Question

The sum of two digit no. Is 12 if the new no. Formed by reversing the digits Is greator than the orginal no. By 54 find the original no.

Answer 1

Correct Question :

• The sum of digits a two-digit number 12 . if the new number formed by reversing the digits is greater than the original number by 54 , find the original number .find the original number.

SoluTion :

$\sf{\blue{\underline{\underline{Given :}}}}$

• Sum of 2 digits of 2 digit number is 12 .
• The number formed by reversing the digits is 54 greater then the original number .

$\sf{\blue{\underline{\underline{To\:Find :}}}}$

• We're supposed to find the original number = ?

$\sf{\blue{\underline{\underline{AnswEr :}}}}$

• Let the original no. be 10x + y

⠀⠀⠀⠀

$:\implies$ x + y = 12 ⠀⠀—eqn(i).

⠀⠀⠀⠀

• Let the number obtain by reversing the digits be 10 y + x.

⠀⠀⠀⠀

According to the question,

⠀⠀⠀⠀⠀⠀⠀

$:\implies$ 10y + x - ( 10x + y ) = 54

$:\implies$ 10y + x - 10x - y = 54

⠀⠀⠀

$:\implies$ 9y - 9x = 54

$:\implies$ 9 ( y - x ) = 54

⠀⠀

$:\implies$ y - x = 6 ⠀⠀⠀—eqn(ii).

⠀⠀⠀⠀

• Adding eq ( i ) and ( ii )

⠀⠀⠀⠀

$:\implies$ x + y + y - x = 12 + 6

⠀⠀

$:\implies$ 2y = 18

⠀⠀

$:\implies$ y = 9

⠀⠀

• Substituting value of x in eqn ( i )

⠀⠀

$:\implies$ x + y = 12

⠀⠀

$:\implies$ x + 9 = 12

⠀⠀

$:\implies$ x = 12 - 9

⠀⠀⠀⠀

$:\implies$ x = 3

⠀⠀

• Finding the no.

⠀⠀

$:\implies$ 10x + y

⠀⠀

$:\implies$ 10 x 3 + 9

⠀⠀

$:\implies$ 30 + 9

⠀⠀

$:\implies$ 39

Therefore,

• Original number is 39.

Answer 2

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

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

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

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

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

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

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

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

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

• 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:-]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\tt{\implies y=3}$

$\sf\large{Hence,}$

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

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

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

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