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Page 1
EXERCISE 9 (A)
(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
Page 2
EXERCISE 9 (A)
(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]
Page 3
EXERCISE 9 (A)
(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
EXERCISE 9 (B)
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
Page 4
EXERCISE 9 (A)
(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
EXERCISE 9 (B)
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
Page 5
EXERCISE 9 (A)
(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
EXERCISE 9 (B)
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.
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