Sum of digits of a two digit number is 8.when we interchange the digit, it is found that the resulting new number is greater than the original number by 18. What is the original number?
Answers
Answer:
let the digit in unit place be y
and that is in tens place be x
the original number is 10x + y
the number obtained by interchanging digits is 10y + x
from first condition,
x + y = 8 .......eqn 1
from second condition,
10x + y = 10y + x -18
18 = 10y - y + x-10x
18 = 9y-9x
-x+y = 2..........eqn2
adding equation first and equation second
x + y = 8
- x + y = 2
2y=10
y=5
put y is equal to 5 in equation 1
by by putting we get value of x is 3
there fore original number is 53
Step-by-step explanation:
Given :-
- Sum of the digit of a two digit number is 8 when we interchange the digits it is found that the resulting new number is greater than the original number by 18.
To find :-
- Two digit number
Solution :-
Let the tens digit be x then ones digit be y
According to the first condition
Sum of the digit of a two digit number is 8
Original number = 10x + y
→ x + y = 8
According to second condition
When we interchange the digits it is found that the resulting new number is greater than the original number by 18.
Reveresd number = 10y + x
→ 10x + y + 18 = 10y + x
→ 10x - x + y - 10y = - 18
→ 9x - 9y = - 18
→ 9(x - y) = - 18
→ x - y = - 2
Add both the equations
→ x + y + x - y = 8 - 2
→ 2x = 6
→ x = 3
Put the value of x in equation (ii)
→ x - y = - 2
→ 3 - y = - 2
→ y = 3 + 2
→ y = 5
Hence,
Tens digit = x = 3
Ones digit = y = 5
Therefore,
Original number = 10x + y = 35
Reversed number = 10y + x = 53