sum of the digits of a two-digit number is 5 when the digits are reversed the resulting number is greater than the original number by 27 find the number
Share
sum of the digits of a two-digit number is 5 when the digits are reversed the resulting number is greater than the original number by 27 find the number
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Let the number be 10x+y
x+y = 5 …(1)
10x+y-(10y+x) = 27, or
10x-x-10y+y= 27, or
9x-9y = 27, or
x-y = 3 …(2)
Add (1) and (2)
2x= 8, or
x = 4 and y = 1
So the number is 41 and its subsidiary is 14 and the difference between them is 27 and the sum of the digits is 5
[tex]\sf{\bold{\green{\underline{\underline{Given}}}}} [/tex]
⠀⠀⠀⠀
______________________
[tex]\sf{\bold{\green{\underline{\underline{To\:Find}}}}} [/tex]
⠀⠀⠀⠀
______________________
[tex]\sf{\bold{\green{\underline{\underline{Solution}}}}} [/tex]
⠀⠀⠀⠀
Let the original number be 10x + y
⠀⠀⠀⠀
Acc. to the first condition :-
⠀⠀⠀⠀
X + Y = 5 --- ( i )
⠀⠀⠀⠀
Acc. to the second condition :-
⠀⠀⠀⠀
Reversed number = 10y + x
⠀⠀⠀⠀
10x + y + 27 = 10y + x
⠀⠀⠀⠀
10y - y - 10x + x = 27
⠀⠀⠀⠀
9y - 9x = 27
⠀⠀⠀⠀
9 ( y - x ) = 27
⠀⠀⠀⠀
y - x = 27 / 9
⠀⠀⠀⠀
y - x = 3 ---- ( ii )
⠀⠀⠀⠀
Adding eq ( i ) and ( ii )
⠀⠀⠀⠀
x + y + y - x = 3 + 5
⠀⠀⠀⠀
y + y = 3 + 5
⠀⠀⠀⠀
2y = 3 + 5
⠀⠀⠀⠀
2y = 8
⠀⠀⠀⠀
y = 8 / 2
⠀⠀⠀⠀
y = 4
⠀⠀⠀⠀
⠀⠀⠀⠀
x + y = 5
⠀⠀⠀⠀
x + 4 = 5
⠀⠀⠀⠀
x = 5 - 4
⠀⠀⠀⠀
x = 1
⠀⠀
⠀⠀⠀⠀
Original no. = 10x + y
⠀⠀⠀⠀
Original no. = 10 × 1 + 4
⠀⠀⠀⠀
Original no. = 10 + 4
⠀⠀⠀⠀
Original no. = 14
______________________
[tex]\sf{\bold{\green{\underline{\underline{Answer}}}}} [/tex]
⠀⠀⠀⠀