chapter 16 maths class 8 deals with Playing with Numbers. Find ncert solutions for class 8 maths chapter 16 here. In Class 8 maths chapter 16, we explored finding factors and multiples and the relationships among natural numbers, whole numbers, integers and rational numbers.
NCERT solutions for Class 8 Chapter 16
Exercises and solutions of chapter 16 maths class 8 available here so students can read online view.
mensuration.in provides all exercises with answers for chapter 8 in class 8 maths.
Find solutions to exercise 16.1 and exercise 16.2 in chapter 8 in class 8 maths.
Ncert Solutions are provided by mensuration.in for all subjects of maths, physics, chemistry, biology for preparation of exams.
Find solutions to ncert maths class 8 chapter 16.
class 8 maths chapter 16 exercise 16.1
Find the values of the letters in each of the following and give reasons for the steps involved.
class 8 maths chapter 16 exercise 16.1 (1)
1. 3 A
+ 2 5
B 2
Solution to class 8 maths chapter 16 exercise 16.1(1)
On putting A = 1, 2, 3, 4, 5, 6, 7 and so on and we get, for A=7
7+5 = 12 in which ones place is 2.
∴ A=7
And putting 2 and carry over 1, we get
B = 6
Hence, A = 7 and B = 6
class 8 maths chapter 16 exercise 16.1 (2)
2. 4 A
+ 9 8
C B 3
Solution to class 8 maths chapter 16 exercise 16.2 (2)
On putting A = 1, 2, 3, 4, 5, and so on and we get, 8 + 5 = 13 in which ones place is 3.
A = 5
And putting 3 and carry over 1, we get B = 4 and C = 1
Hence, A= 5, B = 4 and C=1
class 8 maths chapter 16 exercise 16.1 (3)
3. 1 A
× A
9 A
Solution to class 8 maths chapter 16 exercise 16.2 (3)
On putting A = 1, 2, 3, 4, 5, 6.. and so on and we get, A×A = 6 ×6 = 36 in which ones place is 6.
A = 6
And putting 6 and carry over 3, we get A×1=9.
Hence, A= 6
class 8 maths chapter 16 exercise 16.2 (4)
4. A B
+ 3 7
6 A
Solution to class 8 maths chapter 16 exercise 16.2 (4)
Here, we observe that B = 5 so that 7 +5 = 12.
Putting 2 at ones place and carry over 1 and A = 2, we get
2 + 3+1 = 6
Hence, A= 2 and B = 5
class 8 maths chapter 16 exercise 16.2 (5)
5. A B
× 3
C A B
Solution to class 8 maths chapter 16 exercise 16.2 (5)
Here on putting B = 0, we get 0 ×3 = 0.
And A = 5, then 5 ×3 = 15
A = 5 and C= 1
Hence, A= 5, B = 0 and C= 1
class 8 maths chapter 16 exercise 16.2 (6)
6. A B
× 5
C A B
Solution to class 8 maths chapter 16 exercise 16.2 (6)
On putting B = 0, we get
0 × 5 = 0 and A = 5, then 5 ×5 = 25
A = 5, C = 2
Hence, A= 5, B = 0 and C = 2
class 8 maths chapter 16 exercise 16.2 (7)
7. A B
× 6
B B B
Solution to class 8 maths chapter 16 exercise 16.2 (7)
Here product of B and 6 must be same as ones place digit as B. 6 ×1=6, 6×2=12, 6×3=18, 6×4 = 24
On putting B = 4, we get the ones digit 4 and remaining two B’s value should be 44. For 6×7 = 42+2 = 44
Hence, A=7 and B = 4
class 8 maths chapter 16 exercise 16.2 (8)
8. A 1
+ 1 B
B 0
Solution to class 8 maths chapter 16 exercise 16.2 (8)
On putting B = 9, we get 9 +1 = 10 Putting 0 at ones place and carry over 1, we get
For A = 7, 7+1+1 = 9
Hence A = 7 and B = 9
class 8 maths chapter 16 exercise 16.2 (9)
9. 2 A B
+ A B 1
B 1 8
Solution to class 8 maths chapter 16 exercise 16.2 (9)
On putting B = 7
7+1 =8 Now A = 4, then 4 + 7 = 11
Putting 1 at tens place and carry over 1, we get 2 + 4 + 1 = 7
Hence, A=4 and B = 7
class 8 maths chapter 16 exercise 16.2 (10)
10. 1 2 A
+ 6 A B
A 0 9
Solution to class 8 maths chapter 16 exercise 16.2 (10)
Putting A = 8 and B=1, we get
8+1=9
Now again we add 2+ 8 = 10
Tens place digit is ‘0’ and carry over 1.
Now 1+6+1 = 8 = A
Hence, A= 8 and B = 1
class 8 maths chapter 16 exercise 16.2
class 8 maths chapter 16 exercise 16.2 (1)
1. If 21y5 is a multiple of 9, where y is a digit, what is the value of y?
Solution to class 8 maths chapter 16 exercise 16.2 (1)
Since 21y5 is a multiple of 9.
Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9
∴ 2+1 + y + 5 = 8+y
Given y is a digit, so 0≤y ≤9,
For y = 1, 8+y is divisible by 9.
∴ y=1
class 8 maths chapter 16 exercise 16.2 (2)
2. If 31z5 is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers for the last problem. Why is this so?
Solution to class 8 maths chapter 16 exercise 16.2 (2)
Since 31z5 is a multiple of 9.
Therefore according to the divisibility rule of 9, the sum of all the digits should be multiple of 9.
3+1+z+5=9+z
‘z’ is a digit. 0≤z≤9
For z = 9 and z=0, 9+z is divisible by 9.
∴ z=0 or z=9
Hence 0 and 9 are two possible answers.
class 8 maths chapter 16 exercise 16.2 (3)
3. If 24x is a multiple of 3, where x is a digit, what is the value of x?
(Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers: 0, 3, 6, 9, 12, 15, 18, … . But since x is a digit, it can only be that 6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values.)
Solution to class 8 maths chapter 16 exercise 16.2 (3)
Since 24x is a multiple of 3
Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.
2+4+x=6+x
Since x is a digit.
For x=0 ⇒6+x=6+0=6
For x=3 ⇒6+ x=6+3=9
For x=6⇒6+ x=6+6=12
And for x=9 ⇒6+x=6+9=15
For x=0, x=3, x=6 and x=9
9+z is multiple of 3.
Thus, ‘x’ can have any of four different values.
class 8 maths chapter 16 exercise 16.2 (4)
4. If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Solution to class 8 maths chapter 16 exercise 16.2 (4)
Since 31z5 is a multiple of 3
Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.
Since ‘z’ is a digit.
3+1+z+5=9+z
For z=0 ⇒z+9=0+9=9
For z=3 ⇒z+9=3+9=12
For z=6 ⇒z+9=6+9=15
And For z=9 ⇒z+9=9+9=18
For z=0, z=3, z=6 and z=9
9+z is multiple of 3.
Hence, 0,3, 6 and 9 are four possible answers
By
Sudheer Kumar B