The number consisting of two digit …

Let the ten's digit of required number be x and its unit digit be y respectively.
Required number = 10x + y

According to the question, it is given that a number consisting of two digits is 7 times the sum of its digits.
{tex}\therefore{/tex} 10x + y = 7(x + y)
10x + y = 7x + 7y

10x - 7x - 7y + y = 0
3x -  6y = 0.............(i)

when 27 is subtracted from the number the digits are reversed.
After reversing the digits, the number = 10y + x
{tex}\therefore{/tex}(10x + y) - 27 = 10y + x
10x - x + y - 10y = 27
9x - 9y = 27
x - y = 3.............(ii)
Multiplying (i) by 1 and (ii) by 6, we get
3x -  6y = 0........(iii)
6x - 6y = 18.......(iv)
Subtracting (iii) from (iv), we get
3x = 18
{tex}x = \frac { 18 } { 3 } = 6{/tex}
Put the value of x = 6 in equation (i), we get
3 {tex}\times{/tex} 6 - 6y = 0
18 - 6y = 0
{tex}- 6 y = - 18 \Rightarrow y = \frac { - 18 } { - 6 } = 3{/tex}
Number = 10x + y
= 10 {tex}\times{/tex} 6 + 3
= 60 + 3
= 63
Hence the number is 63.