RBSE Solutions for Class 8 Maths Chapter 4 Mental Exercises Additional Questions is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 4 Mental Exercises Additional Questions.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 8 |
| Subject | Maths |
| Chapter | Chapter 4 |
| Chapter Name | Mental Exercises |
| Exercise | Additional Questions |
| Number of Questions | 39 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 8 Maths Chapter 4 Mental Exercises Additional Questions
I. Objective Type Questions
Question 1.
The expanded form of 37 is –
(a) 37 = 3 + 7
(b) 37 = 10 × 3 + 7
(c) 37 = 7 × 10 + 3
(d) 37 = 37 + 3 + 7
Question 2.
A two digit number ab can be written as –
(a) 10a + b
(b) a + 10b
(c) 10 b + a
(d) a + b
Question 3.
ab + ba = 11 (a + b) is divisible by –
(a) 4
(b) 5
(c) 11
(d) ab
Question 4.
In ab – ba = 9 (a – b), if a = b, the which of the following is true?
(a) ab = 9
(b) ba = 9
(c) ab – ba = 9
(d) ab – ba = 0
Question 5.
Which of the following is divisible by 9?
(a) 11446487
(b) 21436587
(c) 3489
(d) 56789
Question 6.
In a three digit number 24x is divisible by 9, then the value of x will be –
(a) 3
(b) 4
(c) 5
(d) 6
Question 7.
Which of the following is divisible by 2?
(a) 4438
(b) 5791
(c) 233
(d) 147
Answers
1. (b)
2. (a)
3. (c)
4. (d)
5. (b)
6. (a)
7. (a).
II. Fill in the blanks
Question 1.
If ab is a two digit number, then ab = ___+ b
Question 2.
A number is divisible by ___ if it has even digit in it’s ones place.
Question 3.
If 1 is obtained as the remainder(RBSESolutions.com)in the division N ÷ 2, then the digit at the ones place of N is ____
Question 4.
If a three digit number(RBSESolutions.com)24 x is a multiple of 3, then value of x is ___
Question 5.
A number is divisible by____if it has zero(RBSESolutions.com)in its ones place.
Answer
1. 10 a
2. 2
3. 1, 3, 5, 7 or 9
4. 0, 3, 6 or 9
5. 10.
III. Very Short Answer Type Questions
Question 1.
If a two digit number ab can be written as 10a + b, then write down the value of ba.
Answer
10 b + a.
Question 2.
Write the expanded form of a three digit number abc.
Answer
100a +10b + c.
Question 3.
What is the divisibility test for 5?
Answer
A number which has(RBSESolutions.com)either 0 or 5 in it’s unit’s place is divisible by 5.
Question 4.
What is the divisibility test for 9?
Answer
If the sum of the digits of(RBSESolutions.com)a number is divisible by 9, then the number itself is divisible by 9.
Question 5.
In 99a – 99c, if a = c, then what will be the difference?
Answer
Zero.
IV. Short Answer Type Questions
Question 1.
If 31z5 is a multiple of 3, where z is a digit, what might be the values of z?
Solution
Since 31z5 is a multiple of 3.
∴ The sum of the digits (3 + 1 + z + 5) is a multiple of 3.
So, 9 + z is one(RBSESolutions.com)of these numbers 0, 3, 6, 9, 12, …..
When, 9 + z = 0 ⇒ z = – 9 (Negative)
When, 9 + z = 3 ⇒ z = 3 – 9 = -6 (Negative)
When, 9 + z = 6 ⇒ z = 6 – 9 = -3 (Negative)
When, 9 + z = 9 ⇒ z = 0
When, 9 + z = 12 ⇒ z = 3
When, 9 + z = 15 ⇒ z = 6
When, 9 + z = 18 ⇒ z = 9
When, 9 + z = 21 ⇒ z = 12 (Not Possible)
∴ The values of z are 0, 3, 6 or 9.
[∵ Negative values of z are not possible, and z is a digit.]
Questions.
Find the values of the letters in(RBSESolutions.com)each of the following and give reasons for the steps involved.
Question 2.
Solution
∵ A + 5 = 2
∴ A + 5 = 12 or 22 or 32 etc.
∴ A = 12 – 5 = 7 or 22 – 5 = 17
∴ A = 7 (∵ A = 17 is not possible)
∴
Hence, A = 7, B = 6 Ans.
Question 3.
Solution
∵ A + 8 = [1]3
⇒ A = 13 – 8 = 5
Also, 1 + 4 + 9 = [1]4
Obviously B = 4
and C = 1
Thus, A = 5, B = 4 and C = 1 Ans.
Question 4.
Solution
Since the ones digit of A x A = A, so it must be 1, 5 or 6.
∵ 1 x 1 = 1
5 x 5 = [2]5
6 x 6 = [3]6
Now, taking the value of ‘A’ as 1, 5 or 6. When : A = 1,
11 = 9A
When : A = 5
75 = 9A
When : A = 6
96 = 9A
Thus, A = 6 is correct answer.
Question 5.
If a three digit number 4y2 is multiple of 3, and then find all possible values of y where y is one digit number.
Solution
Three digit number = 4y2
Sum of digits = 4 + y + 2 = 6 + y
If the given number is multiple of 3 then 6 + y is divisible by 3.
∴ y = 0, 3, 6, 9
Question 6.
Solve the Cryptarithm : \(\overline { AB } \times \overline { AB } =\overline { ACB } \)
Solution
We have
\(\overline { AB } \times \overline { AB } =\overline { ACB } \) ……..(1)
Here the units digit of B x B is B.
Therefore B = 1 or B = 6
Again \(\overline { AB } \times \overline { AB } =\overline { ACB } \)
⇒ The square of a two digit number is a three digit number. So A can take values 1, 2 and 3.
We find that A = 1, B = 1 satisfies(RBSESolutions.com)equation (1).
For there values of A and B, we have 11 x 11 = 121
∴ C = 2
No other pairs of values of A and B satisfy equation (1).
Therefore A = 1, B = 1 and C = 2 Ans.
Question 7.
Test the divisibility of following numbers by 3 and 9.
(i) 1839
(ii) 236637
Solution
(i) 1839
(a) Divisibility of 3,
Sum of digits = 1 + 8 + 3 + 9 = 21
Which is divisible by 3.
∴ 1839 is(RBSESolutions.com)divisible by 3.
(b) Divisibility of 9.
Sum of digits = 1 + 8 + 3 + 9 = 21
Which is not divisible by 9.
∴ 1839 is not divisible by 9.
(ii) 236637
Divisibility of 3
Sum of digits = 2 + 3 + 6 + 6 + 3 + 7 = 27
Which is divisible by 3 and 9.
∴ 236637 is divisible(RBSESolutions.com)by 3 and 9.
Question 8.
Find A and B. Where A and B are digits.
Solution
∴ A = 9 and B = 7
Question 9.
If a 4-digit number 72×6 is divisible by 3. Find all values of x, where x is a digit.
Solution
Given number is 72×6
Sum of digits = 7 + 2 + x + 6
= 15 + x
(15 + x) is divisible by 3 (given)
This is possible if the value of (15 + x) = 15, 18, 21 or 24 where x is a digit.
15 + x = 15 ⇒ x = 15 – 15 = 0
15 + x = 18 ⇒ x = 18 – 15 = 3
15 + x = 21 ⇒ x = 21 – 15 = 6
15 + x = 24 ⇒ x = 24 – 15 = 9
∴ Value of x = 0, 3, 6 or 9
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