RBSE Solutions for Class 8 Maths Chapter 4 Mental Exercises In Text Exercise is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 4 Mental Exercises In Text Exercise.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Mental Exercises |

Exercise |
In Text Exercise |

Number of Questions |
14 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 8 Maths Chapter 4 Mental Exercises In Text Exercise

**Page No: 44**

Question 1.

Fill in the blanks

Solution

Question 2.

Write the following numbers in generalized form

(i) 10 × 5 + 6

(ii) 8 × 100 + 0 × 10 + 5

(iii) 9 × 100 + 9 × 10 + 9

Solution

(i) 10 × 5 + 6

= 5 × 10 + 6 × 1

= 56

(ii) 8 × 100 + 0 × 10 + 5

= 8 × 100 + 0 × 10 + 5 × 1

= 805

(iii) 9 × 100 + 9 × 10 + 9

= 9 × 100 + 9 × 10 + 9 × 1

= 999

**Page No: 46**

Question 3.

What will be the result if you think the following numbers.

(i) 27

(ii) 67

(iii) 94

Solution

(i) 27

Result 1

Thought Number = 27

Number(RBSESolutions.com)obtained after reversing its digits

= 72

Total of both = 27 + 72 = 99

Dividing received number by 11 = [latex]\frac { 99 }{ 11 }[/latex] = 9

Result = remainder is zero (0) .

Result 2

Thought Ngmber = 27

Number(RBSESolutions.com)obtained after reversing its digits

= 72

Subtracting small number from large number we get = 72 – 27 = 45

Dividing received number by 9 = [latex]\frac { 45 }{ 9 }[/latex] = 5

Result = remainder is zero (0)

(ii) 67 Result 1

Thought Number = 67

Number obtained(RBSESolutions.com)after reversing its digits

= 76

Total of both = 143

Dividing received number by 11 = [latex]\frac { 143 }{ 11 }[/latex] = 13

Result = remainder is zero (0)

Result 2

Thought Number = 67

Number obtained after reversing its digits

= 76

Subtracting small(RBSESolutions.com)number from large number

= 76 – 67 = 9

Dividing received number by 9 = [latex]\frac { 9 }{ 9 }[/latex] = 1

Result = remainder is zero (0)

(iii) 94 Result 1

Thought Number = 94

Number obtained after reversing its digits

= 49

Total of both Number = 94 + 49 = 143

Dividing received(RBSESolutions.com)number by 11 = [latex]\frac { 143 }{ 11 }[/latex] = 13

Result = remainder is zero (0)

Result 2

Thought Number = 94

Number obtained after reversing its digits

= 49

Subtracting small number from large number

= 94 – 49 = 45

Dividing received number by 9 = [latex]\frac { 45 }{ 9 }[/latex] = 5

Result-= remainder is zero (0)

**Page No: 47**

Question 4.

Test if Chhotu thinks the following numbers then what would be the result?

(i) 237

(ii) 119

(iii) 397

(iv) 435

Solution

(i) 237 Result 1,

Number thought by Chhotu = 237

The number obtained after reversing its digits = 732

Subtracting(RBSESolutions.com)small number from large number

= 732 – 237 = 495

Dividing received number by 99

= [latex]\frac { 495 }{ 99 }[/latex] = 5

Result = remainder is zero (0)

Result 2

Number thought by Chhotu = 237

After inter changing the digits we gets these number 723 and 372

Total of these number =

Dividing received number by 37 = [latex]\frac { 1332 }{ 37 }[/latex] = 36

Result = remainder is zero (0)

36 = 3 x 12 = 3 x (2 + 3 + 7)

= 3 x (Sum of digits in given(RBSESolutions.com)number)

(ii) 119 Result 1

Number thought by Chhotu = 119

The number obtained after reversing its digits = 911

Subtracting small(RBSESolutions.com)number from large number = 911 – 119 = 792

Dividing received number by 99 = [latex]\frac { 792 }{ 99 }[/latex] = 8

Result = remainder is zero (0)

Result 2

Number thought by Chhotu =119

After(RBSESolutions.com)interchanging the digits we get these numbers 911 and 191

Sum of these three numbers

= 119 + 911 + 191 = 1221

Dividing received number by 37 = [latex]\frac { 1221 }{ 37 }[/latex] = 33

Result = remainder is zero (0)

Quotient = 33

= 3 x 11

= 3 x (1 + 1 + 9)

= 3 x (Sum of digits in given number)

(iii) 397 Result 1

Number thought by Chhotu = 397

The number obtained after reversing its digits = 793

Subtracting small number from large number

= 793 – 397 = 396

Dividing received number by 99 = [latex]\frac { 396 }{ 99 }[/latex] = 4

Result = remainder is zero (0)

Result 2

Number thought by Chhotu = 397

After(RBSESolutions.com)interchanging the digits we gets these number 739 and 973

Total of these number =

Dividing received number by 37 so we get = [latex]\frac { 2109 }{ 37 }[/latex] = 57

Result = remainder is zero (0)

Quotient = 57

= 3 x 19

= 3 x (3 + 9 + 7)

= 3 x (Sum of digits in given number)

(iv) 435 Result 1

Number thought by Chhotu = 435

The number obtained after reversing its digits = 534

Subtracting small number by large number

= 534 – 435 = 99

On Dividing received number by 99 = [latex]\frac { 99 }{ 99 }[/latex] = 1

Result = remainder is zero (0)

Result 2

Number thought by Chhotu = 435

After interchanging the digits we(RBSESolutions.com)gets these number 543 and 354

Total of these numbers = 435 + 543 + 354

= 1332

Dividing Received number by 37 so we get

= [latex]\frac { 1332 }{ 37 }[/latex] = 36

Result = remainder is zero (0)

Quotient = 36

= 3 x 12

= 3 x (4 + 3 + 5)

= 3 x (Sum of digits in given number)

**Page No: 47**

Question 5.

If N is any Number :

Solution

If N is any number then :

**Page No: 48**

Question 6.

If 79y is divisible by 9, then is it possible more than one value of y?

Solution

Given Number = 79y

Sum of digits = 7 + 9 + y

= 16 + y

Number 16 + y must be divisible by 9.

This is possible only if the value of 16 + y is any one of(RBSESolutions.com)numbers 18, 27, 36… but y is only one digit

∴ 16 + y = 18

⇒ y = 18 – 16 = 2

Also, 16 + y = 27

⇒ y = 27 – 16 = 11

(This value of y is not possible because it has two digits)

Therefore, more(RBSESolutions.com)than one values of y are not possible.

**Page No: 49**

Question 7.

Test the divisibility of 5629003 by 11

Solution

According to the rule of divisibility by 11

= 3 × 1 + 0 × (- 1) + 0 × (1) + 9 (- 1) + 2 (1) + 6 × (- 1) + 5(1)

= 3 + 0 + 0 – 9 + 2 – 6 + 5

= 10 – 15

= – 5

That is not divisible by 11

So, the number 5629003 is not divisible by 11

**Page No: 50**

Question 8.

Find the value of in following addition-operations

Solution

(i) Sum of unit place digits

= * + 7 + 4

= * + 11

= * + 10 + 1

= 10 + (* + 1)

Sum of unit place digit = 7

∴ * + 1 = 7

⇒ * = 7 – 1

⇒ * = 6

∴ Blank digit is 6.

(ii) Sum of unit place digit

= 6 + 7 + 3

= 16

= 10 + 6

Where 1 is tens place digit, therefore, in addition(RBSESolutions.com)operation, it is carried with ten place. Soon adding with ten place digits

= 1 + 5 + 7 + *

= * + 13

Which is equal to 21 (given that)

Therefore * + 13 = 21

= * = 21 – 13

= * = 8

∴ Blank digit is 8

(iii) Sum of unit place digit

= 3 + 7 + 8 = 18 = 10 + 8

Where 1 is tens place digit. It is carried with tens place digits so on adding tens place digits

= 1 + 4 + 5 + 2 = 12

Where 1 is hundredth(RBSESolutions.com)place digit, on adding, it is carry with hundredth place digits = 1 + 4 + * + 1 = 6 + *

Which is equal to 9, Therefore,

6 + * = 9

= * = 9 – 6

= * = 3

∴Blank digit is 3.

(iv) Sum of unit place digits

= 2 + 5 + 9 = 16 = 10 + 6

Where 1 is tens place digit. It is carried with tens place, so on adding with tens place digits

= 1 + 8 + 5 + 9 = *3

⇒23 = *3

Comparing ten’s place digit of both sides

∴ Blank digit = 2

**Page No: 51**

Question 9.

Find the value of * in the following subtraction operations

Solution

(i)

Here, on subtracting unit place digit * from 6, we get 5 which is less than 6

Therefore, 6 – * = 5

⇒ – * = 5 – 6

⇒ – * = – 1

⇒ * = 1

∴ Blank digit is 1

(ii)

On Subtracting * from 4, we get 8 which is greater(RBSESolutions.com)than 4. So, we take one tens (10) from tens place digit 5 and add with unit place digit 4

⇒ (10 + 4) – * = 8

⇒ 14 – * = 8

⇒ – * = 8 – 14

⇒ – * = – 6

⇒ * = 6

Therefore digit is 6.

(iii)

Here we have to subtract 8 from 4 which is less than 8. So, borrow one tens (10) from tens place digit 8 and add with unit place digit 4 and subtract 8

(10 + 4) – 8 = 6

Now, after borrow one tens 10 from tens(RBSESolutions.com)digit 8, there remains 8 – 1 = 7 as tens place

7 – * = 1

⇒ – * = 1 – 7 = – 6

⇒ * = 6

Now we have

Therefore Blank digit is 6.

(iv)

To subtract 6 from unit place digit 3, we take carry one tens (10) from tens place and add with unit place digit 3, then subtract 6. We have

(10 + 3) – 6 = 7

Here, tens place is zero (0), therefore, we(RBSESolutions.com)carry one hundred (100) from hundredth. Place digit 8 and add with tens place digit

= 100 = 10 Tens

10 + 0 = 10 Tens

On transfer one tens to unit place digit, remaining tens are

10 – 1 = 9

Therefore, 9 – * = 6

⇒ – * = 6 – 9 = – 3

⇒ * = 3

Now we have

Therefore, Blank digit is

(v)

To subtract 3 from 2 in unit place, we carry one tens (10) from tens place digit 8 and add with unit place digit 2, then subtract 3

(10 + 2) – 3 = 9

Remaining tens = 8 – 1 = 7

Therefore 7 – 7 = 0

Now, we get 2 after(RBSESolutions.com)subtracting from hundredth place digit 7

7 – * = 2

⇒ – * = 2 – 7 = – 5

⇒ * = 5

Now we have

Therefore, blank digit is 5.

**Page No: 52**

Question 10.

Find the value of Algebraic expression in the following multiplication-operation

Solution.

(i)

Given that, x is written in tens place digit

So, 56 × x5 = 56 × (10x + 5)

= 560x + 280

Now, According to(RBSESolutions.com)question

560x + 280 = 1400

⇒ 560x = 1400 – 280

⇒ 560x = 1120

⇒ [latex]x=\frac { 1120 }{ 560 }=2[/latex]

Now we have

Hence, x = 2

(ii)

Here, x is written in unit place digit

∴ 4x × 37 = (4 × 10 + x) × 37

= (40 + x) × 37

= 40 × 37 + 37x

= 1480 + 37x

But according to question

1480 + 37x = 1554

⇒ 37x = 1554 – 1480

⇒ 37x = 74

[latex]x=\frac { 74 }{ 37 }=2[/latex]

Now we have

Hence x = 2

(iii)

Here, x is in unit place digit

23 × 3x = 23 × (3 × 10 + x)

⇒ 69x = 23 × (30 + x)

⇒ 69x = 690 + 23x

According to question

690 + 23x = 736

⇒ 23x = 736 – 690

⇒ 23x = 46

[latex]x=\frac { 46 }{ 23 }=2[/latex]

Now we have

Hence x = 2

**Page No: 52**

Question 11.

Find the value of x in following division-operation.

Solution.

(i)

We know that

Dividend = divisor × quotient + remainder

⇒ 217 = 27 × x + 1

⇒ 217 = 27x + 1

⇒ 27x = 217 – 1 = 216

⇒ [latex]x=\frac { 216 }{ 27 }=8[/latex]

Hence, x = 8

(ii)

According to division operation rule

Dividend = divisor × quotient + remainder

100 = x6 × 6 + 4

100 = (10x + 6) × 6 + 4

100 = 60x + 36 + 4

100 = 60x + 40

⇒ 100 – 40 = 60x

⇒ 60x = 60

[latex]x=\frac { 60 }{ 60 }=1[/latex]

Hence, x = 1

(iii)

According to division(RBSESolutions.com)operation rule

Dividend = divisor × quotient + remainder

120 = 1x × 9 + 3

⇒ 120 = (10 + x) × 9 + 3

⇒ 120 = 90 + 9x + 3

⇒ 9x = 27

[latex]x=\frac { 27 }{ 9 }=3[/latex]

**Page No: 54**

Question 12.

What is the sum of digits of vertical and horizontal digits? Is this total same?

Solution

Sum of digits of vertical column.

(i) 2 + 9 + 4 = 15

(ii) 7 + 5 + 3 = 15

(iii) 6 + 1 + 8 = 15

Sum of digits of horizontal row

(i) 2 + 7 + 6 = 15

(ii) 9 + 5 + 1 = 15

(iii) 4 + 3 + 8 = 15

Yes, the sum is same.

**Page No: 55**

Question 13.

In a square puzzle of 4 x 4, we change diagonally the digits written in symbols to , to , to and to . Find sum of digits of the vertical(RBSESolutions.com)column and horizontal total of squares.

Solution

Sum of horizontal and vertical squares is 34. Yes this is a 4 × 4 square puzzle.

Question 14.

If 2 to 17 numbers are(RBSESolutions.com)entered in a 4 × 4 square puzzle, what will be the sum?

Solution

A, 4 × 4 square puzzle of numbers 2 to 17 in as under

Sum of horizontal, vertical and diagonal column is 38

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