Question

plz help??????? !!!!?!??!​

Answer 1

$\sf\large\pink{\underbrace{Orginal\: number=39}}$

$\sf\large\underline\purple{Let:-}$

$\sf{\implies Ones\: digit=x}$

$\sf{\implies Tens\: digit=y}$

$\sf{\implies Orginal\: number=10y+x}$

$\sf{\implies Reversed\: number=10x+y}$

$\sf\large\underline\purple{To\: Find:-}$

$\sf{\implies Orginal\: number=?}$

$\sf\large\underline\purple{Solution:-}$

To calculate the original number,at first we have to set up equation with the help of given clue in the question then solve those equation:-]

$\sf\small\underline\green{Given\:in\:case\:(i):-}$

$\sf{\implies Sum\:of\:two\: digit=12}$

$\tt{\implies Sum\:_{(ones+tens)}=12}$

$\tt{\implies x+y=12-----(i)}$

$\sf\small\underline\green{Given\:in\:case\:(ii):-}$

$\sf{\implies Orginal\:_{(number)}+54=reversed\:_{(number)}}$

$\tt{\implies 10y+x+54=10x+y}$

$\tt{\implies 10x-x+y-10y=54}$

$\tt{\implies 9x-9y=54}$

$\tt{\implies Here,\:dividing\:by\:9\:on\:both\:side:}$

$\tt{\implies x-y=6-----(ii)}$

$\sf{\implies In\:eq\:(i)\:and\:(ii)\:adding\:here:}$

$\tt{\implies x+y=12}$

$\tt{\implies x-y=6}$

$\sf{\implies By\: solving\:we\:get\: here:}$

$\tt{\implies 2x=18\implies x=9}$

$\sf{\implies Putting\:the\: value\:of\:x=9\:in\:eq\:(i):}$

$\tt{\implies x+y=12}$

$\tt{\implies 9+y=12}$

$\tt{\implies y=12-9}$

$\tt{\implies y=3}$

$\sf\large{Hence,}$

$\tt{\implies Orginal\:_{(number)}=10y+x}$

$\tt{\implies Orginal\:_{(number)}=10*3+9}$

$\tt{\implies Orginal\:_{(number)}=30+9}$

$\bf{\implies Orginal\:_{(number)}=39}$

Answer 2

$\huge\boxed{\fcolorbox{purple}{ink}{Answer}}$

Let the tens digit is’ y ‘and unit digit is ‘x’

The number is 10y+x

Given x+y=12 ………………..1

10y+x-54=10x+y

-9x+9y=54

-x+y=6 ………2

solving 1 and 2, we get 2y=18

y=9

x+9=12

x=3

The original number is 10(9)+3=93

The reversed number is 39 [93–54=39]