Question

a number of two digits is increased by 45 when the digit are interchanged. if the unit digit is three less than 5 times the ten digit,find the original number​

Answer 1

Answer:

5

Step-by-step explanation:

Let the digits of number be x and y such that the number is 10x+y

Number formed by interchanging the digits = 10y+x

If the digits of a two-digit number are interchanged, the number formed is greater than the the original number by 45

1oy+x = 10x+y +45

9y-9x = 45

y-x = 5

Answer 2

Answer:

27

Step-by-step explanation:

Let the number be N=10a+b , where 'a' is ten's digit and 'b' is unit digit

according to question :-

(10a+b)+45=10b+a

10a-a+b-10b= -45

9a-9b= -45

a-b= -5     . . . . . . .. . (1)

Now,

b=5(a)-3

b-5a=-3

5a-b=3    . . . . . .. . .  . .(2)

(2)-(1)

5a-b-a+b=3+5

4a=8

a=2    .. . . . . . .  put this in (1)

we get 2-b=-5

b=7

put in the number N

N=10(2)+7

N=27 is the original number