Question

a number consists of two digits whose sum is 9.if 27 is subtracted from the number its digits are reversed find the number

Answer 1

Part 1
Initial number
let x be the digit in unit place
therefore the digit in ten's place = 9 - x (* 10)= 90 - 10x

Part 2
final number the digits are reversed,
in unit place = 9 - x
in ten's place = x (* 10)= 10x

Given,
90-10x + x(sum of digits)=9
90 -9x = 9
90 - 9x - 27=10 + 9 - x
90 - 9x - 27=9x + 9
(transpose)
90 - 27 - 9 = 9x + 9x
54= 18x
54/18 = x
3= x

Part 1
initial number unit's place= x=3
ten's place = 9 - x
9-3=6
there fore 63

check:
63-27=36.

Answer 2

Answer:

36 or 63 can be the number

Step-by-step explanation:

Assuming

x as tens digit

y as ones digit

Their sum :

x + y = 9 ..... (i)

Number formed :

10x + y

Interchanging the digits :

10y + x

According to the question :

➡ (10x + y) - (10y + x) = 27

➡ 9x - 9y = 27

➡ 9(x - y) = 27

➡ x - y = 27/9

➡ x - y = 3 ..... (ii)

Subtracting both the equation :

$\bf \: x + y = 9 \\ { \underline{ \bf{x - y = 3}}} \\ \implies \bf \: 2x = 6 \\ \implies \bf \: x = 3$

Substituting the value of x in equation (i) :

➡ x + y = 9

➡ 3 + y = 9

➡ y = 6

Hence

The number can be 10x + y

or, 10(3) + 6

or, 36 either 63