Question

if the sum of two digit number obtain by reversing the order of its digit is 99.if the digit differ by 3.​

Answer 1

Answer:

Step-by-step explanation:

let the digits are x and y

If  xy then= 10x + y

e.g. 32= 10×3+2

so.. reversing of xy is yx and

yx=10y+x

Now..

A/Q : x-y=3 eq1

10x+ y+ 10y+x = 99 eq 2

so from eq 2

10x+y+10y+x= 99

11x+11y=99

x+y=9 ...... eq 3

Now

from eq 1

we have

x-y=3

x=3+y

sub value of x in eq 3

x+y=9

3+y+y=9

2y=6

y=3

Substituting value of y in eq1

x-y=3

x-3=3

x=6

so the no. is 36 or 63

Answer 2

Step-by-step explanation:

let the ones place digit be y and tens place digit be x

and the number be of the form 10x +y

A.T.Q

10y + x + 10x +y= 99_______(1)

11x +11y = 99

x + y = 9

and given that,

x - y = 3

so,

x = 3+y_______-(2)

adding eq. (1) and (2)

we have

2x = 12

x = 6

so , y = 3

so number becomes 63