RBSE Solutions for Class 8 Maths Chapter 4 Mental Exercises Ex 4.2

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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