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 :–
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