Question

Sum of digits of a two digit number is
12. When we interchange the digits, it is
found that the resulting new number is
smaller than the original number by 36.
What is the original number?​

Answer 1

Let the tens digit of the required number be x and the once (unit) digit be y.

Then ,

Required number = 10x + y.

Also,

New number after interchanging the digits = 10y + x.

Now,

According to the question;

The sum of the digits of the required two-digits number is 12.

ie;

=> x + y = 12

=> y = 12 - x --------(1)

Also,

It is given that;

When the digits are interchanged, then the resulting new number is smaller than the original number (required number) by 36.

Thus;

=> 10y + x = 10x + y - 36

=> 10x - x + y - 10y - 36 = 0

=> 9x - 9y - 36 = 0

=> 9(x - y - 4) = 0

=> x - y - 4 = 0

=> x - (12 - x) - 4 = 0. {using eq-(1)}

=> x - 12 + x - 4 = 0

=> 2x - 16 = 0

=> 2x = 16

=> x = 16/2

=> x = 8

Now,

Putting x = 8 in eq-(1) , we get;

=> y = 12 - x

=> y = 12 - 8

=> y = 4

Thus,

Required number = 10x + y

= 10•8 + 4

= 80 + 4

= 84

Hence,

The required number is 84.

Answer 2

$\bold{\underline{\underline{Answer:}}}$

Original Number = 84

$\bold{\underline{\underline{Step\:-\:by\:-\:step\:explanation:}}}$

Given :

• Sum of digits of a two digit number is12.
• On interchange the digits, it is found that the resulting new number is smaller than the original number by 36.

To find :

• The original number

Solution :

Let the digit in the tens place be x.

Let the digit in the units place be y.

•°• Original Number = 10x + y

$\bold{\underline{\underline{As\:per\:the\:first\:condition:}}}$

• Sum of digits of a two digit number is 12.

Constituting it mathematically,

$\rightarrow$ $\bold{x+y=12}$ ---> (1)

$\bold{\underline{\underline{As\:per\:the\:second\:condition:}}}$

• The resulting new number is smaller than the original number by 36, on interchanging.

Reversed Number = 10y + x

Constituting the condition mathematically,

$\rightarrow$ $\bold{10y+x=10x+y-36}$

$\rightarrow$ $\bold{10y-y=10x-x-36}$

$\rightarrow$ $\bold{9y=9x-36}$

$\rightarrow$ $\bold{9x-36=9y}$

$\rightarrow$ $\bold{9x-9y=36}$

$\rightarrow$ $\bold{9(x-y) =36}$

$\rightarrow$ $\bold{x-y ={\dfrac{36}{9}}}$

$\rightarrow$ $\bold{x-y=4}$ ---> (2)

Solve equation 1 and equation 2 simultaneously by elimination method.

Add equation 2 to equation 1,

x + y = 12

x + y = 12 x - y = 8

----------------

2x = 16

$\rightarrow$ $\bold{x={\dfrac{16}{2}}}$

$\rightarrow$ $\bold{x=8}$

Substitute x = 8 in equation 1,

$\rightarrow$ $\bold{x+y=12}$ ---> (1)

$\rightarrow$ $\bold{8+y=12}$

$\rightarrow$ $\bold{y=12-8}$

$\rightarrow$ $\bold{y=4}$

$\bold{\sf{\red{Tens\:digit\:x\:=\:8}}}$

$\bold{\sf{\blue{Units\:digit\:y\:=\:4}}}$

$\bold{\boxed{\sf{\red{Original\:Number\:=\:10x+y\:=10\times\:8+4\:=80+4=84}}}}$