Question

sum of the digit of a two digit number is 11 the given number is less than the number obtained by interchanging the digits by 9 find the number​

Answer 1

Answer:

Two digit number: xy= 10x+y.

Sum of digits: x+y=11

xy is less then yx by 9: xy+9=yx

xy+9=yx

10x+y=10y+x-9

9x-9y=-9

x-y=-1,

So this is the system of equations to be solved:

x + y=11

x - y =-1

x=5, y=6

The number is 56.

Answer 2

Answer :-

$Let  \: x \:  be \: the \: unit \: digit \: of \: the \: number.$

$Then, \: the \: ten's \: digit = 11 - x$

$Therefore, \: Number \: = 10 \: (11 - x) + x$

$= 110 - 10x + x$

$= 110 - 9x$

When digits are interchanged ,

$unit \: digit=11−x  \\ and \: tens \: digit =x$

So, number obtained by interchanging the digits

$= 10x + (11 - x) \\ = 10x - x + 11 \\ = 9x + 11$

As per the question-

$9x + 11 - (110 - 9x) = 9$

$⇒9x + 11 - 110 + 9x = 9$

$⇒18x + 11 - 110 = 9$

$⇒18x = 9 - 11 + 110$

$⇒18x = 108$

$⇒x = \frac{108}{18} = 6$

Hence, unit's digit = 6 and ten's digit = ( 11 - 6 ) = 5

Therefore, the required number is 56.