RBSE Maths Class 8 Chapter 4: Mental Exercise Important Questions and Solutions

RBSE Maths Chapter 4 – Mental Exercise Class 8 Important questions and solutions are provided here. The important questions and solutions of Chapter 4 available at BYJU’S contain step by step, along with concept-based explanations. The important questions which are given here are based on the new pattern prescribed by the RBSE. Students can access the RBSE Class 8 solutions for all the subjects here.

Chapter 4 of the RBSE Class 8 Maths will help the students to solve problems related to the expanded form of numbers, interchanging the two, three-digit numbers, rules of divisibility, to determine the eliminated digits in the four operations.

RBSE Maths Chapter 4: Exercise 4.1 Textbook Important Questions and Solutions

Question 1: If a 3 digit number 24z is divisible by 9, then find the value of z, where z is a digit.

Solution:

Sum of the digits = 2 + 4 + z = 6 + z.

The given number is 24z.

The sum of the digits must be divisible by 9.

This is possible if 6 + z is either 9 or 18.

If 6 + z = 9, then z = 9 – 6 = 3.

If 6 + z = 18, then z = 18 – 6 = 12.

Here, z is a digit, so z ≠ 12.

Hence, z = 3.

Question 2: If a 3 digit number 89m is divisible by 9, then find the value of m.

Solution:

Sum of digits = 8 + 9 + m = 17 + m

The given number is 89m.

The sum of the digits must be divisible by 9.

This is possible if the number is either of them: 18, 27, 36 . . . etc.

But m is a digit.

If 17 + m = 18, then m = 18 – 17 = 1.

If 17 + m = 27, then m = 27 – 17 =10.

But m is a digit, so m ≠ 10.

Hence, m = 1.

Question 3: 31P5 is a multiple of 9 and two values are obtained by P. Why? Where P is a digit.

Solution:

Sum of digits = 3 + 1 + P + 5 = 9 + P.

The given number is 31P5.

It is given that 9 + P is a multiple of 9 which is possible if 9 + P is either 9 or 10.

If 9 + P = 9, then P = 9 – 9 = 0

If 9 + P =18, then P = 18 – 9 = 9

Therefore, P has two values 0 and 9.

Question 4: If a 3-digit number 24q is a multiple of 3, then find the value of q.

Solution:

Sum of digits = 2 + 4 + q = 6 + q.

The given three digit number is 24q.

It is given that 6 + q is a multiple of 3, the possible values of 6 + q are 6, 9, 12, 15, etc.

If 6 + q = 6, then q = 6 – 6 = 0

If 6 + q = 9, then q = 9 – 6 = 3

If 6 + q = 12, then q – 12 – 6 = 6

If 6 + q = 15, then q = 15 – 6 = 9

If 6 + q = 18, then q = 18 – 6 = 12

But q is a digit, so q ≠ 12.

The values of q are 0, 3, 6, 9.

Question 5: Test the divisibility of the given numbers by 3, 9 and 11.

[i] 294

[ii] 4455

[iii] 1041966

Solution:

[i] 294

To check the divisibility by 3.

Sum of digits = 2 + 9 + 4 = 15.

15 is divisible by 3.

Hence, the number 294 is also divisible by 3.

To check the divisibility by 9.

Sum of digits = 2 + 9 + 4 = 15

15 is not divisible by 9.

Hence, the number 294 is not divisible by 9

To check the divisibility by 11.

4 x 1 + 9 x (-1) + 2 x 1 = 4 – 9 + 2 = -3

So, 294 is not divisible by 11.

[ii] 4455

To check the divisibility by 3.

Sum of digits = 4 + 4 + 5 + 5 = 18

18 is divisible by 3.

Hence, 4455 is also divisible by 3.

To check the divisibility by 9.

Sum of digits = 18 which is divisible by 9.

Hence, 4455 is also divisible by 9.

To check the divisibility by 11.

5 x 1 + 5 x (-1) + 4 x 1 + 4 x (-1) = 5 – 5+ 4 – 4 = 0.

Hence, 4455 is divisible by 11.

[iii] 1041966

To check the divisibility by 3.

Sum of digits =1 + 0 + 4 + 1 + 9 + 6 + 6 = 27

27 is divisible by 3.

Hence, 1041966 is also divisible by 3.

To check the divisibility by 9.

27 is divisible by 9.

Hence, 1041966 is divisible by 9.

To check the divisibility by 11.

6 x 1 + 6 x (-1) + 9 x 1 + 1 x (-1) + 4 x 1 + 0 x (-1) + 1 x 1

6 – 6 + 9 – 1 + 4 + 0 + 1 = 20 – 7 = 13

13 is not divisible by 11.

Hence, 1041966 is not divisible by 11.

Question 6: If R = 4 in 31R1, then by the rule of divisibility find that this number is divisible by 11.

Solution:

Given: R = 4.

Hence, the number is 3141.

To check the divisibility by 11.

1 x 1 + 4 x (- 1) + 1 x 1 + 3 x (-1) = 1 – 4 + 1 – 3 = 2 -7 = -5

– 5 is not divisible by 11.

Hence, 3141 is not divisible by 11.

Question 7: If 31R5 is a multiple of 3, then find the value of R, where R is a digit.

Solution:

Given Number = 31R5

Sum of digits = 3 + 1 + R + 5 = 9 + R

9 + R is a multiple of 3 so it is possible if values of 9 + R are 9, 12, 15, 18 etc.

If 9 + R is 9, then R = 0.

If 9 + R is 12, then R = 3.

If 9 + R is 15, then R = 6.

If 9 + R = 18, then R = 9.

If 9 + R = 21, then R = 12

Since R is a digit, R ≠ 12

Therefore, the values of R are 0, 3, 6 and 9.

RBSE Maths Chapter 4: Exercise 4.2 Textbook Important Questions and Solutions

Question 1: Find the values of the alphabets and show the causes of that process.

[i] 5A

+ 34

_____

B2

_____

[ii] 5A

+ 79

_____

CB3

_____

[iii] AB

+ 37

_____

6A

_____

[iv] 5AB

+ AB1

______

B98

______

[v] 12A

+ 6AB

______

A09

______

[vi] 1A

X A

______

9A

______

[vii] AB

X B

_______

CAB

_______

[viii] AB

X 6

________

BBB

________

Solution:

[i] 5A

+ 34

_____

B2

_____

The sum of A and 4 is given as 2.

Hence, the value of A is 8.

8 + 4 = 12

If A = 8, then B = 5 + 3 + 1 [carryover] = 9

[ii] 5A

+ 79

_____

CB3

_____

The sum of A and 9 is given as 3.

Hence, the value of A is 4.

4 + 9 = 13

If A = 4, then CB = 5 + 7 + 1[carryover] = 13.

[iii] AB

+ 37

_____

6A

_____

The sum of A and 3 gives 6.

B + 7 gives A.

So, B = 6.

But 6 + 7 = 13.

In the tens column, 6 is given. So, B = 5.

If B = 5, then 5 + 7 = 12 and 2 in units place.

Therefore, A = 2.

The values of A and B are 2 and 5 respectively.

[iv] 5AB

+ AB1

______

B98

______

The sum of B and 1 is 8.

B + 1 = 8.

B = 8 – 1 = 7.

A + B = 9

A + 7 = 9

A = 2

5 + A = B

5 + 2 = 7

The values of A and B are 2 and 7 respectively.

[v] 12A

+ 6AB

______

A09

______

The sum of 2 and A is 0.

2 + A = 0

Then A = 8 as 2 + 8 = 10. 0 in the units place.

The sum of A and B is 9.

A + B = 9

8 + B = 9

B = 9 – 8 = 1

The values of A and B are 8 and 1 respectively.

[vi] 1A

X A

______

9A

______

The multiplication of A and A gives A in the units place. The product of 6 and 6 gives 36, 6 in the units place.

So, A = 6.

16 x 6 = 96.

[vii] AB

X B

_______

CAB

_______

When B is multiplied by B, the result is B. This holds when A = 2 and B = 5.

25 x 5 = 125.

[viii] AB

X 6

________

BBB

________

When B is multiplied by 6, B is obtained.

It holds when B = 4.

4 x 6 = 24

A x 6 + 2(carry over) = 24

It is true when A = 7.

74 x 6 = 444.

The values of A and B are 7 and 4, respectively.

Question 2: Find the value of x or (*) in the questions given below.

[i] 2 *

+ * 8

+ 95

____

167

____

[ii] 905

+ *12

+ 88*

_____

2100

_____

[iii] 7*3

– 281

______

432

______

[iv] 57

– 3*

______

18

______

[v] 68

x (x)

______

408

______

[vi] 763

x (3x)

_______

25942

_______

[vii] 216 ÷ 2x = 0

Quotient = 8

[viii] 907 ÷ 7x = 19

Quotient = 24

Solution:

[i] 2 *

+ * 8

+ 95

____

167

____

* + 8 + 5 = 7

* = 4

Since 7 is in the units place, 4 + 8 + 5 = 17 [1 is the carryover]

1 + 2 + * + 9 = 16

* = 4.

[ii] 905

+ *12

+ 88*

_____

2100

_____

5 + 2 + * = 0

Since 0 is in the units place, * = 3

5 + 2 + 3 = 10

1 + 9 + * + 8 = 21

* = 3

[iii] 7*3

– 281

______

432

______

* – 8 = 3

11 – 8 = 3

* = 1

[iv] 57

– 3*

______

18

______

7 – * = 8

* = 9

[v] 68

x (x)

______

408

______

(x) x 68 = 408

x = 408 / 68

x = 6

[vi] 763

x (3x)

_______

25942

_______

763 x (3x) = 25942

(30 + x) * 763 = 25942

22890 + 763x = 25942

763x = 25942 – 22890

763x = 3052

x = 3052 / 763

x = 4

[vii] 216 ÷ 2x = 0

Quotient = 8

2x * 8 + 0 = 216

(20 + x) * 8 + 0 = 216

160 + 8x = 216

8x = 216 – 160

8x = 56

x = 56 / 8

x = 7

[viii] 907 ÷ 7x = 19

Quotient = 24

7x * 24 + 19 = 907

(7 + 10x) * 24 + 19 = 907

168 + 240x + 19 = 907

240x + 187 = 907

240x = 720

x = 720 / 240

x = 3

RBSE Maths Chapter 4: Additional Questions and Solutions

Question 1: Write the generalised form of the numbers given below.

[i] 10 x 5 + 6

[ii] 8 x 100 + 0 x 10 + 5

[iii] 9 x 100 + 9 x 10 + 9

Solution:

[i] 10 x 5 + 6

= 5 x 10 + 6 x 1

= 50 + 6

= 56

[ii] 8 x 100 + 0 x 10 + 5

= 8 x 100 + 0 x 10 + 5 x 1

= 800 + 0 + 5

= 805

[iii] 9 x 100 + 9 x 10 + 9

= 9 x 100 + 9 x 10 + 9 x 1

= 900 + 90 + 9

= 999

Question 2: If 79y is divisible by 9, then is it possible that y will have more than one value?

Solution:

Sum of the digits = 7 + 9 + y = 16 + y

The given number is 79y.

It is divisible by 9.

If 16 + y is divisible by 9, then the possible values are 18, 27, 36, etc.

But y is a digit.

If 16 + y = 18, then y = 2

If 16 + y = 27, then y = 11

So, y = 2.

Question 3: Find the value of * in the following problems.

[i] 3 *

+ 57

+ 34

____

127

____

[ii] 56

+ 77

+ *3

_____

216

_____

[iii] 443

+ *57

+ 128

______

928

______

[iv] 82

+ 55

+ 99

______

*36

______

[v] 76

– 5*

_____

25

_____

[vi] 54

– 2*

______

28

______

[vii] 803

– 2*6

_______

567

_______

[viii] 782

– *73

_______

209

_______

[ix] 84

– *8

______

16

______

Solution:

[i] 3 *

+ 57

+ 34

____

127

____

To get 7 in the units place, * = 6

* + 7 + 4 = 7

6 + 7 + 4 = 17

[ii] 56

+ 77

+ *3

_____

216

_____

6 + 7 + 3 = 16 [6 in units place and 1 is the carryover]

1 + 5 + 7 + * = 21

13 + * = 21

* = 21 – 13

* = 8

[iii] 443

+ *57

+ 128

______

928

______

3 + 7 + 8 = 18 [8 in units place and 1 carryover]

1 + 4 + 5 + 2 = 12 [2 in units place and 1 carryover]

1 + 4 + * + 1 = 9

6 + * = 9

* = 9 – 6

* = 3

[iv] 82

+ 55

+ 99

______

*36

______

2 + 5 + 9 = 16 [6 in units place and 1 carryover]

1 + 8 + 5 + 9 = *

23 = *3

* = 2

[v] 76

– 5*

_____

25

_____

6 – * = 5

* = 1

76 – 51 = 25

[vi] 54

– 2*

______

28

______

14 – * = 8

* = 14 – 8

* = 6

14 – 6 = 8

[vii] 803

– 2*6

_______

567

_______

Subtracting 3 from 6 isn’t possible. So, borrowing is done from the next digit which is 0.

After borrowing from 8, in the place of 0, it is 10 and in the place of 3, it is 13, in the place of 8 it is 7.

Hence, it becomes,

13 – 6 = 7

9 – * = 6

9 – 6 = *

* = 3

[viii] 782

– *73

_______

209

_______

Subtracting 2 from 3 isn’t possible. So, borrowing is done from the next digit which is 8.

After borrowing from 8, in the place of 2, it is 12 and in the place of 8, it is 7.

Hence, it becomes,

12 – 3 = 9

7 – 7 = 0

7 – * = 2

* = 5

[ix] 84

– *8

______

16

______

Subtracting 4 from 8 isn’t possible. So, borrowing is done from the next digit which is 8.

After borrowing from 8, in the place of 4, it is 14 and in the place of 8, it is 7.

Hence, it becomes,

14 – 8 = 6

7 – * = 1

* = 6

Question 4: Write a 3 x 3 and a 4 x 4 crossword puzzle by the following digits in such a way that the horizontal, vertical and diagonal squares are the same.

[i] 2, 3, 4, 5, 6, , 8, 9, 10

[ii] 2, 3, 4, 5, 6, , 8, 9, 10, 11, 12, 13, 14, 15, 16, 17

Solution:

[i] Digits along the horizontal row are:

[1] 2 + 7 + 6 = 15

[2] 9 + 5 + 1 = 15

[3] 4 + 3 + 8 = 15

Digits along the vertical columns are:

[1] 2 + 9 + 4 = 15

[2] 7 + 5 + 3 = 15

[3] 6 + 1 + 8 = 15

Digits along the principal diagonal are: 2 + 5 + 8 = 15

Digits along the secondary diagonal are: 6 + 5 + 4 = 15

Total

2

7

6

15

9

5

1

15

4

3

8

15

15

15

15

15

[ii] Digits along the horizontal row are:

[1] 2 + 7 + 6 = 15

[2] 9 + 5 + 1 = 15

[3] 4 + 3 + 8 = 15

Digits along the vertical columns are:

[1] 2 + 9 + 4 = 15

[2] 7 + 5 + 3 = 15

[3] 6 + 1 + 8 = 15

Digits along the principal diagonal are: 2 + 5 + 8 = 15

Digits along the secondary diagonal are: 6 + 5 + 4 = 15

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

The sum of the horizontal, vertical and diagonal columns are the same = 38

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