Question

The sum of the digits of a two
digit number is 12. If the new number formed by reversing the
digits is greater than the orignal
number by 54,find the orignal
number.

I hope someone would answer it as quickly they can.

Pls don't answer this question more points, if you know then only answer it.

Thnxs​

Answer 1

Given :

1. The sum of the digits of a two digit number is 12.
2. When the digits are interchanged the new number formed is greater than original number by 54.

To Find :

• The original number

Solution :

Let the digit at the tens place be x.

Let the digit at the units place be y.

Original Number = (10x+y)

Case 1 :

The sum of digits at the tens place and units place is 12.

Equation :

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

$\sf{x=12-y\:\:\:\:\:(1)}$

Case 2 :

The digits when interchanged of the original number, the new number formed is greater than original number by 54.

Reversed Number = (10y+x)

Equation :

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

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

$\implies$ $\sf{9y=9x+54}$

$\implies$ $\sf{9y=9(12-y) +54}$

$\implies$ $\sf{9y=108-9y+54}$

$\implies$ $\sf{9y+9y=108+54}$

$\implies$ $\sf{18y=162}$

$\implies$ $\sf{y=\dfrac{162}{18}}$

$\implies$ $\sf{y=9}$

Substitute, y = 9 in equation (1),

$\implies$ $\sf{x=12-y}$

$\implies$ $\sf{x=12-9}$

$\implies$ $\sf{x=3}$

$\large{\boxed{\bold{Ten's\:digit\:=\:x\:=\:3}}}$

$\large{\boxed{\bold{Unit's\:digit\:=\:y\:=\:9}}}$

$\large{\boxed{\bold{\red{Original\:Number\:=\:10x+y=10(3)+9=30+9=39}}}}$

Answer 2

Given :

• Sum of digits of a two digit number is 12

• New number formed by reversing the digits is greater than the original number by 54

To Find :

• The Original Number

Solution :

Let the digit at tens place be x and digit at ones place be y

$\star \sf \: Original \: Number = 10x + y$

$\star \sf \: Interchanged\: Number = 10y + x$

Also,It is given that :

$\sf \mapsto \: x + y = 12$

$\sf \mapsto \: x = 12 - y...( i)$

According to the Question :

$\longrightarrow \sf \: 10y + x = 10x + y + 54$

$\longrightarrow \sf \: 10y - y = 10x - x + 54$

$\longrightarrow \sf \: 9y = 9x + 54$

Put the value of x = 12-y from equation i)

$\longrightarrow \sf \: 9y = 9(12 - y) + 54$

$\longrightarrow \sf \: 9y = 108 -9y+ 54$

$\longrightarrow \sf \: 9y + 9y = 162$

$\longrightarrow \sf \: 18y = 162$

$\longrightarrow \sf \: y = \ \cancel\dfrac{162}{18}$

$\longrightarrow \sf \red{y = 9}$

Put the value of y = 9 in equation i)

$\sf \mapsto \: x = 12 - y$

$\sf \longrightarrow\: x = 12 - 9$

$\longrightarrow \sf \red{x= 3}$

Hence,

$\dag \: \sf {\purple{Original \: Number = 10(3) + 9 = \boxed{ \sf \: 39}}}$