RBSE Solutions for Class 8 Maths Chapter 4 Mental Exercises Ex 4.1

RBSE Solutions for Class 8 Maths Chapter 4 Mental Exercises Ex 4.1 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 4 Mental Exercises Exercise 4.1.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 8
Subject	Maths
Chapter	Chapter 4
Chapter Name	Mental Exercises
Exercise	Exercise 4.1
Number of Questions	7
Category	RBSE Solutions

Rajasthan Board RBSE Class 8 Maths Chapter 4 Mental Exercises Exercise 4.1

Question 1.
If 3-digit number 24x is . divisible by 9 then find the value of x. Where x is a digit
Solution
Sum of the digit = 2 + 4 + x = 6 + x
Given Number is 24 x
Sum of the digit must(RBSESolutions.com)be divide by 9 but it
is possible if 6 + x is either 9 or 18
If 6 + x = 9
Then x = 9 – 6 = 3
If 6 + x = 18, then x = 18 – 6 = 12
Here, x is a digit, so x ≠ 12
Hence, x = 3

Question 2.
If 3-digit number 89y is divisible by 9 then find the value of y?
Solution
Given number is 89y
∴Sum of digits = 8 + 9 + y = 17 + y
But 17 + y is(RBSESolutions.com)divisible by 9, this is possible only number may be either 18, 27, 36…etc.
But y is a digit .
If 17 + y = 18, then y = 18 – 17 = 1
If 17 + y = 27, then y = 27 – 17 = 10
But y is a digit so y ≠ 10
therefore y = 1

Question 3.
31M5 is a multiple of 9 and two values are obtained by M. Why? Where M is a digit?
Solution
Given Numbers is 31M5
∴Sum of digits = 3 + 1 + M + 5 = 9 + M
But it is given that 9 + M is a multiple of 9 which is only possible if 9 + M is either 9 or 18.
If 9 + M = 9
Then M = 9 – 9 = 0
If 9 + M = 18
Then M = 18 – 9 = 9
Therefore, M has two values 0 and 9

Question 4.
If 3-digit number 24y is a multiple of 3, then find the value of y?
Solution
Given three digit number = 24y
∴Sum of digits = 2 + 4 + y = 6 + y
But it is given that 6 + y is a multiple of 3,
the possible(RBSESolutions.com)values of 6 + y are 6, 9, 12, 15, …. etc.
If 6 + y = 6, then y = 6 – 6 = 0
If 6 + y = 9, then y = 9 – 6 = 3
If 6 + y = 12, then y – 12 – 6 = 6
If 6 + y = 15, then y = 15 – 6 = 9
If 6 + y = 18, then y = 18 – 6 = 12
But y is a digit so y ≠ 12
therefore, the values of y are 0, 3, 6, 9

Question 5.
Test the divisibility of following numbers by 3, 9 and 11
(i) 294
(ii) 4455
(iii) 1041966
Solution
(i) Given Number = 294
Check the divisibility by 3
Sum of digits = 2 + 9 + 4 = 15
15 is divisible by 3 so number 294 is also divisible by 3
Check the divisibility by 9 .
Sum of digits = 2 + 9 + 4 = 15
∵Sum 15 is not divisible by 9
∴294 is not(RBSESolutions.com)divisible by 9
Check the divisibility by 11
4 x 1 + 9 x (- 1) + 2 x 1 = 4 – 9 + 2 = – 3
∴ 294 is not divisible by 11

(ii) Given Number = 4455
Check the divisibility by 3
Sum of digits = 4 + 4 + 5 + 5 = 18
18 is divisible by 3
∴4455 is divisible by 3
Check the divisibility by 9
Sum of digits 18 is divisible by 9
∴ 4455 is divisible by 9
Divisibility by 11
5 x 1 + 5 x (-1) + 4 x 1 + 4 x (-1)
5 – 5 + 4 – 4 = 0
∴ 4455 is divisible by 11

(iii) Given number = 1041966
Check the divisibility by 3
Sum of digits = 1 + 0 + 4 + 1 + 9 + 6 + 6 = 27
27 is divisible by 3
So,. 1041966 is(RBSESolutions.com)divisible by 3
Divisibility by 9
27 is divisible by 9
∴1041966 is divisible by 9
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
But 13 is not divisible by 11
So, 1041966 is not divisible by 11

Question 6.
If R = 4 in number 31R1 then by the rule of divisibility(RBSESolutions.com)find that this number is divisible by 11.
Solution
Given that R = 4
therefore the number becomes to 3141
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(RBSESolutions.com)divisible by 11
So, 3141 or 31R1 is not divisible by 11.

Question 7.
If 31P5 is a multiple of 3 then find the value of P, where P is a digit?
Solution
Given Number = 31P5
Sum of digits = 3 + 1 + P + 5 = 9 + P
But 9 + P is a multiple of 3 so it is only possible if values of 9 + P are 9, 12, 15, 18, ….etc.
If 9 + P = 9 then P = 0
If 9 + P = 12 then P = 3
If 9 + P = 15 then P = 6
If 9 + P = 18 then P = 9
If 9 + P = 21 then P = 12
But P is a digit so P ≠ 12
Therefore, values of P are 0, 3, 6 and 9

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