 × 12
= 2 × 6
= 3 × 4

∴ Factors of 4 are : 1, 2, 4.
Factors of 8 are : 1, 2, 4, 8.
Factors of 12 are : 1, 2, 3, 4, 6, 12.
Hence, common factors of 4, 8 and 12 are : 1, 2, 4.

(b)
5 = 1 × 5

15 = 1 × 15
= 3 × 5
25 = 1 × 25
= 5 × 5

∴ Factors of 5 are : 1, 5.
Factors of 15 are : 1, 3, 5, 15.
Factors of 25 are : 1, 5, 25.
Hence, common factors of 5, 15 and 25 are : 1, 5.

Question 3.
Find first three common multiples of:
(a) 6 and 8 (b) 12 and 18.
Solution:
(a) Multiples of 6 are ; 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72,……
Multiples of 8 are : 8, 16, 24, 32, 40, 48, 56, 64, 72, 80,
Hence, first three common multiples of 6 and 8 are : 24, 48, 72,………

(b) Multiples of 12 are : 12, 24, 36, 48, 60,
72, 84, 96, 108, 120,
Multiples of 18 are : 18, 36, 54, 72, 90, 108, 126, 144, 162,180, ……….
Hence, first three multiples of 12 and 18 are : 36, 72, 108…….

Question 4.
Write all the numbers less than 100 which are common multiples of 3 and 4.
Solution:
Multiples of 3 less than 100 are : 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99.
Multiples of 4 less than 100 dre : 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96.
Henee, common multiples of 3 and 4 less than 100 are : 12, 24, 36, 48, 60, 72, 84, 96.

Question 5.
Which of the following numbers are co-prime :
(a) 18 and 35
(b) 15 and 37
(c) 30 and 415
(d) 17 and 68
(e) 216 and 215
(f) 81 and 16.
Solution:
(a) 18 = 1 × 18
=2 × 9
= 3 × 6

35 = 1 × 35
=5 × 7

∴ Factors of 18 are : 1, 2, 3, 6, 9, 18.
Factors of 35 are : 1, 5, 7, 35.
∵ Common factor of 18 and 35 is 1 only. Hence, 18 and 35 are co-primes.

(b) 15 = 1 × 15 37 = 1 × 37
= 3 x 5
∴ Factors of 15 are : 1, 3, 5, 15.
Factors of 37 are : 1, 37.
∵ Common factor of 15 and 37 is 1 only.
Hence, 15 and 37 are co-prime.

(c) 30 = 1 × 30
= 2 × 15
= 3 × 10
= 5 × 6

415 = 1 × 415
= 5 × 83

∴ Factors of 30 are : 1, 2, 3, 5, 6, 10, 15, 30.
Factors of 415 are : 1, 5, 83, 415.
∵ Common factor of 30 and 415 are 1, 5.
Hence, 30 and 415 are not co-prime.

(d) 17 = 1 × 17

68 = 1 × 68
= 2 × 34
= 4 × 17
∴ Factors of 17 are : 1, 17.
Factors of 68 are : 1, 2, 4, 17, 34, 68.
∵ Common factor of 17 and 68 are : 1, 17.
Hence, 17 and 68 are not co-prime.

(e) 216 = 1 × 216
= 3 × 72
= 4 × 54
= 6 × 36
= 8 × 27
= 9 × 24
= 12 × 18

215 = 1 × 215
= 5 × 43
∴ Factors of 216 are : 1, 3, 4, 6, 8, 9, 12,
18, 24, 27, 36, 54, 72, 216.
Factors of 215 are : 1, 5, 43, 215.
∵ Common factor of 216 and 215 is 1 only. Henoa 216 and 215 are co-prime.

(f) 81 = 1 × 81
= 3 × 27
=9 × 9

16 = 1 × 16
=2 × 8
=4 × 4

∴ Factors of 81 are : 1, 3, 9, 27, 81.
Factors of 16 are : 1, 2, 4, 8, 16.
∵ Common factor of 81 and 16 is 1 only.
Hence, 81 and 16 are co-prime.

Question 6.
A number is divisible by both 5 and 12. By which other number will that number always be divisible ?
Solution:
∵ 5 and 12 are co-prime.
∴ A number divisible by bpth 5 and 12 will always be divisible by their product i.e. 5 × 12 = 60.

Question 7.
A number is divisible by 12. By what other numbers will that number be divisible ?
Solution:
12 = 1 × 12
= 2 × 6
= 3 × 4
Factors of 12 other than 12 are : 1, 2, 3, 4, 6.
Hence, if a number is divisible by 12, it will also be divisible by 1, 2, 3, 4 and 6.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 3 Playing With Numbers Ex 3.3 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 3 Playing With Numbers Exercise 3.3

Question 1.
Using divisibility tests, determine which of the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 10; by 11 (say yes or no) :
(i) 128 (ii) 990 (ii) 1586 (iv) 275 (iv) 6686
(vi) 639210 (vii) 429714 (viii) 2856 (ix) (x) 40GS39
Solution:

Question 2.
Using divisibility tests, deter¬mine which of the following numbers are divible by 4 ; by 8 :
(a) 572 (b) 726352
(c) 5500 (d) 6000
(e) ^2159 (f) 14560
(g) 21084 (h) 31795072
(i) 1700 (j) 2150
Solution:
(a) 572, since the number formed by the last two digits is divisible by 4, hence it is divisible by 4; 572, since the nurtiber formed by the last three digits is not divisible by 8, hence it is not divisible by 8.
(b) 726352, since the number formed by the last two digits is divisible by 4, hence it is divisible by 4.
726352, since the number formed by the last three digits is divisible by 8, hence it is divisible by 8.
(c) 5500, since the number formed by the last two digits are divisible by 4, hence it is divisible by 4.
5500, since the number formed by the last three digits are not divisible by 8, hence it is not divisible by 8.
(d) 6000, since the number formed by the last two digits are divisible by 4, hence it is divisible by 4.
6000, since the number formed by the last three digits are divisible by 8, hence it is divisible by 8.
(e) 12159, since the number formed by the last two digits are not divisible by 4, hence it is not divisible by 4.
12159, since the number formed by the last three digits are not divisible by 8, hence it is not divisible by 8. .
(f) 14560, since the number formed by the last twp digits are divisible by 4, hence it is divisible by 4.
14560, since the number formed by the last three digits are divisible by 8, hence it is divisible by 8.
(g) 21084, since the number formed by the last two digits are divisible by 4, hence it is divisible by 4.
21084, since the number formed by the last three digits are not divisible by 8, hence it is not divisible by 8.
(h) 31795072, since the number formed by the last two digits are divisible by 4, hence it is divisible by 4.
31795072, since the number formed by the last three digits are divisible by 8, hence it is divisible by 8.
(i) 1700, since the number formed by the last two digits are divisible by 4, hence it is divisible by 4.
1700, since the number formed by the last three digits are not divisible by 8, hence it is not divisible by 8.
(j) 2150, since the number formed by the last two digits are not divisible by 4, hence it is not divisible by 4.
2150, since the number formed by the last three digits are not divisible by 8, hence it is not divisible by 8.

Question 3.
Using divisibility tests, deter¬mine which of the following numbers are
(a) 297144
(b) 1258
(c) 4335
(d) 61233
(e) 901352
(f) 438750
(g) 1790184
(h) 12583
(i) 17852.
(j) 639210
Solution:
(a) 297144, since last digit is 4,
∴ It is divisible by 2.
Sum of the digits = 2 + 9 + 7 + 1 + 4 + 4 = 27, which is a multiple of 3,
∴ It is divisible by 3.
Since 297144 is divisible by both 2 and 3, hence it is divisible by 6.

(b) 1258, since last digit is 8, It is divisi-ble by 2. Sum of the digits =1 + 2 + 5 + 8 = 16, which
is not a multiple of 3, .∴ It is not divisible by 3.
Hence, 1258 is not divisible by 6.

(c) 4335, since last digit is 5,
∴ It is not divisible by 2.
Hence, 4335 is not divisible by 6.

(d) 61233, since last digit is 3,
∴ It is not divisible by 2.
Hence, 61233 is not divisible by 6.

(e) 901352, since last digit is 2,
∴ It is divisi-ble by 2.
Sum of the digits = 9 + 0 + 1 + 3 + 5 + 2 = 20,
which is not a multiple of 3,
∴ It is not divisible by 3.
Hence, 901352 is not divisible by 6.

(f) 438750, since last digit is 0,
∴ It is divisi¬ble by 2.
Sum of the digits = 4 + 3 + 8 + 7 + 5 + 0 = 27, which is a multiple of 3,
∴ It is divisible by 3. Since, 438750 is divisible by both 2 and 3,
hence it is divisible by 6.

(g) 1790184, since last digit is 4,
∴ It is divisible by 2.
Sum of the digits = 1 + 7 + 9 + 0 + 1 + 8 + 4 = 30, which is a multiple of 3,
∴ It is divisible by 3.
Since, 1790184 is divisible by both 2 and 3, hence it is divisible by 6.

(h) 12583, since last digit is 3,
∴ It is not divisible by 2.
Hence, 12583 is not divisible by 6.

(i) 639210, since last digit is 0,
∴ It is divisible by 2.
Sum of the digits = 6 + 3 + 9 + 2 + 1 + 0 = 21, which is a multiple of 3,
∴ It is divisible by 3.
Since 639210 is divisible by both 2 and 3, hence it is divisible by 6.

(j) 17852, since last digit is 2,
∴ It is divisible by 2.
Sum of the digits = 1 + 7 + 8 + 5 + 2 = 23, which is not a multiple of 3,
∴ It is not divisible by 3.
Hence, 17852 is not divisible by 6.

Question 4.
Using divisibility tests, determine which of the following numbers are divisible by 11.
(a) 5545 (b) 10824 (c) 7138965 (d) 70169308 (e) 10000001 if) 901153.
Solution:

We observe that in each case except in (Hi), the difference is either 0 or divisible by 11. Hence, 5445, 10824,70169308, 10000001, 901153 are all divisible by 11, but 7138965 is not divisible by 11.

Question 5.
Write the (a) smallest digit, (6) largest digit in the blank space of each of the following numbers so that the number is divisible by 3 :
(a) ……………….. 6724, (b) 4765 …………….. 2.
Solution:
(a) 2 + 6 + 7 + 2 + 4 = 21, which is a multiple of 3.
8 + 6 + 7 + 2 + 4 = 27, which is a multiple of 3.
Hence, (a) smallest digit is 2 and (b) largest dig^ is 8.
(b) 4 + 7 + 6 + 5 + 0 + 2 = 24, which is a multiple of 3.
4 + 7 + 6 + 5 + 9 + 2 = 33, which is a multiple of 3.
Hence, (a) smallest digit is 0 and (b) largest digit is 9.

Question 6.
Write digit in the blank space of each of the following numbers so that the number is divisible by 11 :
(a) 92 ………….. 389 (6) 8……… 9484.
Solution:

Haryana State Board HBSE 6th Class Maths Solutions Chapter 3 Playing With Numbers Ex 3.1 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 3 Playing With Numbers Exercise 3.1

Question 1.
Write all the factors of the following numbers :
(a) 24 (b) 15 (c) 21
(d) 27 (e) 12 (f) 20
(g) 18 (h) 23 (i) 36
Solution:
(a) 24 = 1 x 24
= 2 x 12
= 3 x 8
= 4 x 6
∴ All the factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.

(b) 15 = 1 x 15
= 3 x 5
∴ All the factors of 15 are 1, 3, 5, 15. Ans.

(c) 21 = 1 x 21
= 3 x 7
∴ All the factors of 21 are 1, 3, 7, 21. Ans.

(d) 27 = 1 x 27
= 3 x 9
∴ All the factors of 27 are 1, 3, 9, 27. Ans.

(e) 12 = 1 x 12
= 2 x 6
= 3 x 4
∴ All the factors of 12 are 1, 2, 3, 4, 6, 12.

(f) 20 = 1 x 20
= 2 x 10 = 4 x 5
∴ All the factors of 20 are 1, 2, 4, 5, 10, 20.

(g) 18 = 1 x 18
= 2 x 9 = 3 x 6
∴ All the factors of 18 are 1, 2, 3, 6, 9, 18.

(h) 23 = 1 x 23
∴ All the factors of 23 are 1, 23.

(i) 36 = 1 x 36
= 2 x 18 = 3 x 12 = 4 x 9 = 6 x 6
∴ All the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.

Question 2.
Write the first five multiples of:
(a) 5 (b) 8 (c) 9
Solution:
(a) The required multiples of 5 are :
5 x 1 = 5
5 x 2 = 10
5 x 3 = 15
5 x 4 = 20
5 x 5 = 25.
i.e., 5, 10, 15, 20 and 25

(b) The required multiples of 8 are :
8 x 1 = 8
8 x 2 = 16
8 x 3 = 24
8 x 4 = 32
8 x 5 = 40.
i.e., 8, 16, 24, 32 and 40

(c) The required multiples of 9 are :
9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45.
i.e., 9, 18, 27, 36 and 45.

Question 3.
Match the items in column 1 with the items in column 2 :
Column 1 — Column 2
(i) 35 — (a) Multiple of 8
(ii) 15 — (b) Multiple of 7
(iii) 16 — (c) Multiple of 70
(iv) 20 — (d) Factor of 30
(v) 25 — (e) Factor of 50
(f) Factor of 20
Solution:
(i) – (b) Multiple of 7
(ii) – (d) Factor of 30
(iii) – (a) Multiple of 8
(iv) – (f) Factor of 20
(v) – (e) Factor of 50

Question 4.
Find all the multiples of 9 upto 100.
Solution:
9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45
9 x 6 = 54
9 x 7 = 63
9 x 8 = 72
9 x 9 = 81
9 x 10 = 90
9 x 11 = 99.
Hence all the multiples of 9 upto 100 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 3 Playing With Numbers Ex 3.2 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 3 Playing With Numbers Exercise 3.2

Question 1.
What is the sum of any two (a) odd numbers and (b) even numbers ?
Solution:
(a) The sum of any two odd numbers is even and
(b) Sum of any two even numbers is also even.

Question 2.
State whether the following statements are true of false :
(a) The sum of three odd numbers is even.
(b) The sum of t wo odd numbers and one even number is even.
(c) The product of three odd numbers is odd.
(d) If an even number is divided by 2, the quotient is always odd.
(e) All prime num bers are odd.
(f) Prime numbers do not. have any factors.
(g) Sum of two prime numbers is always even.
(h) 2 is the only even prime number.
(i) All even numbers are composite numbers.
(j) The product of two even numbers is always even.
Solution:
(a) F (b) T (c) T (d) F (e) F (f) F (g) F (h) T (i) F (j) T.

Question 3.
The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers upto 100.
Solution:
17 and 71; 37 and 73; 79 and 97 are such pairs of prime numbers upto 100.

Question 4.
Write down separately the prime and composite numbers less that 20.
Solution:

Prime numbers less than 20 are: 2, 3, 5, 7, 11, 13, 17, 19.
Composite numbers less than 20 are: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18.

Question 5.
What is the greatest prime number between 1 and 10 ?
Solution:

7 is the greatest prime number between 1 and 10.

Question 6.
Express the following as the sum of two odd primes :
(a) 44
(b) 36
(c) 24
(d) 18
Solution:
(a) 44 = 13 + 31 = 3 + 41
(b) 36 = 13 + 23 = 5 + 31
(c) 24 = 11 + 13 = 5 + 19
(d) 18 = 5 + 13 = 7 + 11.

Question 7.
Give three pairs of prime numbers whose difference is 2.
[Remark : Two prime numbers whose difference is 2 are called twin primes].
Solution:
(i) (3, 5) (ii) (5, 7) (iii) (11, 13).

Question 8.
Which of the following numbers are prime ?
(a) 23
(b) 51
(c) 37
(d) 26.
Solution:
23 and 37 are prime numbers.

Question 9.
Write seven consecutive com¬posite numbers less than 100- so that there is no prime number between them.
Solution:
90, 91, 92, 93, 94, 95 and 96.

Question 10.
Express each of the following numbers as the sum of three odd primes:
(a) 21
(b) 31
(c) 53
(d) 61.
Solution:
(a) 21 = 3 + 5 + 13
(b) 31 = 3 + 5 + 23
(c) 53 = 3 + 7 + 43
(d) 61 = 3 + 5 + 53 = 7 + 13 + 41

Question 11.
Write five pairs of prime numbers below 20 whose sum is divisible by 5.
Solution:
(i) 2 + 3 = 5
(ii) 3 + 7 = 10
(iii) 2 + 13 = 15
(iv) 3 + 17 = 20
(v) 7 + 13 = 20.

Question 12.
Fill in the blanks in the following:
(a) A number which has only two factors is called a ……………
(b) A number which has more than two factors is called a ……………
(c) 1 is neither …………… nor ……………
(d) The smallest prime number is ……………
(e) The smallest composite number is ……………
(f) The smallest even number is ……………
Solution:
(a) Prime number
(b) Composite number.
(c) Prime, Composite
(d) 2
(e) 4
(f) 2.

Haryana State Board HBSE 6th Class Maths Solutions Chapter 2 Whole Numbers Ex 2.3 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 2 Whole Numbers Exercise 2.3

Question 1.
Which of the following will not represent zero:
(a) a + 0
(b) 0 x 0
(c) \(\frac{0}{2}\)
(d) \(\frac{10-10}{2}\)
Solution:
(a) 1 + 0 = 1
(b) 0 x 0 = 0
(c) \(\frac{0}{2}\) = 0
(d) \(\frac{10-10}{2}\) = \(\frac{0}{2}\) = 0
∴ (a) will not represent zero.

Question 2.
If the product of two whole numbers is zero, can we say that one or both of them will be zero ? Justify through e × amples.
Solution:
One of them is zero i.e., 0 × 2 = 0
Both of them are zero i.e., 0 × 0 = 0.

Question 3.
If the product of two whole numbers is 1, can we say that one or both of them will be 1 ? Justify through an e × ample.
Solution:
Both of them will be one i.ef,
1 × 1 = 1.

Question 4.
Find by distributivity method :
(a) 728 × 101
(e) 824 × 25
(e) 504 × 35.
(b) 5437 × 1001
(d) 4275 × 125
Solution:
(a) 728 × 101
728 × (100 + 1)
= 728 × 100 + 728
= 72800 + 728
= 73528

(b) 5437 × 1001
5437 × (1000 + 1)
= 5437000 + 5437
= 5442437

(c) 824 x 25 = 824 x \(\frac{100}{4}\)
\(\frac{824}{4}\) x 100 = 20600

(d) 4275 × 125
= 4275 × (100 + 25)
= 427500 + 4275 × 25
= 427500 + 106875
= 534375

(e) 504 × 35
= (500 + 4) × 35
= 500 × 35 + 4 × 35
= 17500 + 140
= 17640

Haryana State Board HBSE 6th Class Maths Solutions Chapter 2 Whole Numbers Ex 2.2 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 2 Whole Numbers Exercise 2.2

Question 1.
Find the sum by suitable rearrangement :
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647.
Solution:
(a) 837 + 208 + 363
= (837 + 363) + 208 = 1200 + 208 = 1408
(b) 1962 + 453 + 1538 + 647
= (1962 + 1538) + (453 + 647) = 3500 + 1100 = 4600

Question 2.
Find the product by a suitable rearrangement :
(a) 2 × 1768 × 50
(b) 4 × 166 × 25
(c) 8 × 291 × 125
(d) 625 × 279 × 16
(e) 285 × 5 × 60
(f) 125 × 40 × 8 × 25.
Solution:
(a) 2 × 1768 × 50
= (2 × 50) × 1768 = 100 × 1768 = 176800
(b) 4 × 166 × 25
= (4 × 25) × 166 = 100 × 166 = 16600
(c) 8 × 291 × 125
= (8 × 125) × 291 = 1000 × 291 = 291000
(d) 625 × 279 × 16
= (625 × 16) × 279 = 10000 × 279 ‘ = 2790000
(e) 285 × 5 × 60
= 285 × (5 × 60)
= 285 × 300 = 85500
(f) 125 × 40 × 8 × 25
= (125 × 8) × (40 × 25)
= 1000 × 1000 = 1000000

Question 3.
Find the value of the following :
(a) 297 × 17 + 297 × 3
(b) 54279 × 92 + 8 × 54279
(c) 81265 × 169 – 81265 × 69
(d) 3845 × 5 × 782 + 769 × 25 × 218.
Solution:
(a) 297 × 17 + 297 × 3
= 297 × (17 + 3)
= 297 × 20 = 5940 Ans.
(b) 54279 × 92 + 8 × 54279
= 54279 × (92 + 8)
= 54279 × 100 = 5427900
(c) 81265 × 169 – 81265 × 69
= 81265 (169 – 69)
= 81265 × 100 = 8126500
(d) 3845 × 5 × 782 + 769 × 25 × 218
= (769 × 5) × 5 × 782 + 769 × 25 × 218
= 769 × (5 × 5) × 782 + 769 × 25 × 218
= 769 × 25 × 782 + 769 × 25 × 218
= 769 × 25 × (782 + 218)
= 769 × 25 × 1000 = 19225000

Question 4.
Find the product, using suitable properties :
(a) 738 × 103
(b) 854 × 102
(c) 258 × 1008
(d) 1005 × 168.
Solution:
(a) 738 × 103
= 738 × (100 + 3)
= 738 × 100 + 738 × 3
= 73800 + 2214
= 76014

(b) 854 × 102 = 854 × (100 + 2)
= 854 × 100 + 854 × 2
= 85400 + 1708
= 87108.

(c) 258 × 1008
= 258 × (1000 + 8)
= 258 × 1000 + 258 × 8
= 258000 + 2064
= 260064.

(d) 1005 × 168 = (1000 + 5) × 168
= 1000 × 168 + 5 × 168
= 168000 + 840
= 168840.

Question 5.
A ta × i-driver filled his car petrol tank with 40 litres of petrol on Monday. The ne × t day, he filled the tank with 50 litres of petrol. If the petrol costs Rs. 44 per litre, how much did he spend in all on petrol ?
Solution:
Method-1:
Cost of 40 litres of petrol
= Rs. 44 × 40
= Rs. 1760
Cost of 50 litres of petrol
= Rs. 44 × 50
= Rs. 2200
∴ Total cost = Rs. (1760 + 2200)
= Rs. 3960.

Method-2 :
Petrol purchased on Monday = 40 litres
Petrol purchased on ne × t day = 50 litres
Total petrol purchased
= (40 + 50) l = 90 l
∴ Total cost = Rs. 44 × 90
= Rs. 3960.

Question 6.
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 Rs. 15 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 Milk supplied in one day = 68 litres
∴ Milk supplied in one day = (32 + 68)l = 100 l
Hence, total cost of milk per day = Rs. 15 × 100 = Rs. 1500

Question 7.
Match the following :
(i) 425 × 136 = 425 × (6 + 30 + 100) — (a) Commutativity under multiplication.
(ii) 2 × 49 × 50 = 2 × 50 × 49 — (b) Commutativity under addition.
(Hi) 80 + 2005 + 20 = 80 + 20 + 2005 — (c) Distributivity of multiplication over addition.
Solution:
(i)—(c),
(ii)—(a),
(iii)—(b).

Haryana State Board HBSE 6th Class Maths Solutions Chapter 2 Whole Numbers Ex 2.1 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 2 Whole Numbers Exercise 2.1

Question 1.
Write the next three natural numbers after 10,999.
Solution:
After 10,999 next three natural numbers are :
10,999 + 1 = 11,000
11,000 + 1 = 11,001
and 11,001 + 1 = 11,002.

Question 2.
Write the three whole numbers occurring just before 10,001.
Solution:
Before 10,001 three whole numbers are :
10,001 – 1 = 10,000
10,000 – 1 = 9,999
and 9,999 – 1 = 9,998.

Question 3.
Which is the smallest whole number ?
Solution:
Zero is the smallest whole number.

Question 4.
How many whole numbers are there between 32 and 53 ?
Solution:
There are 20 whole numbers between 32 and 53.

Question 5.
Write the successor of:
(a) 24,40,701
(b) 1,00,199
(c) 10,99,999
(d) 23,45,670.
Solution:
(a) Successor of 24,40,701 is 24,40,701 + 1 = 24,40,702.
(b) Successor of 1,00,199 is 1,00,199 + 1 = 1,00,200.
(c) Successor of 10,99,999 is 10,99,999 + 1 = 11,00,000.
(d) Successor of 23,45,670 is 23,45,670 + 1 = 23,45,671.

Question 6.
Write the predecessor of:
(a) 94
(b) 10,000
(c) 2,08,090
(d) 76,54,321.
Solution:
(a) Predecessor of 94 is 94 – 1 = 93.
(b) Predecessor of 10,000 is 10,000 – 1 = 9,999.
(c) Predecessor of 2,08,090 is 2,08,090 – 1 = 2,08,089.
(d) Predecessor of 76,54,321 is 76,54,321 – 1 = 76,54,320.

Question 7.
In each of the following pairs of numbers, state which whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them.
(a) 530, 503
(b) 370, 307
(c) 98765, 56789
(d) 9830415, 10023001.
Solution: (a) 503 is on the left to 530.
∴ 503 < 530 or 530 > 503.
(b) 307 is on the left to 370.
∴ 307 < 370 or 370 > 307. ,
(c) 56789 is on the left to 98765.
∴ 56789 < 98765 or 98765 > 56789.
(d) 9830415 is on the left to 10023001.
∴ 9830415 < 10023001 or 10023001 > 9830415.

Question 8.
Which of the following state¬ments are true (T) and which are false (F)?
(a) Zero is the smallest natural number.
(b) 400 is the predecessor of 399.
(c) Zero is the smallest whole number.
(d) 600 is the successor of 599.
(e) All natural numbers are whole numbers.
(f) All whole numbers are natural numbers.
(g) The predecessor of a two digit number is never a single digit number.
(h) One is the smallest whole number.
(i) The natural number 1 has no predecessor.
(j) The whole number 1 has no predecessor.
(k) The whole number 13 lies between 11 and 12.
(l) The whole number ‘0’ has no predecessor.
(m) The successor of a two-digit number is always a two-digit number.
Solution:
(a) F,
(b) F,
(c) T,
(d) T,
(e) T,
(f) F,
(g) F,
(h) F,
(i) T,
(j) F,
(k) F,
(l) T,
(m) F.