The sum of the digits of a 2 number is 12 . when the digits are reversed the reversed numbers are 54 greater than the original number . find the numbers
Answers
Take the digits to be x and y
10x + y will be the number (x in tens place and y in units place)
sum of digits=
x + y = 12 ...... (i)
Digits reversed = 10y + x (y now in tens place and x in units place)
That is 54 greater than the original number(10x+y)
∴ 10y+x = 10x+y + 54
9y - 9x = 54, and dividing the equation by 9 we get
y - x = 6 ....... (ii)
On solving equations (i) and (ii) (i.e., adding them) we get y = 9 and x = 3.
Hence, the number is 39.
let the unit pace digit be x
and the ten's place digit be y
original no. = 10y + x
In case 1st
x + y = 12 equation no. 1st
In case 2nd
10y + x = 10x + y + 54
10y - y +x -10x = 54
9y-9x = 54 divided by 9 on both side
y - x = 6
this equation can be written
-x +y = 6 equation no. 2nd
by elimination method
from equation 1st and 2nd
x + y = 12
-x + y = 6 by addition
_____________
2y = 18
y = 18/2
y = 9
put the value of y in equation 1st
x + y = 12
x + 9 = 12
x = 12 - 9
x = 3
put the value of x and y to find the original no.
= 10 y + x
= 10×9 + 3
= 93
original no. is 93
Hope! it will be helpful for you