Class 8 ICSE Maths Solutions Chapter 5 Playing with Numbers (Selina Concise)
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Selina Class 8 ICSE Math Book, Chapter 5, Playing with Numbers. Here students can easily find step by step solutions of all the problems for Playing with Numbers, Exercise 5A, 5B, 5C and 5D Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 5 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.
Playing with Numbers Exercise 5(A) Solution :
Question no – (1)
Solution :
(i) Let, a × b = 73
and, b × a = 37
Now, The quotient of 73 + 37 when divided by 11,
= (7 + 3)
= 10
Therefore, the quotient is 10.
(ii) Let, a × b = 73
and, b × a = 37
Now, The quotient of 73 + 37 When divided by 10 than,
= 73 + 37/7 + 3
= 11
Hence, the quotient will be 11.
Question no – (2)
Solution :
(i) Here, Let,
ab = 94; ba = 49
a = 9; b = 4
∴ The quotient of (94 + 49) when divided by 11 is,
= 94 + 49/11
= 13
Thus, the quotient will be 13.
(ii) Here Let,
ab = 94; ba = 49
a = 9; b = 4
∴ The quotient of (94 + 49) when divided by 11 is,
= 94 + 49/13
= 11
Therefore, the quotient will be 11.
Question no – (3)
Solution :
(i) The quotient of 73 – 37 when divided by 7 that is,
= (7 – 3)
= 4
Hence, the quotient will be 4.
(ii) The quotient of 73 – 37 when divided by 4 that is 9.
= 73 – 37/7 – 3
= 36/4
= 9
Question no – (4)
Solution :
Given in the question, a = b
Prof = abc = bac
∴ abc = 100 + 10 b + c ….(i)
bac = 100 + 10 a + c ….(iv)
Subtract (iv) from (iii)
abc – bac = 0
∴ abc = bac…(Proved)
Question no – (5)
Solution :
(i) Let, ab = 94
ba = 49
∴ The quotient of 94 – 49 is when divided by 9 is,
= (9 – 4)
= 5
Therefore, the quotient will be 5.
(ii) Let, ab = 94
ba = 49
∴ The quotient of 94 – 49 when divided by 5 is,
= (94 – 49/9 – 4)
= 9
Therefore, the quotient will be 9.
Question no – (6)
Solution :
Here, abc = 100 a + 10 b + c …..(i)
bca = 100 b + 10 c + a ….(ii)
cab = 100 c + 10 a + b …..(iii)
Adding it,
abc + bca + cab = 11 a + 111b + 111 c
= 111 ( a + b + c)
= 3 × 37 (a + b + c)
Now, 527 + 752 + 275, to check
Here, a = 5, b = 2, c = 7
∴ 527 + 752 + 275 = 3 × 37 (5 + 2 + 7)
= 3 × 37 × 14
Therefore, it Divisible by 14
Question no – (8)
Solution :
In the given question, , c > a
Proof – cba = 100 c + 10 b + a
∴ abc = 100 a + 10 b + c
Subtracting, (ii) from (i)
= cba – abc = 100 c + 10 b + a – 100 a – 10 b – c
= cba – abc = 99 c – 99 a
∴ 99 (c – a)…(Proved)
Playing with Numbers Exercise 5(B) Solution :
Question no – (1)
Solution :
Given in the question,
3 A + 25 = B 2
Now, 5 + 2 = 7
7 + 5 = 12
∴ 37 + 25 = 62
∴ A = 7; B = 6
Question no – (2)
Solution :
Given, 9 8 + 4 A = C B 3
∴ A = 5, so, 8 + 5 = 13
9 + 4 + 1 = 14
∴ A = 5, B = 4, C = 1
∴ 98 + 45 = 143
Question no – (3)
Solution :
Given in the question, A 1+ 1 B = B 0
Here, B = 9 so, 9 + 1 = 10
= 9 – 2 = 7
= A
Therefore, 71 + 19 = 90
Question no – (4)
Solution :
Given, 2 A B + A B 1 = B 1 8
Here, 8 – 1 = 7 = B
7 + A = 11
A = 11 -7
= 4
∴ 247 + 471 = 718
Question no – (5)
Solution :
Given, 1 2 A + 6 A B = A 0 9
Now, A + B = 9
= 2 + A = 10
= A = 8
∴ 9 – 8 = 1 = B
∴ 128 + 681 = 809
Question no – (6)
Solution :
Given, 1 A × A = 9A
Here, A = 6
∴ 16 × 6 = 96
Question no – (7)
Solution :
Given, AB + 6 = BBB
Here, B = 4
∴ 6 × 4 = 24
6 × A + 2
= 4
= A = 7
∴ 74 × 6 = 444
Question no – (9)
Solution :
Given, AB × 5 = CAB
B = 0 or 5 × 0 = 0
Now, A × 5 = A
5 × 5 = 25
C = 2
∴ 50 × 5 = 250
Question no – (10)
Solution :
Given, 8 A 5 + 9 4 A = 1 A 3 3
Now, 5 + A = 33
A + 4 = 13
A = 13 – 4
= 8
∴ 885 + 948 = 1833
Question no – (11)
Solution :
As per the question,
6 A B 5 + D 5 8 C = 9 3 5 1
C + 5 = 11
= C = 11 – 5
= 6
∴ 8 + B + 1 = 15
B = 15 – 9
= 6
A + 5 + 1
= 13
= A = 13 – 6
= 7
And, 6 + D + 1 = 9
= D = 9 – 7
= 2
∴ 6765 + 2586 = 9351
Playing with Numbers Exercise 5(C) Solution :
Question no – (1)
Solution :
We know that, a number having its unit digit 0, 2, 4, 6, 8 divisible by 2.
So that, here (i) 192 and (ii) 1660 are divisible by 2.
Question no – (2)
Solution :
We know, a number is divisible by 3, when the sum of the digits of number is divisible by 3.
Hence, here (i) 261 and 777 is divisible by 3.
Question no – (3)
Solution :
A number is divisible by 4, if the number formed by the last two digits is divisible by 4.
That’s why, (i) 360, (iii) 5348 and (iv) 7756 are divisible by 4
Question no – (4)
Solution :
As we know, a number having its unit digit 0 or 5 is divisible by 5,
So, we can say that, here only (i) 3250 and (ii) 39255 is divisible by 5.
Question no – (5)
Solution :
Here, A number having its unit digit ‘0’ is divisible by 10.
Thus, we can say that, here (i) 5100 and (iii) 3400 is divisible by 10.
Question no – (6)
Solution :
A number is divisible by 11 it the digit of the sum of digits at the odd place and the sum of digits at sum place is zero or eleven.
Therefore, given in the question only (i) 2563 is divisible by 11.
Playing with Numbers Exercise 5(D) Solution :
Question no – (1)
Solution :
Given, 1 x 5 divisible by 3
= 1 + x + 5 = 3
= 6 + x = 0, 3, 6, 9
= x = – 6, 0, 3, 6, 9
Therefore, the value of x is 0, 3, 6, 9
Question no – (2)
Solution :
Given, 31×5 divisible by 3
= 3 + 1 + x + 5 is multiple of 3
= 9 + x
= 0, 3, 6, 9
= x = 0, 3, 6, 9, – 9
Hence, the value of x is 0, 3, 6, 9
Question no – (3)
Solution :
Given, 28×6 a multiple of 3
= 2 + 8 + x + 6 is multiple of ‘3’
= 16 + x = 0, 3, 6, 9, 12, 15, 18
= x = -18, -5, -2, 0, 2, 5, 8
Thus, the value of x is 2, 5, 8.
Question no – (4)
Solution :
Given, 24x divisible by 6
= 2 + 4 + x is multiple of ‘6’
= 6 + x = 0, 6, 12
= x = -6, 0, 6
Therefore, the value of x is 0, 6
Question no – (5)
Solution :
Given, 3×26 a multiple of 6
= 3 + x + 2 + 6 is multiple of 3
= 11 + x = 0, 3, 6, 9, 12, 15, 18, 21
= x = -11, -8, -5, -2, 1, 4, 7, 10.
Therefore, the value of x is 1, 4, 7
Question no – (6)
Solution :
Given, 42×8 divisible by 4?
∴ 4 + 2 + x + 8 is multiple of 2
= 14 + x = 0, 2, 4, 6, 8
= x = -8, -6, -4, -2, 2, 4, 6, 8
Therefore, the value of x is 2, 4, 6, 8
Question no – (7)
Solution :
Given, 9142x a multiple of 4
= 9 + 1 = 4 + 2 + x is multiple of 4
= 16 + x = 0, 4, 8, 16, …….
= x = – 8, – 4, 0, 4, 8
∴ The value x is = 4, 8
Question no – (8)
Solution :
Given, 7×34 a multiple by 9
∴ 7 + x + 3 + 4 is multiple of 9
= 14 + x = 0, 9, 18, 27, …..
= x = -1, 4, 13
Therefore, the value of x will be 4
Next Chapter Solution :
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