RBSE Maths Chapter 4 – Mental Exercise Class 8 Important questions and solutions are provided here. The important questions and solutions of Chapter 4 available at BYJU’S contain step by step, along with concept-based explanations. The important questions which are given here are based on the new pattern prescribed by the RBSE. Students can access the RBSE Class 8 solutions for all the subjects here.
Chapter 4 of the RBSE Class 8 Maths will help the students to solve problems related to the expanded form of numbers, interchanging the two, three-digit numbers, rules of divisibility, to determine the eliminated digits in the four operations.
RBSE Maths Chapter 4: Exercise 4.1 Textbook Important Questions and Solutions
Question 1: If a 3 digit number 24z is divisible by 9, then find the value of z, where z is a digit.
Solution:
Sum of the digits = 2 + 4 + z = 6 + z.
The given number is 24z.
The sum of the digits must be divisible by 9.
This is possible if 6 + z is either 9 or 18.
If 6 + z = 9, then z = 9 – 6 = 3.
If 6 + z = 18, then z = 18 – 6 = 12.
Here, z is a digit, so z ≠ 12.
Hence, z = 3.
Question 2: If a 3 digit number 89m is divisible by 9, then find the value of m.
Solution:
Sum of digits = 8 + 9 + m = 17 + m
The given number is 89m.
The sum of the digits must be divisible by 9.
This is possible if the number is either of them: 18, 27, 36 . . . etc.
But m is a digit.
If 17 + m = 18, then m = 18 – 17 = 1.
If 17 + m = 27, then m = 27 – 17 =10.
But m is a digit, so m ≠ 10.
Hence, m = 1.
Question 3: 31P5 is a multiple of 9 and two values are obtained by P. Why? Where P is a digit.
Solution:
Sum of digits = 3 + 1 + P + 5 = 9 + P.
The given number is 31P5.
It is given that 9 + P is a multiple of 9 which is possible if 9 + P is either 9 or 10.
If 9 + P = 9, then P = 9 – 9 = 0
If 9 + P =18, then P = 18 – 9 = 9
Therefore, P has two values 0 and 9.
Question 4: If a 3-digit number 24q is a multiple of 3, then find the value of q.
Solution:
Sum of digits = 2 + 4 + q = 6 + q.
The given three digit number is 24q.
It is given that 6 + q is a multiple of 3, the possible values of 6 + q are 6, 9, 12, 15, etc.
If 6 + q = 6, then q = 6 – 6 = 0
If 6 + q = 9, then q = 9 – 6 = 3
If 6 + q = 12, then q – 12 – 6 = 6
If 6 + q = 15, then q = 15 – 6 = 9
If 6 + q = 18, then q = 18 – 6 = 12
But q is a digit, so q ≠ 12.
The values of q are 0, 3, 6, 9.
Question 5: Test the divisibility of the given numbers by 3, 9 and 11.
[i] 294
[ii] 4455
[iii] 1041966
Solution:
[i] 294To check the divisibility by 3.
Sum of digits = 2 + 9 + 4 = 15.
15 is divisible by 3.
Hence, the number 294 is also divisible by 3.
To check the divisibility by 9.
Sum of digits = 2 + 9 + 4 = 15
15 is not divisible by 9.
Hence, the number 294 is not divisible by 9
To check the divisibility by 11.
4 x 1 + 9 x (-1) + 2 x 1 = 4 – 9 + 2 = -3
So, 294 is not divisible by 11.
[ii] 4455To check the divisibility by 3.
Sum of digits = 4 + 4 + 5 + 5 = 18
18 is divisible by 3.
Hence, 4455 is also divisible by 3.
To check the divisibility by 9.
Sum of digits = 18 which is divisible by 9.
Hence, 4455 is also divisible by 9.
To check the divisibility by 11.
5 x 1 + 5 x (-1) + 4 x 1 + 4 x (-1) = 5 – 5+ 4 – 4 = 0.
Hence, 4455 is divisible by 11.
[iii] 1041966To check the divisibility by 3.
Sum of digits =1 + 0 + 4 + 1 + 9 + 6 + 6 = 27
27 is divisible by 3.
Hence, 1041966 is also divisible by 3.
To check the divisibility by 9.
27 is divisible by 9.
Hence, 1041966 is divisible by 9.
To check the 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
13 is not divisible by 11.
Hence, 1041966 is not divisible by 11.
Question 6: If R = 4 in 31R1, then by the rule of divisibility find that this number is divisible by 11.
Solution:
Given: R = 4.
Hence, the number is 3141.
To check the 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 divisible by 11.
Hence, 3141 is not divisible by 11.
Question 7: If 31R5 is a multiple of 3, then find the value of R, where R is a digit.
Solution:
Given Number = 31R5
Sum of digits = 3 + 1 + R + 5 = 9 + R
9 + R is a multiple of 3 so it is possible if values of 9 + R are 9, 12, 15, 18 etc.
If 9 + R is 9, then R = 0.
If 9 + R is 12, then R = 3.
If 9 + R is 15, then R = 6.
If 9 + R = 18, then R = 9.
If 9 + R = 21, then R = 12
Since R is a digit, R ≠ 12
Therefore, the values of R are 0, 3, 6 and 9.
RBSE Maths Chapter 4: Exercise 4.2 Textbook Important Questions and Solutions
Question 1: Find the values of the alphabets and show the causes of that process.
[i] 5A
+ 34
_____
B2
_____
[ii] 5A
+ 79
_____
CB3
_____
[iii] AB
+ 37
_____
6A
_____
[iv] 5AB
+ AB1
______
B98
______
[v] 12A
+ 6AB
______
A09
______
[vi] 1A
X A
______
9A
______
[vii] AB
X B
_______
CAB
_______
[viii] AB
X 6
________
BBB
________
Solution:
[i] 5A+ 34
_____
B2
_____
The sum of A and 4 is given as 2.
Hence, the value of A is 8.
8 + 4 = 12
If A = 8, then B = 5 + 3 + 1 [carryover] = 9
[ii] 5A+ 79
_____
CB3
_____
The sum of A and 9 is given as 3.
Hence, the value of A is 4.
4 + 9 = 13
If A = 4, then CB = 5 + 7 + 1[carryover] = 13.
[iii] AB+ 37
_____
6A
_____
The sum of A and 3 gives 6.
B + 7 gives A.
So, B = 6.
But 6 + 7 = 13.
In the tens column, 6 is given. So, B = 5.
If B = 5, then 5 + 7 = 12 and 2 in units place.
Therefore, A = 2.
The values of A and B are 2 and 5 respectively.
[iv] 5AB+ AB1
______
B98
______
The sum of B and 1 is 8.
B + 1 = 8.
B = 8 – 1 = 7.
A + B = 9
A + 7 = 9
A = 2
5 + A = B
5 + 2 = 7
The values of A and B are 2 and 7 respectively.
[v] 12A+ 6AB
______
A09
______
The sum of 2 and A is 0.
2 + A = 0
Then A = 8 as 2 + 8 = 10. 0 in the units place.
The sum of A and B is 9.
A + B = 9
8 + B = 9
B = 9 – 8 = 1
The values of A and B are 8 and 1 respectively.
[vi] 1AX A
______
9A
______
The multiplication of A and A gives A in the units place. The product of 6 and 6 gives 36, 6 in the units place.
So, A = 6.
16 x 6 = 96.
[vii] ABX B
_______
CAB
_______
When B is multiplied by B, the result is B. This holds when A = 2 and B = 5.
25 x 5 = 125.
[viii] ABX 6
________
BBB
________
When B is multiplied by 6, B is obtained.
It holds when B = 4.
4 x 6 = 24
A x 6 + 2(carry over) = 24
It is true when A = 7.
74 x 6 = 444.
The values of A and B are 7 and 4, respectively.
Question 2: Find the value of x or (*) in the questions given below.
[i] 2 *
+ * 8
+ 95
____
167
____
[ii] 905
+ *12
+ 88*
_____
2100
_____
[iii] 7*3
– 281
______
432
______
[iv] 57
– 3*
______
18
______
[v] 68
x (x)
______
408
______
[vi] 763
x (3x)
_______
25942
_______
[vii] 216 ÷ 2x = 0
Quotient = 8
[viii] 907 ÷ 7x = 19
Quotient = 24
Solution:
[i] 2 *+ * 8
+ 95
____
167
____
* + 8 + 5 = 7
* = 4
Since 7 is in the units place, 4 + 8 + 5 = 17 [1 is the carryover]
1 + 2 + * + 9 = 16
* = 4.
[ii] 905+ *12
+ 88*
_____
2100
_____
5 + 2 + * = 0
Since 0 is in the units place, * = 3
5 + 2 + 3 = 10
1 + 9 + * + 8 = 21
* = 3
[iii] 7*3– 281
______
432
______
* – 8 = 3
11 – 8 = 3
* = 1
[iv] 57– 3*
______
18
______
7 – * = 8
* = 9
[v] 68x (x)
______
408
______
(x) x 68 = 408
x = 408 / 68
x = 6
[vi] 763x (3x)
_______
25942
_______
763 x (3x) = 25942
(30 + x) * 763 = 25942
22890 + 763x = 25942
763x = 25942 – 22890
763x = 3052
x = 3052 / 763
x = 4
[vii] 216 ÷ 2x = 0Quotient = 8
2x * 8 + 0 = 216
(20 + x) * 8 + 0 = 216
160 + 8x = 216
8x = 216 – 160
8x = 56
x = 56 / 8
x = 7
[viii] 907 ÷ 7x = 19Quotient = 24
7x * 24 + 19 = 907
(7 + 10x) * 24 + 19 = 907
168 + 240x + 19 = 907
240x + 187 = 907
240x = 720
x = 720 / 240
x = 3
RBSE Maths Chapter 4: Additional Questions and Solutions
Question 1: Write the generalised form of the numbers given below.
[i] 10 x 5 + 6
[ii] 8 x 100 + 0 x 10 + 5
[iii] 9 x 100 + 9 x 10 + 9
Solution:
[i] 10 x 5 + 6= 5 x 10 + 6 x 1
= 50 + 6
= 56
[ii] 8 x 100 + 0 x 10 + 5= 8 x 100 + 0 x 10 + 5 x 1
= 800 + 0 + 5
= 805
[iii] 9 x 100 + 9 x 10 + 9= 9 x 100 + 9 x 10 + 9 x 1
= 900 + 90 + 9
= 999
Question 2: If 79y is divisible by 9, then is it possible that y will have more than one value?
Solution:
Sum of the digits = 7 + 9 + y = 16 + y
The given number is 79y.
It is divisible by 9.
If 16 + y is divisible by 9, then the possible values are 18, 27, 36, etc.
But y is a digit.
If 16 + y = 18, then y = 2
If 16 + y = 27, then y = 11
So, y = 2.
Question 3: Find the value of * in the following problems.
[i] 3 *
+ 57
+ 34
____
127
____
[ii] 56
+ 77
+ *3
_____
216
_____
[iii] 443
+ *57
+ 128
______
928
______
[iv] 82
+ 55
+ 99
______
*36
______
[v] 76
– 5*
_____
25
_____
[vi] 54
– 2*
______
28
______
[vii] 803
– 2*6
_______
567
_______
[viii] 782
– *73
_______
209
_______
[ix] 84
– *8
______
16
______
Solution:
[i] 3 *+ 57
+ 34
____
127
____
To get 7 in the units place, * = 6
* + 7 + 4 = 7
6 + 7 + 4 = 17
[ii] 56+ 77
+ *3
_____
216
_____
6 + 7 + 3 = 16 [6 in units place and 1 is the carryover]
1 + 5 + 7 + * = 21
13 + * = 21
* = 21 – 13
* = 8
[iii] 443+ *57
+ 128
______
928
______
3 + 7 + 8 = 18 [8 in units place and 1 carryover]
1 + 4 + 5 + 2 = 12 [2 in units place and 1 carryover]
1 + 4 + * + 1 = 9
6 + * = 9
* = 9 – 6
* = 3
[iv] 82+ 55
+ 99
______
*36
______
2 + 5 + 9 = 16 [6 in units place and 1 carryover]
1 + 8 + 5 + 9 = *
23 = *3
* = 2
[v] 76– 5*
_____
25
_____
6 – * = 5
* = 1
76 – 51 = 25
[vi] 54– 2*
______
28
______
14 – * = 8
* = 14 – 8
* = 6
14 – 6 = 8
[vii] 803– 2*6
_______
567
_______
Subtracting 3 from 6 isn’t possible. So, borrowing is done from the next digit which is 0.
After borrowing from 8, in the place of 0, it is 10 and in the place of 3, it is 13, in the place of 8 it is 7.
Hence, it becomes,
13 – 6 = 7
9 – * = 6
9 – 6 = *
* = 3
[viii] 782– *73
_______
209
_______
Subtracting 2 from 3 isn’t possible. So, borrowing is done from the next digit which is 8.
After borrowing from 8, in the place of 2, it is 12 and in the place of 8, it is 7.
Hence, it becomes,
12 – 3 = 9
7 – 7 = 0
7 – * = 2
* = 5
[ix] 84– *8
______
16
______
Subtracting 4 from 8 isn’t possible. So, borrowing is done from the next digit which is 8.
After borrowing from 8, in the place of 4, it is 14 and in the place of 8, it is 7.
Hence, it becomes,
14 – 8 = 6
7 – * = 1
* = 6
Question 4: Write a 3 x 3 and a 4 x 4 crossword puzzle by the following digits in such a way that the horizontal, vertical and diagonal squares are the same.
[i] 2, 3, 4, 5, 6, , 8, 9, 10
[ii] 2, 3, 4, 5, 6, , 8, 9, 10, 11, 12, 13, 14, 15, 16, 17
Solution:
[i] Digits along the horizontal row are: [1] 2 + 7 + 6 = 15 [2] 9 + 5 + 1 = 15 [3] 4 + 3 + 8 = 15Digits along the vertical columns are:
[1] 2 + 9 + 4 = 15 [2] 7 + 5 + 3 = 15 [3] 6 + 1 + 8 = 15Digits along the principal diagonal are: 2 + 5 + 8 = 15
Digits along the secondary diagonal are: 6 + 5 + 4 = 15
Total |
|||
2 |
7 |
6 |
15 |
9 |
5 |
1 |
15 |
4 |
3 |
8 |
15 |
15 |
15 |
15 |
15 |
Digits along the vertical columns are:
[1] 2 + 9 + 4 = 15 [2] 7 + 5 + 3 = 15 [3] 6 + 1 + 8 = 15Digits along the principal diagonal are: 2 + 5 + 8 = 15
Digits along the secondary diagonal are: 6 + 5 + 4 = 15
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
16 |
17 |
The sum of the horizontal, vertical and diagonal columns are the same = 38