sum of the digits of a two digit no. is 9. the no. obtained by interchanging the digits exceeds the given number by 27. find the given number.
Answers
Answer:36 and 63
Given:
Sum of two digits number=9
Number obtained after interchanging exceed by 27
To find:
The numbers
Solution:-
Let the tens digital number be 10x and unit digit number be y
Original number=10x+y
After interchanging It will be 10y+x
Sum of two digit is 9
I.e,x+y=9
y=9-x
Now according to the question
10x+y=10y+x+27
As y=9-x
10x+9-x=10(9-x)+x+27
10x+9-x=90-10x+x+27
9x+9=117-9x
18x=108
x=108/18
x=6
•Y=9-x
=9-6
=3
----------
The numbers are
10x+y=10×6+3
=63
10y+x=10×3+6
=36
Answer :
The numbers are 36 and 63
____________________________
To find :-
The numbers .
____________________________
Solution :-
Let, the first number at unit place is x tens place is y
_____________________[Assume]
• So, original number is 10y + x .
After interchanging digits number will be 10x + y.
Number obtained by interchanging the digits exceeds the given number by 27.
________________________[Given]
According to given condition,
10y + x + 27 = 10x + y
=> 10y - y + x - 10x + 27 = 0
=> 9y - 9x + 27 = 0
=> y - x + 3 = 0 ..... (I) ......[Divided by 9]
Also,
The sum of two digits is 9
i.e., x + y = 9
=> y = 9 - x ......(II)
put the value of y in eqn (I), we get
(9 - x) - x + 3 = 0
=> -2x + 12 = 0
=> 2x = 12
=> x = 6
Now, putting the value of x in eqn (II),
=> y = 9 - 6
=> y = 3
Hence, the original number is 10y + x
= 10(3) + 6
= 30 + 6
= 36
&
Interchanged number is 10x + y
= 10(6) + 3
= 60 + 3
= 63
_______________________________
.°. The numbers are 36 and 63.
_______________________[Answer]