Question

find the sum of a 2-digit number and the number obtained by reversing its digit if the sum of two digits of the number is 9.​

Answer 1

⠀⠀${ \huge \bf{ \mid{ \overline{ \underline{ \pink{QUESTION}}}} \mid}} </p><p>$

find the sum of a 2-digit number and the number obtained by reversing its digit if the sum of two digits of the number is 9.

⠀⠀⠀⠀${ \huge{ \mathbf{ \overbrace{ \underbrace{ \purple{ANSWER}}}}}}$

$\large\underline{ \underline{ \green{ \bold{solution}}}} = >$

let the two digits be x and y.

• tens digit=y
• ones digit=x

$\large\underline{ \underline{ \green{ \bold {given}}}} = >$

⠀⠀⠀⠀${ \boxed{ \sf{x + y = 9}}}$

so,

the number will be

⠀⠀⠀⠀${ \boxed{ \sf{10y + x}}}$

so,

sum of the numbers

⠀⠀⠀⠀⟹10x+y+10y+x=0

⠀⠀⠀⠀⟹11x+11y=0

⠀⠀⠀⠀⟹11(x+y)

we know that sum of x and y =9

so,

⠀⠀⠀⠀⟹11(9)

⠀⠀⠀⠀=99

⠀⠀${ \large{ \red{ \fbox{number = 99}}}}$