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

We hope the given RBSE Solutions for Class 8 Maths Chapter 4 Mental Exercises Ex 4.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 4 Mental Exercises Exercise 4.1, drop a comment below and we will get back to you at the earliest.

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