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

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 :–

Return to –  ML Aggarwal Maths Solutions for ICSE Class -8

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