Playing with Numbers Class 8 RS Aggarwal Exe-4A Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-4. We provide step by step Solutions of council prescribe textbook / publication to develop skill and confidence. Visit official Website **CISCE** for detail information about ICSE Board Class-8 Mathematics.

## Playing with Numbers Class 8 RS Aggarwal Exe-4A Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-4

Board | ICSE |

Publications | Goyal Brothers Prakshan |

Subject | Maths |

Class | 8th |

writer | RS Aggarwal |

Book Name | Foundation |

Ch-4 | Playing with Numbers |

Exe-4A | Two and Three Digit Numbers |

Edition | 2024-2025 |

### Two and Three Digit Numbers

Playing with Numbers Class 8 RS Aggarwal Exe-4A Goyal Brothers Prakashan ICSE Foundation Maths Solutions Ch-4.

**Page- 58**

**Exercise- 4A**

Two and Three Digit Numbers

**Que-1: In a two-digit number the units digit is four times the tens and the sum of the digits is 10. Find the number.**

**Solution- **Let the unit digit be x and tens digit be y. So,

x = 4y ……………………(1) and

x+y =10 ……………………(2)

Substituting x= 4y in eqution (2) gives

4y +y = 10

or, 5y = 10

or, y = 2

So, x = 4y = 4 X 2 = 8

Hence, the number is 28. **Ans**.

**Que-2: The units digit of a two-digit number is 3 and seven times the sum of the digits is the number itself. Find the number.**

**Solution- **Unit digit of a two digit no = 3

Let digit at tens place be x

Two digit No. = 10 * tens place + 1 * ones place

= 10x+3

7(x + 3) =10x + 3

7x + 21 = 10x+3

21-3 = 10x-7x

x = 18/3

x = 6

Two digit no = 10*6+3

= 60+3

= 63 **Ans**.

**Que-3: In a two-digit number, the digit at the units place is thrice the digit in the tens place. The number exceeds the sum of its digits by 27. Find the number.**

**Solution- **Digit in the tens place = x then digit at unit place be 3x

number is = 10x+3x = 13x……. (1)

sum of digits = x+3x = 4x

given the number exceeds sum of it’s digits by 27.

Then, 13x = 4x+27

(13-4)x = 27

9x = 27

x = 3

Then the number become from (1)

13x = 13×3 = 39

The number is 39 **Ans.**

**Que-4: A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, its digits are reversed. Find the number.**

**Solution- **Let the digits be x and y.

Thus, number = 10x+y Now,

(10x+y) – 4(x+y) = 3…..(i) and

(10x+y) + 18 = 10y+x…..(ii)

Solving (i)

10x+y-4x-4y = 3

or,6x-3y = 3

or,2x-y = 1….(1) Solving (ii)

10x+y-10y-x = -18

or, 9y-9x = 18

or, y-x = 2…(2)

(1) + (2)

2x-y+y-x = 1+2

or, x = 3

Thus, y = x+2 (from (2))

y = 5

Therefore the number= 10x+y = 10×3+5 = 35 **Ans.**

**Que-5: The sum of a digits of a two-digit number is 13. The number obtained by interchanging its digits exceeds the given number by 9. Find the original number.**

**Solution- **Let the unit digit be ‘x’ and tens digit be ‘y’

So, x + y = 13 ……………(1)

The original number is 10y +x

if the digits are interchanged, the number becomes 10x+y

Now, 10x+y = 10y+x +9

or, 10x+y-10y-x = 9

or, 9x -9y = 9

or, 9(x-y) = 9

or, x – y = 1 ……………………(2)

Adding equtions (1) and (2) gives

2x = 14

or, x =7

So, y = 13-x = 13-7 =6

Hence, the original number = 67 **Ans**.

**Que-6: The difference between a 2-digit number and the number obtained by interchanging its digit is 63. What is the difference between the digits of the number ?**

**Solution- **Assume that the ten’s digit be x and one’s digit be y.

So, original number is 10x + y.

As per given condition, the difference between two digit number and the number obtained by interchanging it’s digits is 63.

As the original number is 10x + y. Then it’s interchange will be 10y + x.

Let’s say that original number is bigger then the smaller one.

Therefore,

Bigger number – Smaller number = 63

10x + y – (10y + x) = 63

10x + y – 10y – x = 63

9x – 9y = 63

Take 9 as common,

9(x – y) = 9(7)

x – y = 7 **Ans.**

**Que-7: In a 3-digit number, the hundreds digit is twice the tens digit while the units digit is thrice the tens digit. Also, the sum of its digits is 18. Find the number.**

**Solution- **ten digit number = x

hundred digit number = 2x

unit digit number = 3x

number is 100 × (2x) + 10 x + 3x

sum of digits = 2x + x + 3x = 18

6x = 18

x = 3

Number = 100 × (2×3) + 10×3 + 3 × 3 = 600+ 30 +9

= 639 **Ans.**

**— : End of Playing with Numbers Class 8 RS Aggarwal Exe-4A Goyal Brothers Prakashan ICSE Foundation Maths :–**

**Return to :- ICSE Class -8 RS Aggarwal Goyal Brothers Math Solutions**

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