RBSE Solutions for Class 8 Maths Chapter 4 Mental Exercises Ex 4.2 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 4 Mental Exercises Exercise 4.2.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Mental Exercises |

Exercise |
Exercise 4.2 |

Number of Questions |
2 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 8 Maths Chapter 4 Mental Exercises Exercise 4.2

Question 1.

Find the value of alphabets and show the causes of that process.

Solution

(i) ∵ 2 is obtained from A + 4

∴ A = 8

8 + 4 = 12 = 10 + 2

(ii) ∵ 3 is obtained from A + 9

∴ A = 4

4 + 9 = 13 – 10 + 3

(iii) In the column of tens, 6 is obtained from A + B

Therefore, 3 is added in 3.

3 is obtained(RBSESolutions.com)from B + 7

so B should equal to 6

But

therefore in the column of tens

1 + 3 + 3 = 7 when 6 is given

∴ B = 5

∴ A = 2

Now sum of tens place digit

= 1 + 2 + 3 = 6 which is given

∴ A = 2

B = 5

(iv) B + 1 = 8

⇒ B = 8 – 1 = 7

A + B = 9

⇒ A + 7 = 9

⇒ A = 9 – 7 = 2

5 + A = B

⇒ 5 + 2 = B

⇒ B = 7

(v) zero (0) is obtained from 2 + A

A = 8

A + B = 9

⇒ 8 + B = 9

⇒ B = 1

(vi)

Value of A is to be find where on multiplying A with A there(RBSESolutions.com)must be A at unit place in product which may be 6

∵ 6 x 6 = 36

∴ A = 6

(vii)

It is true for B = 5 and A = 2

∴ A = 2, B = 5, C = 1

(viii)

∵ B x 6 = B

It is true for B = 4

4 x 6 = 24

A x 6 + 2 = 4

It is true for A = 7

∴ A = 7, B = 4

Question 2.

Find the value of x or (*) in following questions

Solution

(i) 7 is obtained from * + 8 + 5

∴ * = 4

4 + 8 + 5 = 17

10 + 7

↓

1 Tens

6 is obtained from 1 + 2 + * + 9

∴ * = 4

Now, 1 + 2 + 4 + 9 = 16

(ii) Zero (0) is obtained(RBSESolutions.com)from 5 + 2 + *

∴ * = 3

5 + 2 + 3 = 10 = 1 Tens

1 + 0 + 1 + 8 = 10 = 1 Hundred

1 + 9 + * + 8 = 21

⇒ * + 18 = 21

⇒ * = 21 – 18 = 3

(iii) 3 – 1 = 2

3 is obtained from * – 8

∴ * = 1

(iv) 8 is obtained from 7 – *

∴ * – 9

(v) x × 68 = 408

\(x=\frac { 408 }{ 68 }=6\)

(vi) 3x × 763 = 25942

(30 + x) × 763 = 25942

⇒ 22890 + 763x = 25942

⇒ 763x = 25942 – 22890

= 3052

\(x=\frac { 3052 }{ 763 }=4\)

(vii) We know that

2x × 8 + 0 = 216

⇒ (20 + x) × 8 = 216

⇒ 160 + 8x = 216

⇒ 8x = 216 – 160 = 56

\(x=\frac { 56 }{ 8 }=7\)

(viii) Dividend = Divisor x Quotient + Remainder

x7 × 24 + 19 = 907

⇒ (10x + 7) × 24 + 19 = 907

⇒ 240x + 168 + 19 = 907

⇒ 240x + 187 = 907

⇒ 240x = 907 – 187 = 720

\(x=\frac { 720 }{ 240 }=3\)

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