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Frank Learning Maths Class 8 Solutions Chapter 8 Playing With Numbers
Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Frank Learning Maths Class 8 Book, Chapter 8 Playing With Numbers. Here students can easily find step by step solutions of all the problems for Playing With Numbers. Here students will find solutions for Exercise 8.1, 8.2 and 8.3. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.
Playing With Numbers Exercise 8.2 Solution
Question no – (1)
Solution :
Given, 5n6 is multiple of 3
∴ The sum of the digit is a multiple of 3,
∴ 5 + n + 6 = 0, 3, 6, 9, 12, 15, 18
∴ 11 + n = 0, 3, 6, 9, 12, 15, 18
∴ The values of n,
= 11 + n = 12, 15, 18
∴ n = 1, 4, 7
Therefore, the value of n will be 1, 4, 7
Question no – (2)
Solution :
Given, 597n8 is divisible by 9 if the sum of the digits are divisible by 9
∴ 5 + 9 + 7 + n + 8 = 0, 9, 18, 27, 36, 45
= 29 + n = 36
∴ n = 7
Therefore, the value of n will be 7
Question no – (3)
Solution :
If 814y is divisible by 6, and the 814 y is divisible by 2, 3 both 814 y is divisible by 2
= y should be an even digit,
∴ y = 0, 2, 4, 6, 8 ….(1)
814 y is divisible y 2 then 8 + 1 + 4 + y = sum of the digit divisible by 3,
= 13 + y = 0, 3, 9, 12, 15, 18, 21 ….(3.)
∴ 13 + y = 15, 21
∴ Values of y = 2, 8
Hence, the values of y will be 2, 8
Question no – (4)
Solution :
If 7216a is divisible by 6 then it is divisible by 2, 3 both
∴ 7216a divisible by 2 if,
a = 0, 2, 4,8, 10 ….(1)
∴ 7216a is divisible by 3 if,
7 + 2 + 1 + 6 + a = 0, 3, 6, 9, 12, 15, 18
16 + a = 0, 6
∴ Values of a = 0, 6
Therefore, the values of ‘a’ will be 0, 6
Question no – (5)
Solution :
If 36×2 is divisible by 4, where x is a digit, then the value of x will be 1, 3, 5, 7, 9
Step by Step Solution :
If 36×2 be divisible by 4, then the number (x2) must be divisible by 4
And this will happen if x = 1, 3, 5, 7, 9
∵ We know, A number is divisible by 4 if the last two digit of the number is divisible by
4 or the whole number is divisible by 4.
Question no – (6)
Solution :
If 148y is divisible 4, where y is a digit, then the value of will be 0, 4 or 8
Step by Step Solution :
If 148y be divisible by 4 the number (8y) must be divisible by 4.
And, it will happen if y = 0, 4, 8
Because we know, a number is divisible by 4 if the last two digit of the number is divisible by 4 or the whole number is divisible by 4.
Question no – (7)
Solution :
(a) 48, 962 is divided by 3 and by 9 if sum of the digits divided by 3 and 9
∴ 4 + 8 + 9 + 6 + 2 = 29
∴ Remainder will be = 2, 2
(b) 3, 75, 273
Divided by 5 then remainder = 3
Divided by 2 then remainder = 1
(c) 78,562
Divided by 3 then remainder = 1
Divided by 4 then remainder = 2
(d) 73, 6843 is divided by 2 and by 10 than,
Divided by 2 then remainder = 1
Divided by 10 then remainder = 3
(e) 843294 is divided by 5 and by 10 than,
Divided by 5 then remainder = 4
Divided by 10 then remainder = 4
Playing With Numbers Exercise 8.3 Solution
Question no – (1)
Solution :
(a)
A B A
+ 1 B 2
———————–
6 6 6
∴ A + 2 = 6 than – A = 4
∴ B + B = 6 than – B = 8
∴ A + 1 = 6
(b)
3 A 5
+ 6 2 B
———————–
B 7 A
3 + 6 = B than B = 9
∴ A + 2 = 7 than
∴ 5 + B = A than A = 4
(c)
1 A 8
+ B B B
———————–
A B 5
8 + B = 5 than B = 7
∴ A + B = B then A = 9
∴ A + 7 = 7
(d)
2 A B
+ 5 9 B
———————–
A 7 A
2 + 5 = A then
∴ A + 9 = 7 than A = 8
∴ B + B = A than B = 4
(e)
5 A
× A
———————–
B B A
∴ A should be = 6
∴ B should be = 3
(f)
7 A
× 3
———————–
B B A
∴ A should be = 5
∴ B should be = 2
(g)
A B
× 9
———————–
6 A B
∴ A should be = 7
∴ B should be = 5
(h)
A B
× 6
———————–
B B B
∴ A should be = 7
∴ B should be = 4
(i)
A B
× 5
———————–
C A B
∴ A should be = 7
∴ B should be = 5
∴ C should be = 3
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