# ML Aggarwal Playing with Numbers Exe-5.1 Class 8 ICSE Ch-5 Maths Solutions

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

(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

Sum of two-digit number 34 and the number

obtained by reversing the digit 43 = 34 + 43

= 77

= 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.

Difference of two digit number 73 and the number

obtained by reversing the digits is = 73 – 37 = 36

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

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

Difference of 3-digit number 843 and the number

obtained by reversing the digit is 348

= 843 – 348 = 495

495 ÷ 99 = 5

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.

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.

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.

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.

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.

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.

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 :–