ML Aggarwal Playing with Numbers Exe-5.1 Class 8 ICSE Ch-5 Maths Solutions. We Provide Step by Step Answer of Exe-5.1 Questions for Playing with Numbers as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.
ML Aggarwal Playing with Numbers Exe-5.1 Class 8 ICSE Maths Solutions
| Board | ICSE |
| Publications | Avichal Publishig Company (APC) |
| Subject | Maths |
| Class | 8th |
| Chapter-5 | Playing with Numbers |
| Writer | ML Aggarwal |
| Book Name | Understanding |
| Topics | Solution of Exe-5.1 Questions |
| Edition | 2023-2024 |
Playing with Numbers Exe-5.1
ML Aggarwal Class 8 ICSE Maths Solutions
Page-83
Question 1. Write the following numbers in the generalized form:
(i) 89
(ii) 207
(iii) 369
Answer :
(i) 89 = 8 × 10 + 9
(ii) 207 = 2 × 100 + 0 × 10 + 7 × 1
(iii) 369 = 3 × 100 + 6 × 10 + 9 × 1
Question 2. Write the quotient, when the sum of a 2-digit number 34 and number obtained by reversing the digits is divided by
(i) 11
(ii) sum of digits
Answer :
Sum of two-digit number 34 and the number
obtained by reversing the digit 43 = 34 + 43
= 77
(i) 77 ÷ 11 = 7
(ii) 77 ÷ (Sum of digit) = 77 ÷ (4 + 3)
= 77 ÷ 7
= 11
Playing with Numbers Exe-5.1
ML Aggarwal Class 8 ICSE Maths Solutions
Page-84
Question 3. Write the quotient when the difference of a 2-digit number 73 and number obtained by reversing the digits is divided by
(i) 9
(ii) the difference of digits.
Answer :
Difference of two digit number 73 and the number
obtained by reversing the digits is = 73 – 37 = 36
(i) divide by 9
So, 36 ÷ 9 = 4
(ii) a difference of digits.
36 ÷ (7 – 3) = 36 + 4
= 9
Question 4. Without actual calculation, write the quotient when the sum of a 3-digit number abc and the number obtained by changing the order of digits cyclically i.e. bca and cab is divided by
(i) 111
(ii) (a + b + c)
(iii) 37
(iv) 3
Answer :
Sum of 3-digit number abc and the number
obtained by changing the order of digits i.e. bca and cab.
∴ abc + bca + cab
= 100a + 10b + c + 1006 + 10c + a + 100c + 10a + b
= 111a + 111b + 111c = 111 (a + b + c)
(i) When divided by 111
111 (a + b + c) ÷ 111 = a + b + c
(ii) When divided by (a + b + c)
111 (a + b + c) ÷ (a + b + c) = 111
(iii) When divided by 37
111 (a + b + c) ÷ 37 = 3(a + b + c)
(iv) When divided by 3
111 (a + b + c) ÷ 3 = 37(a + b + c)
Question 5. Write the quotient when the difference of a 3-digit number 843 and number obtained by reversing the digits is divided by
(i) 99
(ii) 5
Answer :
Difference of 3-digit number 843 and the number
obtained by reversing the digit is 348
= 843 – 348 = 495
(i) Divided by 99, we get
495 ÷ 99 = 5
(ii) Divided by 5, we get
495 ÷ 5 = 99
Question 6. The sum of digits of a 2-digit number is 11. If the number obtained by reversing the digits is 9 less than the original number, find the number.
Answer :
Sum of two digit number = 11
Let unit’s digit be ‘x’
and tens digit be ‘y’,
then x + y = 11 … (i)
and number = x + 10y
Unit digit be ‘y’
and tens digit be ‘x’
and number = y + 10x + 9
y + 10x + 9 = x + 10y
10x + y – 10y – x = -9
9x – 9y = -9
x – y = -1 … (ii)
Adding (i) and (ii), we get
2x = 10
x = 10/2
= 5
∴ y = 1 + 5 = 6
By substituting the vales of x and y, we get
Number = x + 10y
= 5 + 10 × 6
= 5 + 60
= 65
∴ The number is 65.
Question 7. If the difference of a two-digit number and the number obtained by reversing the digits is 36, find the difference between the digits of the 2-digit number.
Answer :
Let us consider the unit digit be ‘x’
and tens digit be ‘y’
So, the number is = x + 10y
By reversing the digits
Unit digit be ‘y’
and tens digit be ‘x’
The number is y + 10x = 36
Now by equating both numbers,
x + 10y – y – 10x = 36
-9x + 9y = 36
(y – x) = 36/9
y – x = 4
∴ The difference between digits of the 2-digit number is y – x = 4.
Question 8. If the sum of two-digit number and number obtained by reversing the digits is 55, find the sum of the digits of the 2-digit number.
Answer :
Consider unit digit be ‘x’
and tens digit be ‘y’
So the number is x + 10y
By reversing the digits
Unit digit be ‘y’
and tens digit be ‘x’
The number is y + 10x = 55
Now by equating both numbers,
x + 10y + y + 10x = 55
11x + 11y = 55
11(x + y) = 55
x + y = 55/11
x + y = 5
∴ Difference of the digits of the number is 5.
Question 9. The middle digit of three digit number is 0 and the some of the other two digits is 11. if the number obtained by reversing the digits exceeds the original number by 495, find the number.
Answer :
Let the number in 100s digit be x,
and the number in 1s digit be y,
Then,the number would be x0y : 100x+y
Condition 1: Sum of the digits : x+y=11…..(i)
If it is reversed ,it would be y0x : 100y+x,
Given The difference between them: 100x+y -(100y+x) = -495
100x+y-100y-x =-495
99x-99y = -495 (divided by 99)
x-y=-5…..(ii)
Adding (i) and (ii), we get,
(x+y)+(x-y)=11+(-5)
x+y+x-y=11-5
2x=6
∴ x=3
Substituting value of x in (i),
x+y=11
3+y=11
y=11-3
∴ y=8
∴The number is 308
Question 10. In a 3-digit number, unit’s digit, ten’s digit and hundred’s digit are in the ratio 1 : 2 : 3. If the difference of the original number and the number obtained by reversing the digits is 594, find the number.
Answer :
Ratio in the digits of a three digit number = 1 : 2 : 3
Let us consider unit digit be ‘x’
Tens digit be ‘2x’
and hundreds digit be ‘3x’
So the number is x + 10 × 2x + 100 × 3x
= x + 20x + 300x
= 321x
By reversing the digits,
Unit digit be ‘3x’
Ten’s digit be ‘2x’
Hundreds digit be ‘x’
So the number is 3x + 10 × 2x + 100 × x
= 3x + 20x + 100x
= 123x
According to the Question,
321x – 123x = 594
198x = 594
x = 594/198
= 3
∴ The number is = 321x
= 321 × 3 = 963
Question 11. In a 3-digit number, the unit’s digit is one more than the hundred’s digit and ten’s digit is one less than the hundred’s digit. If the sum of the original 3-digit number and numbers obtained by changing the order of digits cyclically is 2664, find the.
Answer :
Consider the hundreds digit be ‘x’
Unit digit be ‘x + 1’
and ten’s digit be ‘x – 1’
So the number = (x + 1) + 10(x – 1) + 100 × x
= x + 1 + 10x – 10 + 100x
= 111x – 9
By reversing the digits,
Unit digit be ‘x – 1’
Tens digit be ‘x’
Hundred digit be ‘x + 1’
So the number = x – 1 + 10x + 100x + 100
= 111x + 99
and sum of original 3-digit number = x + 10(x + 1) + 100(x – 1)
= x + 10x + 10 + 100x – 100
= 111x – 90
Now according to the condition,
111x – 9 + 111x + 99 + 111x – 90 = 2664
333x + 99 – 99 = 2664
333x = 2664
x = 2664/333
= 8
∴ The number = 111x – 9
= 111(8) – 9
= 888 – 9
= 879
— : End of ML Aggarwal Playing with Numbers Exe-5.1 Class 8 ICSE Maths Solutions :–
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