Question

The sum of the digits of a 2 digit number is 10. If the digits change their
places , the new number
obtained is 18 less than the original number. What is the original number
?​

Answer 1

Answer:

64

Step-by-step explanation:

let the tens digit = x and ones digit =y

ATQ

10x+ y is the real no.

x +y = 10~(1)

if the digits interchange their positions then

10y + x = ( 10x + y ) -18

9x - 9y = 18

x - y = 2~(2)

adding 1 and 2

2x = 12

x=6

now y = 4

so the real no. is 64

Answer 2

Answer:

Step-by-step explanation:x+y=10

y=10-x

10y+x+18=10x+y

10(10-x)+x+18=10x+10-x

100-9x+18-10=9x

108=9x+9x

108=18x

x=6

y=10-x

y=4

original no=10x+y

10*6+4

64