Playing with Numbers ICSE Class-6th Concise Maths Selina Solutions Ch-9

Playing with Numbers ICSE Class-6th Concise Selina Mathematics Solutions Chapter-9. We provide step by step Solutions of Exercise / lesson-9 Playing with Numbers  for ICSE Class-6 Concise Selina Mathematics. Our Solutions contain all type Questions of Exe-9 A, Exe-9 B and Exe-9 C to develop skill and confidence. Visit official Website  CISCE for detail information about ICSE Board Class-6 .

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


 –: Select Topics :–

 

Exe-9 A,

Exe-9 B,

Exe-9 C,

 


Solved Questions of Exercise- 9 A Playing with Numbers for ICSE Class-6th Concise Maths

(Using BODMAS)

Question -1.

19 – (1 + 5) – 3

Answer-1

19 – (1 + 5) – 3
= 19 – 6 – 3
= 19 – 9 = 10

Question -2.

30 x 6 + (5 – 2)

Answer-2

30 x 6 + (5 – 2)
= 30 x 6 – 3
= 30 x 2 = 60

Question- 3.

28 – (3 x 8) + 6
Answer-3

28 – (3 x 8) – 6
= 28 – 24 – 6
= 28 – 4 = 24

Question -4.

9 – [(4 – 3) + 2 x 5]
Answer-4
9 – [(4 – 3) + 2 x 5]
= 9 – [1 + 10]
= 9 – 11 = -2

Question -5.

[18 – (15 – 5) + 6]
Answer-5

[18 -(15 -5) + 6]
= [18 – 3 + 6]
= [18 + 3] = 21

Question -6.

[(4 x 2) – (4 + 2)] + 8
Answer-6

[(4 x 2) – (4 – 2)] + 8
= 8 – 2 + 8
= 16 – 2 = 14

Question -7.

48 + 96 – 24 – 6 x 18
Answer-7
48 + 96 – 24 – 6 x 18
= 48 + 4 – 6 x 18
= 48 + 4 – 108
= 52 – 108 = -56

Question- 8.

22 – [3 – {8 – (4 + 6)}]
Answer-8

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

Question -9.

selina-concise-mathematics-class-6-icse-solutions-playing-with-numbers-A-9
Answer-9

= 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)}
Answer-10

60 – {16 + (4 x 6 – 8)}
= 60 – {16 + (24 – 8)}
= 60 – {16 + 16}
= 60 – 1 = 59

Question- 11.

selina-concise-mathematics-class-6-icse-solutions-playing-with-numbers-A-11
Answer-11

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)}]
Answer-12

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


Concise Maths Selina Solutions Playing with Numbers Exercise – 9 B ICSE Class-6th 

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.
Answer-1

(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
Answer-2

(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
Answer-3

(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, 9

Question -4.

The product of two numbers is 36 and their sum is 13. Find the numbers.
Answer-4

Since, 36 = 1 x 36, 2 x 18, 3 x 12, 6 x 6 or 4 x 9,

Sum of 1 + 36,= 37

Sum of 2 + 18,= 20

Sum of 3 + 12,= 15

Sum of 6 + 6,= 12

Sum of 4 + 9,= 13

Clearly, numbers are 4 and 9

Question -5.

The product of two numbers is 48 and their sum is 16. Find the numbers.
Answer-5

Since, 48 = 1 x 48, 2 x 24, 3 x 16, 6 x 8 , 4 x 12

Sum of 1 + 48,= 49

Sum of 2 + 24,= 26

Sum of 3 + 16= 19

Sum of 6 + 8 = 14

Sum of 12 + 4,= 16

Clearly, numbers are 4 and 12

Question- 6.

Write two numbers which differ by 3 and whose product is 54.
Answer-6

Since, 54 = 1 x 54, 2 x 27, 3 x 18, 6 x 9

Difference of 54 – 1 = 53

Difference of 27 – 2 = 25

Difference of 18 – 3 = 15

Difference of 9 – 6 = 3 

Clearly, numbers are 6 and 9.

Question- 7.

Without making any actual division show that 7007 is divisible by 7.
Answer-7

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.
Answer-8

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
Answer-9

(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
Answer-10

(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


 Exercise – 9 C  ICSE Class-6th Concise Selina Maths Solution

Question- 1.

find which of the following numbers are divisible by 2 :
(i) 352
(ii) 523
(iii) 496
(iv) 649
Answer-1

(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
Answer-2

(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

Answer-3

(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
Answer-4

(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
Answer-5

(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

Answer-6

if A number which is divisible by 2 and 3 or both then this given number is also divisible by 6

Therefore check if given number divisible by 2 as well as 3
(i) 324
The given number = 324

Sum of its digits =3 + 2 + 4 = 9
Which is divisible by 3

324 is even hence it is divisible by 2

Therefore 324 is divisible by 2 as well as 3

Hence 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

2010 is even hence it is divisible by 2

Therefore 2010 is divisible by 2 as well as 3

Hence 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.

Hence not divisible by 2
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 not divisible by 3.
The given number is not divisible by 6

Question- 7.

Find which of the following numbers are divisible by 5 :
(i) 5080
(ii) 66666
(iii) 755
(iv) 9207

Answer-7

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

Answer-8

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

Answer-9

A number is divisible by 11, if the difference of sum of 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

Answer-10

A number is divisible by 15, if it given number is divisible by 3 as well as 5

(i) 960
960 is divisible by 3

960 is also divisible by  5.

Therefore 960 is divisible by 15

(ii) 8295

8295 is divisible by 3

8295 is also divisible by  5.

Therefore 8295 is divisible by 15

(iii) 10243

10243 is not divisible by 3

10243 is also not divisible by  5.

Therefore 8295 is not divisible by 15

(iv) 5013

5013 is divisible by 3

5013 is also not divisible by  5.

Therefore 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
Answer-11

(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
Answer-12

(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
Answer-13

(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.
Answer-14

(i) False
(ii) True
(iii) True
(iv) True

— End of Playing with Numbers Solutions :–

Return to –  Concise Selina Maths Solutions for ICSE Class -6

Thanks

Share with your friends

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.