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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 5 Natural Numbers and Whole Numbers

November 10, 2023May 18, 2022 by Raju

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 5 Natural Numbers and Whole Numbers (Including Patterns)

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 5 Natural Numbers and Whole Numbers (Including Patterns)

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Natural Numbers and Whole Numbers Exercise 5A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the blanks :
(i) Smallest natural number is …………..
(ii) Smallest whole number is …………
(iii) Largest natural number is …………
(iv) Largest whole number is ………..
(v) All natural numbers are …………
(vi) All whole numbers are not …………
(vii) Successor of 4099 is …………..
(viii) Predecessor of 4330 is …………….
Solution:
(i) Smallest natural number is 1
(ii) Smallest whole number is 0
(iii) Largest natural number is can not be obtained
(iv) Largest whole number is can not be obtained
(v) All natural numbers are whole numbers
(vi) All whole numbers are not natural numbers
(vii) Successor of 4099 is 4099 + 1 = 4100
(viii) Predecessor of 4330 is 4330 – 1 = 4329

Question 2.
Represent the following whole numbers on a number line :
0, 3, 5, 8, 10
Solution:
Number line used to represent whole numbers 0, 3, 5, 8, 10 is as given below:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 5 Natural Numbers and Whole Numbers image - 1

Question 3.
State, true or false :
(i) Whole number are closed for addition.
(ii) If a and b are any two whole numbers, then a + b is not a whole number.
(iii) If a and b are any two whole numbers, then a + b = b + a
(iv) 0 + 18= 18 + 0
(v) Addition of whole numbers is associative.
(vi) 10 + 12 + 16 = (10 + 12) + 16 = 10 + (12 + 16)
Solution:
(i) True
(ii) True
(iii) True
(iv) True
(v) True
(vi) True

Question 4.
Fill in the blanks :
(i) 54 + 234 = 234 + …………
(ii) 332 + 497 = ………… + 332
(iii) 286 + 0 = ………..
(iv) 286 x 1 = ……….
(v) a + (b + c) = (a + ………) + c
Solution:
(i) 54 + 234 = 234 + 54
(ii) 332 + 497 = 497 + 332
(iii) 286 + 0 = 286
(iv) 286 x 1 = 286
(v) a + (b + c) = (a + b) + c

Question 5.
By re-arranging the given numbers, evaluate :
(i) 237 + 308 + 163
(ii) 162 + 253 +338 + 47
(iii) 21 + 22 + 23 + 24 + 25 + 75 + 76 + 77 + 78 + 79
(iv) 1 + 2 + 3 + 4 + 596 + 597 + 598 + 599
Solution:
(i) 237 + 308 + 163
= (237 + 163) + 308 (by associative law)
= 400 + 308 = 708
(ii) 162 + 253 + 338 + 47
= (162 + 338) + (253 + 47) (By associative law)
= 500 + 300 = 800
(iii) 21 + 22 + 23 + 24 + 25 + 75 + 76 + 77 + 78 + 79
= (21 + 79) + (22 + 78) + (23 + 77) + (24 + 76) + (25 + 75) (By associative law)
= 100 + 100+ 100+ 100 + 100 = 500
(iv) 1 + 2 + 3 + 4 + 596 + 597 + 598 + 599
= (1 + 599) + (2 + 598) + (3 + 597) + (4 + 596) (By associative law)
= 600 + 600 + 600 + 600 = 2400

Question 6.
Is a + b + c = a + (b + c) = (b + a) + c ?
Solution:
Yes, because any set of three whole numbers if the sum of any two whole numbers is added to the third whole number, then whatever be their order, the sum will remain same.

Question 7.
Which property of addition is satisfied by :
(i) 8 + 7 = 15
(ii) 3+ (5 + 4) = (3 + 5)+ 4
(iii) 8 x (8 + 0) = 8 x 8 + 8 x 0
(iv) (7 + 6) x 10 = 7 x 10 + 6 x 10
(v) (15 – 12) x 18 = 15 x 18 – 12 x 18
(vi) 16 + 0= 16
(vii) 23 + (-23) = 0
Solution:
(i) 8 + 7=15
The property used as closure property satisfied by 15.
(ii) 3+ (5 + 4) = (3 + 5)+ 4 3 +(5+ 4) = 3 + 9=12 and (3+ 5)+ 4 = 8 + 4=12
The property used as Associative law of addition satisfied by 12.
(iii) 8 x (8 + 0) = 8 x 8 + 8 x 0
8 x (8 + 0) = 8 x 8 = 64
and 8 x 8 + 8 x 0 = 64 + 0 = 64
The property used as Distributive property over addition satisfied by 64.
(iv) (7 + 6) x 10 = 7 x 10 + 6 x 10 (7 + 6) x 10= 13 x 10= 130
and 7 x 10 + 6 x 10 = 70 + 60 = 130
The property used as Distributive over addition satisfied by 130.
(v) (15 – 12) x 18= 15 x 18- 12 x 18 (15 – 12) x 18 = 3 x 18 = 54
and 15 x 18 – 12 x 18 = 270 – 216 = 54
The property used as Distributive over subtraction satisfied by 54.
(vi) 16 + 0=16
The property used as Existence of identity satisfied by 16
(vii) 23 + (-23) = 0
The property used as Additive inverse satisfied by 0.

Question 8.
State, True or False :
(i) The sum of two odd numbers is an odd number.
(ii) The sum of two odd numbers is an even number.
(iii) The sum of two even numbers is an even number.
(iv) The sum of two even numbers is an odd number.
(v) The sum of an even number and an odd number is odd number.
(vi) Every whole number is a natural number.
(vii) Every natural number is a whole number.
(viii) Every whole number + 0 = The whole number itself.
(ix) Every whole number x 1 = The whole number itself.
(x) Commutativity and associativity are properties of natural numbers and whole numbers both.
(xi) Commutativity and associativity are properties of addition for natural numbers and whole numbers both.
(xii) If x is a whole number then -x is also a whole number.
Solution:
(i) False.
The sum of two odd numbers is an even number.
(ii) True.
(iii) True.
(iv) False.
The sum of two even numbers is an even number.
(v) True.
(vi) False.
Every natural number is a whole number.
(vii) True.
(viii) True.
(ix) True.
(x) True.
(xi) True.
(xii) True.

Natural Numbers and Whole Numbers Exercise 5B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Consider two whole numbers a and b such that a is greater than b.
(i) Is a – b a whole number ? Is this result always true ?
(ii) b-a a whole number ? Is this result always true ?
Solution:
Let us take a as 2 and has 1
(i) a – b = 2 – 1 = 1,
Yes, a – b is a whole number and the result will always remain the same.
(ii) b – a = 1 – 2 = -1
No, a – b can never be a whole number. Yes, the result always be true.

Question 2.
Fill in the blanks :
(i) 8 – 0 = ……….. and 0 – 8 = ……….
8 – 0 ≠ 0 – 8, this shows subtraction of whole numbers is not ………..
(ii) 5 – 10 = ………., which is not a …………
=> Subtraction of ……….. is not closed.
(iii) 7 – 18 = ……….. and (7 – 18) – 5 = …………..
18 – 5 = …………. and (7 – 18) – 5 = ………….
Is (7 – 18) – 5 = 7 – (18 – 5) ?
=> Subtraction of whose numbers is not ………….
Solution:
(i) 8 – 0 = 8 and 0 – 8 = -8
8 – 0 & 0 – 8, this shows subtraction of whole numbers is not commutative
(ii) 5 – 10 = -5, which is not a whole number
=> Subtraction of whole numbers is not associative.
(iii) 7 – 18 = -11 and (7 – 18) – 5 = -16
18 – 5 = 13 and (7 – 18) – 5 = -6
Is (7 – 18) – 5 = 7 – (18 – 5) = ?
No, (7- 18) – 5 ≠ 7 – (18 – 5)
=> Subtraction of whole number is not associative.

Question 3.
Write the identify number, if possible for subtraction.
Solution:
It is not possible to identify the number.

Question 4.
Write the inverse, if possible for subtraction of whole numbers ?
Solution:
The inverse does not exit.

Question 5.
12 x (9 – 6) = ………….. = ………….
12 x 9 – 12 x 6 = …………. = …………….
Is 12 x (9 – 6) = 12 x 9 – 12 x 6 ? ………….
Is this type of result always true ? ……………
Name the property used here …………..
Solution:
12 x (9-6)= 12 x 3 = 36
12 x 9- 12 x 6 = 108 – 72 = 36
Is 12 x (9 – 6) = 12 x 9 – 12 x 6 ? =
Is this type of result always true ? Yes
Name the property used here Distributive property.

Question 6.
(16 – 8) x 24 = ………….. = …………..
16 x 24 – 8 x 24 = …….. – ……….. = ………..
Is (16 – 8) x 24 = 16 x 24 – 8 x 24 ? …………..
Is the type of result always true ? ………….
Name the property used here ………………
Solution:
(16 – 8) x 24 = 8 x 24 = 192
16 x 24 – 8 x 24 = 384 – 192 = 192
Is (16 – 8) x 24 = 16 x 24 – 8 x 24 ? Yes.
Is the type of result always true ? Yes
Name the property used here Distributivity.

Natural Numbers and Whole Numbers Exercise 5C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the blanks :
(i) 42 x o = ……….
(ii) 592 x 1 = ………..
(iii) 328 x 573 = ……….. x 328
(iv) 229 x ………… = 578 x 229
(v) 32 x 15 = 32 x 6 + 32 x 7 + 32 x ……….
(vi) 23 x 56 = 20 x 56 + ……… x 56
(vii) 83 x 54 + 83 x 16 = 83 x ( ……….. ) = 83 x ………….. = …………
(viii) 98 x 273 – 75 x 273 = ( …………. ) x 273 = ………… x 273
Solution:
(i) 42 x 0 = 0
(By closure property 0)
(ii) 592 x 1 = 592
(By closure property 1)
(iii) 328 x 573 = 573 x 328
(By commutative law of multiplication)
(iv) 229 x 578 = 578 x 229
(By commutative law of multiplication)
(v) 32 x 15 = 32 x 6 + 32 x 7 + 32 x 2
(By commutative law of multiplication)
(vi) 23 x 56 = 20 x 56 + 3 x 56
(By Distributive law of multiplication)
(vii) 83 x 54 + 83 x 16 = 83 x (54 + 16) = 83 x 70 = 5810
(viii) 98 x 273 – 75 x 273 = (98 – 75) x 273 = 23 x 273

Question 2.
By re-arranging the given numbers, evaluate :
(i) 2 x 487 x 50
(ii) 25 x 444 x 4
(iii) 225 x 20 x 50 x 4
Solution:
(i) 2 x 487 x 50
2 x 50=100
2 x 487 x 50 = (2 x 50) x 487 = 100 x 487 = 48700
(ii) 25 x 444 x 4
25 x 4 = 100
25 x 444 x 4 = (25 x 4) x 444 = 100 x 444 = 444000
(iii) 225 x 20 x 50 x 4
= (225 x 4) x (20 x 50) = 900 x 1000 = 900, 000

Question 3.
Use distributive law to evaluate :
(i) 984 x 102
(ii) 385 x 1004
(iii) 446 x 10002
Solution:
(i) 984 x 102
= 984 x (100 + 2)
= 984 x 100 x 984 x 2
= 98400 + 1968 = 100,368
(ii) 385 x 1004
= 385 x (1000 + 4)
= 385 x 1000 x 385 x 4
= 385000 + 1540 = 386540
(iii) 446 x 10002
= 446 x (10000 + 2)
= 446 x 10000 x 446 x 2
= 4460000 + 892 = 4460892

Question 4.
Evaluate using properties :
(i) 548 x 98
(ii) 924 x 988
(iii) 3023 x 723
Solution:
(i) 548 x 98
= (500 + 40 + 8) x 98
= 500 x 98 + 40 x 98 + 8 x 98
= 49000 + 3920 + 784 = 53704
(ii) 924 x 988
= (900 + 20 + 4) x 988
= 900 x 988 + 20 x 988 + 4 x 988
= 889200 + 19760 + 3952 = 912912
(iii) 3023 x 723
= (3000 + 20 + 3) x 723
= 3000 x 723 + 20 x 723 + 3 x 723
= 2169000 + 14460 + 2169 = 2185629

Question 5.
Evaluate using properties :
(i) 679 x 8 + 679 x 2
(ii) 284 x 12 – 284 x 2
(iii) 55873 x 94 + 55873 x 6
(iv) 7984 x 15 – 7984 x 5
(v) 8324 x 1945 – 8324 x 945
(vi) 3333 x 987 + 13 x 3333
Solution:
(i) 679 x 8 + 679 x 2
= 679 x (8 + 2) (using distributivity)
= 679 x 10 = 6790
(ii) 284 x 12-284 x 2
= 284 x (12-2) (using distributivity)
= 284 x 10 = 2840
(iii) 55873 x 94 + 55873 x 6
= 55873 x (94 + 6) (using distributivity)
= 55873 x 100 = 5587300
(iv) 7984 x 15 – 7984 x 5
= 7984 x (15 – 5) (using distributivity)
= 7984 x 10 = 79840
(v) 8324 x 1.945 – 8324 x 945
= 8324 x (1945 – 945) (using distributivity)
= 8324 x 1000 = 8324000
(vi) 3333 x 987 + 13 x 3333
= 3333 x (987 + 13) (using distributivity)
= 3333 x 1000 = 3333000

Question 6.
Find the product of the :
(i) greatest number of three digits and smallest number of five digits.
(ii) greatest number of four digits and the greatest number of three digits.
Solution:
(i) Greatest number of three digits = 999
and Smallest number of five digits = 10000
Required product = 999 x 10000 = 9990000
(ii) Greatest number of four digits = 9999
and Greatest number of three digits = 999
Required product = 9999 x 999
= 9999 x (1000 – 1)
= 9999 x 1000 – 9999 x 1 (using distributivity)
= (10000- 1) x 1000 – (10000- 1) x 1
= 10000 x 1000 – 1 x 1000 – 10000 + 1
= 10000000 – 1000 – 10000 + 1
= 10000001 – 11000
= 9989001

Question 7.
Fill in the blanks :
(i) (437 + 3) x (400 – 3) = 397 x ……….
(ii) 66 + 44 + 22 = 11 x (………..) = 11 x ……….
Solution:
(i) (437 + 3) x (400 – 3) = 397 x 440
(ii) 66 + 44 + 22 = 11 x (6 + 4 + 2) = 11 x 12 = 132

Question 8.
Evaluate :
(i) 355 x 18
(ii) 6214 x 12
(iii) 15 x 49372
(iv) 9999 x 8
Solution:
(ï) 355 x 18
(300 + 50 + 5) x 18
300 x 18 + 50 x 18 + 5 x 18
= 5400 + 900 + 90
= 6390
(ii) 6214 x 12
=(6000 + 200 + 10 + 4) x 12
=6000 x 12+ 200 x 12 + 10 x 12 +4 x 12
= 72000 + 2400 + 120 + 48
= 74568
(iii) 15 x 49372
= 15 x (40000 + 9000 + 300 + 70 + 2)
= 15 x 40000 + 15 x 9000 + 15 x 300+ 15 x 70 + 15 x 2
= 600000 + 135000 + 4500 + 1050 + 30
= 740580
(iv) 9999 x 8
= (9000 + 900 + 90 + 9) x 8
=9000 x 8+900 x 8 + 90 x 8 + 9 x 8
= 72000 + 7200 + 720 + 72
= 79992

Natural Numbers and Whole Numbers Exercise 5D – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Show that :
(i) division of whole numbers is not closed
(ii) any whole number divided by 1, always gives the number itself.
(iii) every non-zero whole number divided by itself gives 1 (one).
(iv) zero divided by any non-zero number is zero only.
(v) a whole number divided by 0 is not defined.
For each part, given above, give an example.
Solution:
(i) For example
5 and 8 are whole numbers, but 5 + 8 is not a whole number
Closure property does not exist for division.
(ii) For example
2 + 1 =2, 18 + 1 = 18, 129 + 1 = 129
The above statement is true.
(iii) For example
2 ÷ 2 = 1, 128 ÷ 128 = 1, 256 ÷ 256 = 1
The above statement is true.
(iv) For example
0 ÷ 138 = 0, 0 ÷ 2028 = 0, 0 ÷ 15140 = 0
The above statement is true.
(v) For example
3 ÷ 0 = Not defined or 19 ÷ 0 is not defined
The above statement is true.

Question 2.
If x is a whole number such that x ÷ x = x, state the value of x.
Solution:
Since, we know that any number divided by 1, always gives the number itself.
x can be any number 1, 2, 3, 4, 5, 6, ………… and so on.

Question 3.
Fill in the blanks :
(i) 987 + 1 = …………
(ii) 0 + 987 = ……….
(iii) 336 – (888 + 888) = ……….
(iv) (23 + 23) – (437 + 437) = ………
Solution:
(i) 987 ÷ 1 = 987
(ii) 0 ÷ 987 = 0
(iii) 336 – (888 + 888) = 335
(iv) (23 + 23) – (437 + 437) = 0

Question 4.
Which of the following statements are true ?
(i) 12 ÷ (6 x 2) = (12 ÷ 6) x (12 ÷ 2)
(ii) a ÷ (b – c) = \(\frac { a }{ b } -\frac { a }{ c }\)
(iii) (a – b) ÷ c = \(\frac { a }{ c } -\frac { b }{ c }\)
(vi) (15 – 13) ÷ 8 = (15 ÷ 8) – (13 ÷ 8)
(v) 8 ÷ (15 – 13) = \(\frac { 8 }{ 15 } -\frac { 8 }{ 13 }\)
Solution:
(iii) and (iv) are true.

Natural Numbers and Whole Numbers Exercise 5E – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Find the difference between the largest number of four digits and the smallest number of six digits.
Solution:
Largest number of four digits = 9999 and, smallest number of six digits = 100000
Difference = 100000 – 9999 = 90001

Question 2.
Find the difference between the smallest number of eight digits and the largest number of five digits.
Solution:
Smallest number of eight digits = 10000000
and, largest number of five digits = 99999
Difference = 10000000 – 99999 = 9900001

Question 3.
The product of two numbers is 528. If the product of their unit’s digits is 8 and the product of their ten’s digits is 4 ; find the numbers.
Solution:
Product of unit’s digit = 8 = 2 x 4
Unit’s digits are 2 and 4
Thus, the numbers are either 24 and 22
24 x 22 = 528
The required numbers are 528

Question 4.
Does there exist a number a such that a ÷ a = a ?
Solution:
Yes, A number a is 1
a ÷ a = a
1 ÷ 1 = 1

Question 5.
Divide 5936 by 43 to find the quotient and remainder. Also, check your division by using the formula,
dividend = divisor x quotient + remainder.
Solution:
On dividing 5936 by Divisor 43 gives quotient 138 and remainder 2.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 5 Natural Numbers and Whole Numbers image - 2
Now, to verify,
Divided = divisor x quotient + remainder
5936 =43 x 138 + 2
= 43 x (100 + 38) + 2
= 4300 + 1634 + 2
= 5936
Hence verified.

Natural Numbers and Whole Numbers Exercise 5F – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
(i) 1 x 9 + 1 = 10
12 x 9 + 2 = 110
123 x 9 + 3 = 1110
(ii) 9 x 9 + 7 = 88
98 x 9 + 6 = 888
987 x 9 + 5 = 8888
(iii) 1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
(iv) 111 ÷ 3 = 37
222 ÷ 6 = 37
333 ÷ 9 = 37
Solution:
(i) 1 x 9 + 1 = 10
12 x 9 + 2 = 110
123 x 9 + 3 = 111O
1234 x 9 + 4 = 11110
12345 x 9 + 5 = 111110
123456 x 9 + 6 = 1111110
(ii) 9 x 9 + 7 = 88
98 x 9 + 6 = 888
987 x 9 + 5 = 8888
9876 x 9 + 4 = 88888
98765 x 9 + 3 = 888888
987654 x 9 + 2 = 8888888
(iii) 1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + = 98765
123456 x 8 + 6 = 987654
(iv) 111 ÷ 3 = 37
222 ÷ 6 = 37
333 ÷ 9 = 37
444 ÷ 12 =37
555 ÷ 15 = 37
666 ÷ 18 =37

Question 2.
Complete each of the following magic squares :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 5 Natural Numbers and Whole Numbers image - 3
Solution:

Question 3.
See the following pattern carefully :
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 5 Natural Numbers and Whole Numbers image - 6
(i) If n denotes the number of figure and S denotes the number of matches ; find S in terms of n.
(ii) Find how many matches are required to make the :
(1) 15th figure
(2) 40th figure
(iii) Write a discretion of the pattern in words.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 5 Natural Numbers and Whole Numbers image - 7
(ii) (1) 15th figure has = 3 x 15 + 4 = 49 matches
(2) 40th figure has = 3 x 40 + 4 = 124 matches
(iii) It can easily be observed that each time the figure (n) is increased by 4, the number of matches (S) increased by 3.

Question 4.
(i) In the following pattern, draw the next two figures.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 5 Natural Numbers and Whole Numbers image - 8
(ii) Construct a table to describe the figures in the above pattern.
(iii) If n denotes the number of figure and L denotes the number of matches, find L in terms of n.
(iv) Find how may matches are required to make the :
(1) 12th figure
(2) 20th figure
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 5 Natural Numbers and Whole Numbers image - 9
(iii) L = 2n
(iii) (1) 12th figure
12th figure has = 2 x 12 = 24
(2) 20th figure = 2 x 20 = 40

Question 5.
In each of the following patterns, construct next figure.
(i) In each case, if n denotes the number of figure and F denotes the number of matchsticks used, find F in terms of n.
(ii) Also find, in each case, how many matches are required to make the : 16th figure and 30th figure.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 5 Natural Numbers and Whole Numbers image - 10

Solution:

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 6 Negative Numbers and Integers

November 10, 2023May 18, 2022 by Raju

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 6 Negative Numbers and Integers

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 6 Negative Numbers and Integers

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

Negative Numbers and Integers Exercise 6 – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the blanks :
(i) Negative of -20 is ……….
(ii) Negative of 0 is …………
(iii) Negative of 8 is ………..
(iv) If 10 represents gain of ₹ 10, then -10 represents …………..
(v) If going south is negative then going north is …………
(vi) Because 5 < 7, therefore -5 ……….. -7. (vii) If 3 > -2, then 3 is on the ………….. of -2.
(viii) If -8 < -6, then -8 is on the …………. of-6.
Solution:
(i) 20
(ii) 0
(iii) -8
(iv) loss of ₹ 10
(v) positive
(vi) less than
(vii) right side
(viii) left side

Question 2.
Use a number line to write the following integers in ascending (increasing) order :
(i) -5, 8, 0, -9, 4, -14 and 12
(ii) -6, 7, 0, -9, 5 and 9
Solution:
(i) -5, 8, 0, -9, 4, -14 and 12
Draw a number line for integers,, as shown below, and mark on it all the given integers.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 6 Negative Numbers and Integers image - 1
Clearly, the given integers in the ascending order are :
-14 < -9 < -5 < 0 < 4 < 8 < 12
(ii) -6, 7, 0, -9, 5 and 9
Draw a number line for integers, as shown below, and mark on it all the given integers.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 6 Negative Numbers and Integers image - 2
Clearly, the given integers in the ascending order are :
-9 < -6 < 0 < 5 < 7 < 9

Question 3.
Use a number line to write the following integers in descending (decreasing) order :
(i) -10, 0, 3, -4, 12, 11, -1 and 5
(ii) -4, 3, -8, -12, -7 and 6.
Solution:
(i) -10, 0, 3,-4, 12, 11,-1 and 5
Draw a number line for integers, as shown below, and mark on it all the given integers.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 6 Negative Numbers and Integers image - 3
Clearly, the given integers in the descending order are :
6 > 3 > 0 > -7 > -8 > -10 > – 12
(ii) -4, 3, -8, -12, -7 and 6.
Draw a number line for integers, as shown below, and mark on it all the given integers.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 6 Negative Numbers and Integers image - 4
Clearly, the given integers in the descending order are :
6 > 3 > -4 > -7 > -8 > -12

Question 4.
Add:
(i) 13 and 15
(ii) -13 and 15
(iii) 13 and -15
(iv) -13 and -15
Solution:
(i) 13 and 15
13 + 15 = 28
(ii) -13 and 15
(-13) + 15 = 2
(iii) 13 and -15
= (13)+ (-15)
= 13 – 15 = -2
(iv) -13 and-15
= (-13) + (-15)
= -13 – 15 = -28

Question 5.
Add:
(i) 259 from 214
(ii) -528 and -243
(iii) -623 and 326
(iv) 258 and -473
(v) -622 and -254
(vi) 257 and -254
Solution:
(i) 259 from 214
= 259 + 214 = 473
(ii) -528 and-243
= -528 + (-243)
= -528 – 243
= -771
(iii) -623 and 326
= -623 + 326 = -297
(iv) 258 and —473
= 258 +(-473)
= 258 – 473
= -215
(v) -622 and-254
= -622 +(-254)
= -622 – 254
= -876
(vi) 257 and-254
= 257 +(-254)
= 257 – 254
= 3

Question 6.
Subtract :
(i) 5 from 8
(ii) -5 from 8
(iii) 4 from -7
(iv) -8 from -2
(v) -3 from 12
(vi) -6 from -3
Solution:
(i) 5 from 8
= 8 – 5 = 3
(ii) -5 from 8
= 8 – (-5)
= 8 + 5 = 13
(ii) 4 from -7
= -7 – 4
= -11
(iv) -8 from -2 = -2 – (-8)
= -2 + 8
= 6
(v) -3 from 12
= -12 – 3 = -15
(vi) -6 from -3
= -3 – 3 (-6)
= – 3 + 6 = + 3

Question 7.
Subtract:
(i) -123 from 453
(ii) -78 from -12
(iii) 329 and -124
(iv) -222 from 0
Solution:
(i) -123 from 453
= + 453 – (-123)
= 453 + 123 = 576
(ii) -78 from -12
= -12 – (-78)
= -12 + 78 = 66
(iii) 329 and-124
= + 329 – 124 = 205
(iv) -222 from 0
= 0 – (-222)
= 0 + 222 = 222

Question 8.
Using a number line, find the integer which is :
(i) 3 more than -1
(ii) 5 less than 2
(iii) 5 more than -9
(iv) 4 less than -4
(v) 7 more than 0
(vi) 7 less than -8
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 6 Negative Numbers and Integers image - 5

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line

November 10, 2023May 18, 2022 by Raju

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS

Number Line : A Number line is used to represent numbers, such as : fractions, whole numbers, integers, etc.
Using A Number Line to Compare Numbers : Out of any two numbers, marked on a number line, the number which is on the right of the other number is greater and the number which is on the left of the other number is lesser (smaller).

Number Line Exercise 7A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the blanks, using the following number line :

(i) An integer, on the given number line, is ………… than every number on its left.
(ii) An integer, on the given number line, is greater than every number to its …………..
(iii) 2 is greater than – 4 implies 2 is to the ………….. of – 4.
(iv) -3 is ………….. than 2 and 3 is ………. than – 2.
(v) – 4 is ………….. than -8 and 4 is …………… than 8.
(vi) 5 is …………. than 2 and -5 is …………… than – 2.
(vii) -6 is …………. than 3 and the opposite of -6 is ………… than opposite of 3.
(viii) 8 is …………. than -5 and -8 is ……….. than -5.
Solution:
(i) An integer, on the given number line, is greater than every number on its left.
(ii) An integer, on the given number line, is greater than every number to its left.
(iii) 2 is greater than – 4 implies 2 is on the right of – 4.
(iv) – 3 is less than 2 and 3 is greater than -2.
(v) – 4 is greater than -8 and 4 is less than 8.
(vi) 5 is greater than 2 and – 5 is less than – 2.
(vii) -6 is less than 3 and the opposite of -6 is greater than opposite of 3.
(viii) 8 is greater than -5 and -8 is less than -5.

Question 2.
In each of the following pairs, state which integer is greater :
(i) -15, -23
(ii) -12, 15
(iii) 0, 8
(iv) 0, -3
Solution:
(i) -15, -23
-15 is greater than -23 as -15 lies on the right side of-23 on the number line
(ii) -12, 15
15 is greater than than -12 as 15 lies on the right side of -12 on the number line
(iii) 0, 8 8 > 0
(iv) 0, -3 0 > – 3

Question 3.
In each of the following pairs, which integer is smaller :
(i) o, -6
(ii) 2, -3
(iii) 15, -51
(iv) 13, 0
Solution:
(i) 0, -6
-6 < 0
(ii) 2, -3
-3 < 2
(iii) 15, -51
-51 < 15
(iv) 13, 0
0 < 13

Question 4.
In each of the following pairs, replace * with < or > to make the statement true:
(i) 3 * 0
(ii) 0 * -8
(iii) -9 * -3
(iv) 3 * 3
(v) 5 * -1
(vi) -13 * 0
(vii) -8 * -18
(viii) 516 * -316
Solution:
(i) 3 > 0
(ii) 0 > -8
(iii) -9 < -3
(iv) -3 < 3
(v) 5 > -1
(vi) -13 < 0
(vii) -8 > -18
(viii) 516 > -316

Question 5.
In each case, arrange the given integers in ascending order using a number line.
(i) – 8, 0, – 5, 5, 4, – 1
(ii) 3, – 3, 4, – 7, 0, – 6, 2
Solution:
(i) – 8, 0, – 5, 5, 4, – 1
Draw a number line and mark the numbers on it. Arranging in ascending order, as shown -8,-5,-1, 0, 4, 5 as on the number line
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line image - 2
(ii) 3, -3, 4, -7, 0, -6, 2
Draw the number line and mark the numbers on it. Arranging in ascending order as shown on the number line.
-7, -6, -3, 0, 2, 3, 4

Question 6.
In each case, arrange the given integers in descending order using a number line.
(i) -5, -3, 8, 15, 0, -2
(ii) 12, 23, -11, 0, 7, 6
Solution:
(i) -5, -3, 8, 15, 0, -2
Draw the number line and mark these numbers on it. Arranging in descending order 15, 8, 0 -2, -3, -5 as shown on the number line
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line image - 4
(ii) 12, 23, -11, 0, 7, 6
Draw a number line and mark these numbers on it. Arranging in descending order. 23, 12, 7, 6, 0, -1 as shown on the number line

Question 7.
For each of the statements, given below, state whether it is true or false :
(i) The smallest integer is 0.
(ii) The opposite of -17 is 17.
(iii) The opposite of zero is zero.
(iv) Every negative integar is smaller than 0.
(v) 0 is greater than every positive integer.
(vi) Since, zero is neither negative nor positive ; it is not an integer.
Solution:
(i) False
(ii) True
(iii) True
(iv) True
(v) False
(vi) False

Number Line Exercise 7B – Selina Concise Mathematics Class 6 ICSE Solutions

Use a number line to evaluate each of the following :
Question 1.
(i) (+ 7) + (+ 4)
(ii) 0 + (+ 6)
(iii) (+ 5) + 0
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line image - 6

Question 2.
(i) (-4) + (+5)
(ii) 0 + (-2)
(iii) (-1) + (-4)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line image - 7

Question 3.
(i) (+ 4) + (-2)
(ii) (+3) + (-6)
(iii) 3 + (-7)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line image - 8

Question 4.
(i) (-1) + (-2)
(ii) (-2) + (-5)
(ii) (-3) + (-4)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line image - 10

Question 5.
(i) (+ 10) – (+2)
(ii) (+8)- (-5)
(iii) (-6) – (+2)
(iv) (-7) – (+5)
(v) (+4) – (-2)
(vi) (-8) – (-4)
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line image - 11

Question 6.
Using a number line, find the integer which is :
(i) 3 more than -1
(ii) 5 less than 2
(iii) 5 more than -9
(iv) 4 less than -4
(v) 7 more than 0
(vi) 7 less than -8
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line image - 14

Number Line Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the blanks :
(i) 5 is …………… than -2 and -5 is ………… than 2.
(ii) -3 is ………… than 0 and 3 is …………. than 0.
(iii) on a number line, if x is to the left of y, then x is ………… than y.
(iv) on a number line if x is to the right of y, then y is …………. than x.
Solution:
(i) 5 is greater than -2 and -5 is less than 2.
(ii) -3 is less than 0 and 3 is greater than 0.
(iii) On a number line, if x is to the left of y, then x is less than y.
(iv) On a number line, x is to the right of y, then y is less than x.

Question 2.
Using a number line, write the numbers -15, 7, 0, -8 and -3 in ascending order of value.
Solution:
On the given number line, we mark the numbers -15, 7, 0, -8 and -3 on it, we see that

We see that -15 < -8 < -3 < 0 < 7
-15, -8, -3, 0, 7 are in ascending order

Question 3.
Using a number line, write the numbers 8, -6, 2 -12, 0, 15 and -1 in descending order of value.
Solution:
On the given number line, we mark the numbers 8, -6, 2, -12, 0, 15 and -1 on it

We see that
15 > 8 > 2 > 0 > -1 > -6 > -12
15, 8, 2, 0, -1, -6, -12 are in descending order

Question 4.
Using a number line, evaluate :
(i) (+5) + (+4)
(ii) (+6) + (+8)
(iii) (-3) + (+5)
(iv) (-3) + (+7)
(v) (+6) + (-2)
(vi) (-3) + (+3)
(vii) (-5) + (-5)
(viii) (-7) + (-1)
(ix) (+6) – (+2)
(x) (+5) – (-3)
(xi) (+4) – (-1)
(xii) (-7) – (-2)
Solution:
(i) (+5) + (+4)
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line image - 18
First of all, we move 5 units to the right of zero then for (+4), move 4 units right of 5, then we reach at 9, then
(+5) + (+4) = +9
(ii) (+6) + (+8)
selina-concise-mathematics-class-6-icse-solutions-number-line-R-4.1
First of all, we move 6 units to the right of zero then for (+8), we move 8 units to the right of (+6)
Then we reach at +14, then
(+6) + (+8) = +14
(iii) (-3) + (+5)
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line image - 20
First of all for (-3) we move, 3 units to the left of zero, then move (+5) units to the right of 5, then we reach at (+2), then
(-3) + (+5) = -3 + 5 = 2
(iv) (-3) + (+7)

First of all, we move for (-3) 3 unit to the left of zero and then for (+7), we move 7 units to the right of (-3) reaching +4 Then (-3) + (+7) = +4
(v) (+6) + (-2)
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line image - 22
First of all, we move for (+6), 6 units to the right of zero and then for (-2), move 2 units to the left of 6, then we reach 4 Then (+6) + (-2) = 6 – 2 = 4
(vi) (-3) + (+3)

First of all for (-3), we move 3 units left of zero and then for (+3) we move 3 unit right of (-3) reaching at 0
So, (-3) + (+3) = 0
(vii) (-5) + (-5)
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line image - 24
First of all for -5, we move 5 units to left of zero and then for (-5), we move 5 units to left of (-5) reaching at -10
(-5) +(-5) = -10
(viii) (-7) + (-1)

First of all for -7, we move 7 units left of zero and then for (-1) we move 1 unit left of -7 reaching -8
(-7) + (-1) = -8
(ix) (+6) – (+2)
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line image - 26
First of all for (+6) we move 6 units right of 0 and then for (+2), we move 2 units left of 6 reaching 4
(+6)-(+2) = 6 – 2 = 4
(x) (+5) – (-3)

Mark the points (+5) and (-3) on the same number line. We see that the position of (-3) is 8 units from (+5) to its right 3.
(+5) – (-3) = 5 + 3 = 8
(xi) (+4) – (-1)

Mark the points (+4) and (-1) on the same number line, we see that the position of (-1) is 5 units from (+4) to its right
(+4) – (-1) = 4 + 1 = 5
(xii) (-7) – (-2)
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 7 Number Line image - 29
Mark the points (-7) and (-2) on the same number line, we see that (-2) is 5 units on the left (-2)
-7 – (-2) = -7 + 2 = -5

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM

November 10, 2023May 18, 2022 by Raju

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM

ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions

Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

IMPORTANT POINTS

Factor of a given number is that number by which the given number can be divided completely.

Prime Numbers :
A Natural number, which is divisible by 1 (one) and itself only is called a prime number.
Highest Common Factor :
H.C.F. stands for Highest Common Factor and H.C.F. of two or more given numbers is the greatest number (factor) which divides each given number completely.
Lowest Common Factor :
L.C.M. stands for Lowest Common Multiple. The L.C.M. of two or more given numbers is the lowest (smallest) number which is exactly divisible by each of the given numbers.

HCF and LCM Exercise 8A – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Write all the factors of :
(i) 15
(ii) 55
(iii) 48
(iv) 36
(v) 84
Solution:
(i) Factors of 15 = F₁₅ = 1, 3, 5 and 15
(ii) Factors of 55 = F₅₅ = 1, 5, 11 and 55
(iii) Factors of 48 = F₄₈ = 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48
(iv) Factors of 36 = F₅₆ = 1, 2, 3, 4, 6, 9, 12, 18 and 36.
(v) Factors of 84 = F₈₄ = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84.

Question 2.
Write all prime numbers :
(i) less than 25
(ii) between 15 and 35
(iii) between 8 and 76
Solution:
(i) 2, 3, 5, 7, 11, 13, 17, 19 and 23
(ii) 17, 19, 23, 29 and 31
(iii) 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71 and 73.

Question 3.
Write the prime-numbers from :
(i) 5 to 45
(ii) 2 to 32
(iii) 8 to 48
(iv) 9 to 59
Solution:
(i) 7, 11, 13, 17, 19, 23, 29, 31, 37, 41 and 43.
(ii) 3, 5, 7, 11, 13, 17, 19, 23 29 and 31.
(iii) 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47.
(iv) 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 and 53.

Question 4.
Write the prime factors of:
(i) 16
(ii) 27
(iii) 35
(iv) 49
Solution:
(i) Prime factors of 16 = 2
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 1
(ii) Prime factors of 27 = 3

(iii) Prime factors of 35 = 5, 7

(iv) Prime factors of 49 = 7

Question 5.
If P_n means prime factors of n, find:
(i) p₆
(ii) P₂₄
(iii) p₅₀
(iv) P₄₂
Solution:
(i) F₆ = 1, 2, 3, 6
P.F₆ (Prime factor of 6) = 2 and 3.
(ii) F₂₄ = 1, 2, 3, 4, 6, 8, 12, 24
P.F₂₄ = 2 and 3.
(iii) F₅₀ = 2, 5, 5
P.F₅₀ = 2 and 5.
(iv) F₄₂ = 1, 2, 3, 6, 7, 14, 21, 42
P.F₄₂ = 2, 3 and 7.

HCF and LCM Exercise 8B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Using the common factor method, find the H.C.F. of :
(i) 16 and 35
(ii) 25 and 20
(iii) 27 and 75
(iv) 8, 12 and 18
(v) 24, 36, 45 and 60
Solution:
(i) F₁₆ = 1, 2, 4, 8, 16
F₃₅ = 1, 5, 7, 35
Common factors between 16 and 35 = 1
H.C.F. of 16 and 35 = 1
(ii) F₂₅ = 1, 5, 25
F₂₀ = 1, 2, 4, 5, 10, 20
Common factors between 25 and 20 = 1, 5
H.C.F. of 25 and 20 = 5
(iii) F₂₇ = 1, 3, 9, 27
F₇₅ = 1, 3, 5, 15, 25, 75
Common factors between 27 and 75 = 1, 3
H.C.F. of 27 and 75 = 3
(iv) F₈ = 1, 2, 4, 8
F₁₂ = 1, 2, 3, 4, 6, 12
F₁₈ = 1, 2, 3, 6, 9, 18
Common factors between 8, 12 and 18 = 1, 2
H.C.F. of 8, 12 and 18 = 2
(v) F₂₄ = 1, 2, 3, 4, 6, 8, 12, 24
F₃₆ = 1, 2, 3, 4, 6, 12, 18, 36
F₄₅ = 1, 3, 5, 9, 15, 45
F₆₀ = 1, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Common factor between 24, 36, 45 and 60 = 1, 3
H.C.F. of 24, 36, 45 and 60 = 3

Question 2.
Using the prime factor method, find the H.C.F. of:
(i) 5 and 8
(ii) 24 and 49
(iii) 40, 60 and 80
(iv) 48, 84 and 88
(v) 12, 16 and 28
Solution:
(i) Prime factor of 5 = 5
Prime factor of 8 = 2 x 2 x 2
No common prime factor
H.C.F. of 5 and 8 = 1
(as both the number are co-prime)
(ii) Prime factor of 24 = 2 x 2 x 2 x 3
Prime factor of 49 = 7 x 7
No common prime factor, number are co-prime.
H.C.F. of 24 and 49 = 1.
(iii) Prime factor of 40 = 2 x 2 x 2 x 5
Prime factor of 60 = 2 x 2 x 3 x 5
Prime factor of 80 = 2 x 2 x 2 x 2 x 5
Common prime factor = 2 x 2 x 5
H.C.F. of 40, 60 and 80 = 2 x 2 x 5 = 20
(iv) Prime factor of 48 = 2 x 2 x 2 x 2 x 3
Prime factor of 84 = 2 x 2 x 3 x 7
Prime factor of 88 = 2 x 2 x 2 x 11
Common prime factor of 48, 84 and 88 = 2 x 2
H.C.F. of 48, 84 and 88 = 2 x 2 = 4
(v) Prime factor of 12 = 2 x 2 x 3
Prime factor of 16 = 2 x 2 x 2 x 2
Prime factor of 28 = 2 x 2 x 7
Common prime factor between 12, 16 and 28 = 2 x 2
H.C.F. of 12, 16 and 28 = 2 x 2 = 4

Question 3.
Using the division method, find the H.C.F. of the following :
(i) 16 and 24
(ii) 18 and 30
(iii) 7, 14 and 24
(iv) 70,80,120 and 150
(v) 32, 56 and 46
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 5

Question 4.
Use a method of your own choice to find the H.C.F. of :
(i) 45, 75 and 135
(ii) 48, 36 and 96
(iii) 66, 33 and 132
(iv) 24, 36, 60 and 132
(v) 30, 60, 90 and 105
Solution:
(i) Factor of 45 = F₄₅ = 3 x 3 x 5
Factor of 75 = F₇₅ = 3 x 5 x 5
and Factor of 135 = F₁₃₅ = 3 x 3 x 3 x 5
Now the common factors of 45, 75 and 135 = 3 and 5
H.C.F. = 3 x 5 = 15
(ii) Factor of 48 = F₄₈ = 2 x 2 x 2 x 2 x 3
Factor of 36 = F₃₆ = 2 x 2 x 3 x 3
and factor of 96 = 2 x 2 x 2 x 2 x 2 x 3
Now the common factor of 48, 36 and 96 = 2, 2 and 3
H.C.F. = 2 x 2 x 3 = 12
(iii) Factor of 66 = F₆₆ = 2 x 3 x 11
Factor of 33 = F₃₃ = 3 x 11
and factor of 132 = F₁₃₂ = 2 x 2 x 3 x 11
Now the common factor of 66, 33 and 132 = 3 and 11
H.C.F. = 3 x 11 =33
(iv) Factor of 24 = F₂₄ = 2 x 2 x 2 x 3
Factor of 36 = F₃₆ = 2 x 2 x 3 x 3
Factor of 60 = F₆₀ = 2 x 2 x 3 x 5
and Factor of 132 = F₁₃₂ = 2 x 2 x 3 x 11
Now the common factors of 24, 36, 60 and 132 = 2, 2 and 3
H.C.F. = 2 x 2 x 3 = 12
(v) Factor of 30 = F₃₀ = 2 x 3 x 5
Factor of 60 = F₆₀ = 2 x 2 x 3 x 5
Factor of 90 = F₉₀ = 2 x 3 x 3 x 5
and factor of 105 = F₁₀₅ = 3 x 5 x 7
Now the common factor of 30, 60, 90 and 105 = 3 and 5
H.C.F. = 3 x 5 = 15

Question 5.
Find the greatest number that divides each of 180, 225 and 315 completely.
Solution:
The greatest number that divides 180, 225 and 315 will be HCF of 180, 225, 315
Let us first find HCF of 180 and 225
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 9

Question 6.
Show that 45 and 56 are co-prime numbers.
Solution:
The HCF of two co-prime numbers is always HCF of 45 and 56
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 10

Question 7.
Out of 15, 16, 21 and 28, find out all the pairs of co-prime numbers.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 11

From above it is clear that 15 and 16 are co-prime because common factor is 1
Hence pairs 15 and 16, 16, 21, 15, 28 are co-prime number.

Question 8.
Find the greatest no. that will divide 93, 111 and 129, leaving remainder 3 in each case.
Solution:
Since Remainder is 3 in each case numbers are
93 – 3 = 90
111 – 3 = 108
129 – 3 = 126
Required number will be HCF of90,108 and 126 HCF of 90 and 108
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 13

HCF and LCM Exercise 8C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Using the common multiple method, find the L.C.M. of the following :
(i) 8, 12 and 24
(ii) 10, 15 and 20
(iii) 3, 6, 9 and 12
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 15

Question 2.
Find the L.C.M. of each the following groups of numbers, using
(i) the prime factor method and
(ii) the common division method :
(i) 18, 24 and 96
(ii) 100, 150 and 200
(iii) 14, 21 and 98
(iv) 22, 121 and 33
(v) 34, 85 and 51
Solution:
(i) L.C.M. of 18, 24 and 96
(i) By prime factors
Prime factors of 18 = 2 x 3 x 3
Prime factors of 24 = 2 x 2 x 2 x 3
Prime factors of 96 = 2 x 2 x 2 x 2 x 2 x 3
L.C.M. = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288
By common division method
L.C.M. of 18, 24 and 96 = 2 x 2 x 2 x 3 x 3 x 4 = 288
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 16

Question 3.
The H.C.F. and the L.C.M. of two numbers are 50 and 300 respectively. If one of the numbers is 150, find the other one.
Solution:
H.C.F. = 50
L.C.M. = 300
Product of L.C.M. and H.C.F. = 300 x 50 = 15000
One number = 150
The other number
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 19

Question 4.
The product of two numbers is 432 and their L.C.M. is 72. Find their H.C.F.
Solution:
Product of two numbers = Product of their L.C.M. and H.C.F.
Here, product of two number = 432
L.C.M. = 72
H.C.F. = \(\frac { 432 }{ 72 }\) = 6

Question 5.
The product of two numbers is 19,200 and their H.C.F. is 40. Find their L.C.M.
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 20

Question 6.
Find the smallest number which, when divided by 12, 15, 18, 24 and 36 leaves no remainder
Solution:
The least number which is exactly divisible by each given number is their L.C.M.
Required number L.C.M. of 12, 15, 18, 24 and 36.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 21

Question 7.
Find the smallest number which, when increased by one is exactly divisible by 12, 18, 24, 32 and 40
Solution:
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 22

Question 8.
Find the smallest number which, on being decreased by 3, is completely divisible by 18, 36, 32 and 27.
Solution:
LCM of 18, 36, 32 and 27
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 23

HCF and LCM Revision Exercise – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Find the H.C.F. of :
(i) 108, 288 and 420
(ii) 36, 54 and 138
Solution:
(i) H.C.F. of 108, 288, 420 = 12
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 24

Question 2.
Find the L.C.M. of:
(i) 72, 80 and 252
(ii) 48, 66 and 120
Solution:
L.C.M. 72, 80, 252
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 25

Question 3.
State true or false : Give an example.
(i) H.C.F. of two prime numbers is 1.
(ii) H.C.F. of two co-prime numbers is 1.
(iii) L.C.M. of two prime numbers is equal to their product.
(iv) L.C.M. of two co-prime numbers is equal to their product.
Solution:
(i) True : Because the prime numbers have no common factor except 1.
(ii) True : Becuase co-prime numbers have no common factor except 1.
(iii) True : Because the prime number have no common factor except 1.
(iv) True : Because co-prime numbers have no common factor except 1.

Question 4.
The product of two numbers is 12096 and their H.C.F. is 36. Find their L.C.M.
Solution:
We know that
Product of two numbers = Product of their H.C.F. and L.C.M.
=> 12096 = 36 x L.C.M.
=> L.C.M. = \(\frac { 12096 }{ 36 }\) = 336

Question 5.
The product of the H.C.F. and the L.C.M. of two numbers is 1152. If one number is 48, find the other one.
Solution:
We know that:
Product of two numbers = Product of their H.C.F. and L.C.M.
=> 1st number x 2nd number = Product of their H.C.F. and L.C.M.
=> 48 x 2nd number = 1152
=> 2nd number = \(\frac { 1152 }{ 48 }\) = 24

Question 6.
(i) Find the smallest number that is completely divisible by 28 and 42.
(ii) Find the largest number that can divide 28 and 42 completely.
Solution:
(i) We know that the least number which is divisible by 28 and 42 is their L.C.M.
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 27
L.C.M. of 28 and 42 = 2 x 2 x 3 x 7 = 84
(ii) We know that the largest number which can divide 28 and 42 completely will be their H.C.F.

H.C.F. of 28 and 42 = 14

Question 7.
Find the L.C.M. of 140 and 168. Use the L.C.M. obtained to find the H.C.F. of the given numbers.
Solution:
Numbers are 140 and 168
L.C.M. of 140 and 168
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 29

Question 8.
Find the H.C.F. of 108 and 450 and use the H.C.F. obtained to find the L.C.M. of the given numbers.
Solution:
Numbers are given : 108 and 450
H.C.F. of 108 and 450= 18
Selina Concise Mathematics Class 6 ICSE Solutions Chapter 8 HCF and LCM image - 30

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Selina Concise Mathematics Class 6 ICSE Solutions Chapter 9 Playing with Numbers

November 10, 2023May 18, 2022 by Raju

Selina Concise Mathematics Class 6 ICSE Solutions Chapter 9 Playing with Numbers

Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 9 Playing with Numbers

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Selina Class 6 Maths ICSE Solutions Physics Chemistry Biology Geography History & Civics

Playing with Numbers Exercise 9A – Selina Concise Mathematics Class 6 ICSE Solutions

(Using BODMAS)
Question 1.
19 – (1 + 5) – 3
Solution:
19 – (1 + 5) – 3
= 19 – 6 – 3
= 19 – 9 = 10

Question 2.
30 x 6 + (5 – 2)
Solution:
30 x 6 + (5 – 2)
= 30 x 6 – 3
= 30 x 2 = 60

Question 3.
28 – (3 x 8) + 6
Solution:
28 – (3 x 8) – 6
= 28 – 24 – 6
= 28 – 4 = 24

Question 4.
9 – [(4 – 3) + 2 x 5]
Solution:
9 – [(4 – 3) + 2 x 5]
= 9 – [1 + 10]
= 9 – 11 = -2

Question 5.
[18 – (15 – 5) + 6]
Solution:
[18 -(15 -5) + 6]
= [18 – 3 + 6]
= [18 + 3] = 21

Question 6.
[(4 x 2) – (4 + 2)] + 8
Solution:
[(4 x 2) – (4 – 2)] + 8
= 8 – 2 + 8
= 16 – 2 = 14

Question 7.
48 + 96 – 24 – 6 x 18
Solution:
48 + 96 – 24 – 6 x 18
= 48 + 4 – 6 x 18
= 48 + 4 – 108
= 52 – 108 = -56

Question 8.
22 – [3 – {8 – (4 + 6)}]
Solution:
22 – [3 – {8 – (4 + 6)}]
= 22 – [3 – {8 – 10}]
= 22 – [3 + 2]
= 22 – 5 = 17

Question 9.

Solution:
= 34 – [29 – {30 + 66 + (24 – 2)}]
= 34 – [29 – {30 + 66 + 22}]
= 34 – [29 – {30 + 3}]
= 34 – [29 – 33]
= 34 – [-4]
= 34 + 4 = 38

Question 10.
60 – {16 + (4 x 6 – 8)}
Solution:
60 – {16 + (4 x 6 – 8)}
= 60 – {16 + (24 – 8)}
= 60 – {16 + 16}
= 60 – 1 = 59

Question 11.

Solution:
25 – [12 – {5 + 18 + ( 4 – 5 – 3)}]
= 25 – [12 – {5 + 18 + (4 – 2)}]
= 25 – [12 – {5 + 18 + 2}]
= 25 – [12 – {5 + 9}]
= 25 – [12 – 14]
= 25 – [-2]
= 25 + 2 = 27

Question 12.
15 – [16 – {12 + 21 ÷ (9 – 2)}]
Solution:
15 – [16 – {12 + 21 ÷ (9 – 2)}]
= 15 – [16 – {12 + 21 ÷ 7}]
= 15 – [16 – {12 + 3}]
= 15 – [16 – 15]
= 15 – 1 = 14

Playing with Numbers Exercise 9B – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
Fill in the blanks :
(i) On dividing 9 by 7, quotient = …………. and remainder = ……….
(ii) On dividing 18 by 6, quotient = …………. and remainder = ………….
(iii) Factor of a number is ………….. of …………..
(iv) Every number is a factor of …………….
(v) Every number is a multiple of …………..
(vi) …………. is factor of every number.
(vii) For every number, its factors are ………… and its multiples are …………..
(viii) x is a factor of y, then y is a ………… of x.
Solution:
(i) On dividing 9 by 7, quotient = 1 and remainder = 3
(ii) On dividing 18 by 6, quotient = 3 and remainder = 0
(iii) Factor of a number is an exact division of the number
(iv) Every number is a factor of itself
(v) Every number is a multiple of itself
(vi) One is factor of every number.
(vii) For every number, its factors are finite and its multiples are infinite
(viii) x is a factor of y, then y is a multiple of x.

Question 2.
Write all the factors of :
(i) 16
(ii) 21
(iii) 39
(iv) 48
(v) 64
(vi) 98
Solution:
(i) 16
All factors of 16 are : 1, 2, 4, 8, 16
(ii) 21
All factors of 21 are : 1, 3, 7, 21.
(iii) 39
All factors of 39 are : 1, 3, 13, 39
(iv) 48
All factors of 48 are : 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
(v) 64
All factors of 64 are : 1, 2, 4, 8, 16, 32, 64
(vi) 98
All factors of 98 are : 1, 2, 7, 14, 49, 98

Question 3.
Write the first six multiples of :
(i) 4
(ii) 9
(iii) 11
(iv) 15
(v) 18
(vi) 16
Solution:
(i) 4
Multiples of 4 =1 x 4, 2 x 4, 3 x 4, 4 x 4, 4 x 5, 4 x 6
First six multiples of 4 are : 4, 8, 12, 16, 20, 24
(ii) 9
Multiples of 9 = 1 x 9, 2 x 9, 3 x 9, 4 x 9, 5 x 9, 6 x 9
First six multiples of 9 are : 9, 18, 27, 36, 45, 54
(iii) 11
Multiples of 11 = 1 x 11, 2 x 11, 3 x 11, 4 x11, 5 x 11, 6 x 11
First six multiples of 11 are : 11, 22, 33, 44, 55, 66
(iv) 15
Multiples of 15 = 1 x 15, 2 x 15, 3 x 15, 4 x 15, 5 x 15, 6 x 15
First six multiples of 15 are : 15, 30, 45, 60, 75, 90
(v) 18
Multiples of 18 = 1 x 18, 2 x 18,3 x 18, 4 x 18, 5 x 18, 6 x 18
First six multiples of 18 are : 18, 32, 54, 72, 90, 108
(vi) 16
Multiples of 16 = 1 x 16, 2 x 16, 3 x 16,4 x 16, 5 x 16, 6 x 16
First six multiples of 16 are : 16, 32, 48, 64, 80, 96

Question 4.
The product of two numbers is 36 and their sum is 13. Find the numbers.
Solution:
Since, 36 = 1 x 36, 2 x 18, 3 x 12, 4 x 9, 6 x 6
Clearly, numbers are 4 and 9

Question 5.
The product of two numbers is 48 and their sum is 16. Find the numbers.
Solution:
Since, 48 = 1 x 48, 2 x 24, 3 x 16, 4 x 12, 6 x 8
Clearly, numbers are 4 and 12.

Question 6.
Write two numbers which differ by 3 and whose product is 54.
Solution:
Since, 54 = 1 x 54, 2 x 27, 3 x 18, 6 x 9
Clearly, numbers are 6 and 9.

Question 7.
Without making any actual division show that 7007 is divisible by 7.
Solution:
7007
= 7000 + 7
= 7 x (1000+ 1)
= 7 x 1001
Clearly, 7007 is divisible by 7.

Question 8.
Without making any actual division, show that 2300023 is divisible by 23.
Solution:
2300023 = 2300000 + 23
= 23 x (100000 + 1)
= 23 x 100001
Clearly, 2300023 is divisible by 23.

Question 9.
Without making any actual division, show that each of the following numbers is divisible by 11.
(i) 11011
(ii) 110011
(iii) 11000011
Solution:
(i) 11011 = 11000+ 11
= 11 x (1000+ 1)
= 11 x 1001
Clearly, 11011 is divisible by 11.
(ii) 110011
= 110000+ 11
= 11 x (10000+ 1)
= 11 x 10001
Clearly, 110011 is divisible by 11.
(iii) 11000011
= 11000000+ 11
= 11 x (1000000+ 1)
= 11 x 1000001
Clearly, 110000 is divisible by 11.

Question 10.
Without actual division, show that each of the following numbers is divisible by 8 :
(i) 1608
(ii) 56008
(iii) 240008
Solution:
(i) 1608
= 1600 + 8
= 8 (200 + 1)
= 8 x 201
Clearly, 1608 is divisible by 8.
(ii) 56008
= 56000 + 8
= 8 x (7000 + 1)
= 8 x 7001
Clearly, 56008 is divisible by 8.
(iii) 240008
= 240000 + 8
= 8 x (30000 + 1)
= 8 x 30001
Clearly, 240008 is divisible by 8.

Playing with Numbers Exercise 9C – Selina Concise Mathematics Class 6 ICSE Solutions

Question 1.
find which of the following numbers are divisible by 2 :
(i) 352
(ii) 523
(iii) 496
(iv) 649
Solution:
(i) 352
The given number = 352
Digit at unit’s place = 2
It is divisible by 2
(ii) 523
The given number = 523
Digit at unit’s place = 3
It is not divisible by 2
(iii) 496
The given number = 496
Digit at unit’s place = 6
It is divisible by 2
(iv) 649
The given number = 649
Digit at unit’s place = 9
It is not divisible by 2

Question 2.
Find which of the following number are divisible by 4 :
(i) 222
(ii) 532
(iii) 678
(iv) 9232
Solution:
(i) 222
The given number = 222
The number formed by ten’s and unit’s digit is 22, which is not divisible by 4.
222 is not divisible by 4
(ii) 532
The given number = 532
The number formed by ten’s and unit’s digit is 32, which is divisible by 4.
532 is divisible by 4
(iii) 678
The given number = 678
The number formed by ten’s and unit’s digit is 78, which is not divisible by 4
678 is not divisible by 4
(iv) 9232
The given number = 9232
The number formed by ten’s and unit’s digit is 32, which is divisible by 4.
9232 is divisible by 4.

Question 3.
Find the which of the following numbers are divisible by 8 :
(i) 324
(ii) 2536
(iii) 92760
(iv) 444320
Solution:
(i) 324
The given number = 324
The number formed by hundred’s, ten’s and unit’s digit is 324, which is not divisible by 8
324 is not divisible by 8
(ii) 2536
The given number = 2536
The number formed by hundred’s, ten’s and unit’s digit is 536, which is divisible by 8
2536 is divisible by 8
(iii) 92760
The given number = 92760
The number formed by hundred’s, ten’s and unit’s digit is 760, which is divisible by 8
92760 is divisible by 8
(iv) 444320
The given number = 444320
The number formed by hundred’s, ten’s and unit’s digit is 320, which is divisible by 8
444320 is divisible by 8.

Question 4.
Find which of the following numbers are divisible by 3 :
(i) 221
(ii) 543
(iii) 28492
(iv) 92349
Solution:
(i) 221
Sum of digits = 2 + 2 + 1 = 5
Which is not divisible by 3
221 is not divisible by 3.
(ii) 543
Sum of digits = 5 + 4 + 3 = 12
Which is divisible by 3
543 is divisible by 3
(iii) 28492
The given number = 28492
Sum of its digits = 2 +8+4 + 9 + 2 = 25
Which is not divisible by 3
28492 is divisible by 3.
(iv) 92349
The given number = 92349
Sum of its digits = 0 + 2 + 3 + 4 + 9 = 27
Which is divisible by 3
92349 is divisible by 3.

Question 5.
Find which of the following numbers are divisible by 9 :
(i) 1332
(ii) 53247
(iii) 4968
(iv) 200314
Solution:
(i) 1332
The given number = 1332
Sum of its digits = 1 + 3 + 3+ 2 = 9
Which is divisible by 9
1332 is divisible by 9
(ii) 53247
The given number = 53247
Sum of its digits = 5 + 3 + 2 + 4 + 7 = 21
Which is not divisible by 9
53247 is not divisible by 9
(iii) 4968
The given number = 4968
Sum of its digits = 4 + 9 + 6 + 8 = 27
Which is divisible by 9
4968 is divisible by 9
(iv) 200314
The given number = 200314
Sum of its digits = 2 + 0 + 0 + 3 + 1 + 4 = 10
Which is not divisible by 9

Question 6.
Find which of the following number are divisible by 6 :
(i) 324
(ii) 2010
(iii) 33278
(iv) 15505
Solution:
A number which is divisible by 2 and 3 or both then the given number is divisible by 6
(i) 324
The given number = 324
Sum of its digits =3 + 2 + 4 = 9
Which is divisible by 3
The given number is divisible by 6
(ii) 2010
The given number = 2010
Sum of its digits = 2 + 0 + 1 + 0 = 3
Which is divisible by 3
The given number is divisible by 6
(iii) 33278
The given number = 33278
Sum of its digits =3 + 3 + 2 + 7 + 8 = 23
Unit digit is 3 which is odd.
The given number is not divisible by 6.
(iv) 15505
The given number = 15505
Sum of its digits = 1 + 5 + 5 + 0 + 5 = 16
which is divisible by 2.
The given number is divisible by 6.

Question 7.
Find which of the following numbers are divisible by 5 :
(i) 5080
(ii) 66666
(iii) 755
(iv) 9207
Solution:
We know that a number whose units digit is 0 or 5, then the number is divisible by 5.
(i) 5080
Here, unit’s digit 0 5080 is divisible by 5.
(ii) 66666
Here, unit’s digit is 6.
66666 is not divisible by 5.
(iii) 755
Here, unit’s digit is 5.
755 is divisible by 5.
(iv) 9207
Here, unit’s digit is 7
9207 is not divisible by 5.

Question 8.
Find which of the following numbers are divisible by 10 :
(i) 9990
(ii) 0
(iii) 847
(iv) 8976
Solution:
We know that a number is divisible by 10 if its ones digit is 0.
(i) 9990
Here, unit’s digit is 0
9990 is divisible by 10.
(ii) 0
Here, unit’s digit is 0
0 is divisible by 10.
(iii) 847
Here, unit’s digit is 7
847 is not divisible by 10.
(iv) 8976
Here, unit’s digit is 6
8976 is not divisible by 10.

Question 9.
Find which of the following numbers are divisible by 11 :
(i) 5918
(ii) 68,717
(iii) 3882
(iv) 10857
Solution:
A number is divisible by 11, if the difference of sumof its digits in odd places from the right side and the sum of its digits in even places from the right side is divisible by 11.
(i) 5918
Sum of digits at odd places = 5 + 1=6 and,
sum of digits at even places = 9 + 8= 17
Their difference = 17 – 6 = 11 Which is divisible by 11
5918 is divisible by 11.
(ii) 68, 717
Sum of digits at odd places = 6 + 7 + 7 = 20
and, sum of digits at even places = 8 + 1 =9
Difference = 20 – 9 = 11
which is divisible by 11
68717, is divisible by 11.
(iii) 3882
Sum of digits at odd places = 3 + 8 = 11 and,
Sum of digits at even places = 8 + 2 = 10
Difference = 11 – 10 = 1 Which is not divisible by 11
3882 is not divisible by 11.
(iv) 10857
Sum of digits at odd places =1 + 8 + 7 = 16
and, Sum of digits at even places = 0 + 5 = 5
Difference = 16 – 5 = 11
which is divisible by 11
10857 is divisible by 11.

Question 10.
Find which of the following numbers are divisible by 15 :
(i) 960
(ii) 8295
(iii) 10243
(iv) 5013
Solution:
A number is divisible by 15, if it is divisible by both 3 and 5
(i) 960
960 is divisible by both 3 and 5.
960 is divisible by 15
(ii) 8295
8295 is divisible by both 3 and 5.
8295 is divisible by 15
(iii) 10243
10243 is not divisible by both 3 and 5
10243 is not divisible by 15
(iv) 5013
5013 is divisible by both 3 but is not divisible by 5.
5013 is not divisible by 15.

Question 11.
In each of the following numbers, replace M by the smallest number to make resulting number divisible by 3 :
(i) 64 M 3
(ii) 46 M 46
(iii) 27 M 53
Solution:
(i) 64 M 3
The given number = 64 M 3
Sum of its digit = 6 + 4 + 3 = 13
The number next to 13 which is divisible by 3 is 15
Required smallest number =15 – 13 = 2
(ii) 46 M 46
The given number = 46 M 46
Sum of its digits = 4 + 6 + 4 + 6 = 20
The number next to 20 which is divisible by 3 is 21
Required smallest number = 21 – 20 = 1
(iii) 27 M 53
The given number = 27 M 53
Sum of its digits = 2 + 7 + 5 + 3 = 18
which is divisible by 3
Required smallest number = 0

Question 12.
In each of the following numbers replace M by the smallest number to make resulting number divisible by 9.
(i) 76 M 91
(ii) 77548 M
(iii) 627 M 9
Solution:
(i) 76 M 91
The given number = 76 M 91
Sum of its given digits = 7 + 6 + 9 + 1 = 23
The number next to 23, which is divisible by 9 is 27
Required smallest number = 27 – 23 = 4
(ii) 77548 M
The given number = 77548 M
Sum of its given digits = 7 + 7 + 5 + 4 + 8 = 31
The number next to 31, which is divisible by 9 is 36.
Required smallest number = 36 – 31 = 5
(iii) 627 M 9
The given number = 627 M 9
Sum of its given digits = 6 + 2 + 7 + 9 = 24
The number next to 24, which is divisible by 9 is 27
Required smallest number = 27 – 24 = 3

Question 13.
In each of the following numbers, replace M by the smallest number to make resulting number divisible by 11.
(i) 39 M 2
(ii) 3 M 422
(iii) 70975 M
(iv) 14 M 75
Solution:
(i) 39 M 2
The given number = 39 M 2
Sum of its digits in odd places = 3 + M
Sum of its digits in even place = 9 + 2 = 11
Their Difference = 11 – (3 + M)
11 – (3 + M) = 0 11 – 3 = M M = 8
(ii) 3 M 422
The given number = 3 M 422
Sum of its digits in odd places = 3 + 4 + 2 = 9
Sum of its digit in even places = M + 2
Difference of the two sums = 9 – (M + 2)
9 – (M + 2) = 0
9 – 2 = M
M = 7
(iii) 70975 M
The given number = 70975 M
Sum of its digits in odd places = 0 + 7 + M = 7 + M
Sum of its digit in even places = 5 + 9 + 7 = 21
Difference of the two sums = 21 – (7 + M)
=> 21 – (7 + M) = 0
=> 21 = 7 + M
=> M = 14
Since, M cannot be two digit number M = 14 – 11 = 3
(iv) 14 M 75
The given number = 14 M 75
Sum of its digit in odd places = 1 + M + 5 = M + 6
Sum of its digit in even places = 4 + 7 = 11
11 – (M + 16) = 0
11 = M + 6
11 – 6 = M
M = 5

Question 14.
State, true or false :
(i) If a number is divisible by 4. It is divisible by 8.
(ii) If a number is a factor of 16 and 24, it is a factor of 48.
(iii) If a number is divisible by 18, it is divisible by 3 and 6.
(iv) If a divide b and c completely, then a divides (i) a + b (ii) a – b also completely.
Solution:
(i) False
(ii) True
(iii) True
(iv) True