Question

the sum of digit of a two digit number is 11?if the number obtained by reversing the digit is 9 less than the original number. find the number​

Answer 1

Given :

• The sum of digit of a two digit number is 11.
• The number obtained by reversing the digit is 9 less than original number.

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 the digit at tens place and at the units place is 11.

Equation :

$\longrightarrow$ $\sf{x+y=11}$

$\sf{x=11-y\:\:\:(1)}$

Case 2 :

The number when reversed is 9 less than the original number.

Reversed Number = 10y + x.

Equation :

$\longrightarrow$ $\sf{10y+x=10x+y-9}$

$\longrightarrow$ $\sf{10y-y=10x-x-9}$

$\longrightarrow$ $\sf{9y=9x-9}$

$\longrightarrow$ $\sf{9x-9=9y}$

$\longrightarrow$ $\sf{9x-9y=9}$

From equation (1), x = 11-y,

$\longrightarrow$ $\sf{9(11-y) -9y=9}$

$\longrightarrow$ $\sf{99-9y-9y=9}$

$\longrightarrow$ $\sf{-18y=9-99}$

$\longrightarrow$ $\sf{-18y=-90}$

$\longrightarrow$ $\sf{18y=90}$

$\longrightarrow$ $\sf{y=\dfrac{90}{18}}$

$\longrightarrow$ $\sf{y=5}$

Substitute, y = 5 in equation (1),

$\longrightarrow$ $\sf{x=11-y}$

$\longrightarrow$ $\sf{x=11-5}$

$\longrightarrow$ $\sf{x=6}$

$\large{\boxed{\bold{Ten's\:digit\:=\:x\:=\:6}}}$

$\large{\boxed{\bold{Unit's\:digit\:=\:y\:=\:5}}}$

$\large{\boxed{\bold{Original\:Number\:=\:10x+y\:=\:10(6)+5=60+5=65}}}$

Answer 2

Given :

• Sum of digit of a two digit number = 11

• Number obtained by reversing the digit is 9 less than the original number

To Find :

• The number

Solution :

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

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

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

Here, It is given that :

$\sf \longrightarrow \: x + y = 11...(i)$

According to the Question :

$\longrightarrow$ $\sf{10y+x=10x+y-9}$

$\longrightarrow$ $\sf{10y-y=10x-x-9}$

$\longrightarrow$ $\sf{9y=9x-9}$

$\longrightarrow$ $\sf{9x-9=9y}$

$\longrightarrow$ $\sf{9x-9y=9}$

From equation (1), x = 11-y,

$\longrightarrow$ $\sf{9(11-y) -9y=9}$

$\longrightarrow$ $\sf{99-9y-9y=9}$

$\longrightarrow$ $\sf{-18y=9-99}$

$\longrightarrow$ $\sf{-18y=-90}$

$\longrightarrow$ $\sf{18y=90}$

$\longrightarrow$ $\sf{y=\cancel\dfrac{90}{18}}$

$\sf \longrightarrow \: y = \boxed{ \red{ \sf \: 5}}$

Putting the value of y = 5 in equation i)

$\sf \longrightarrow \: x + y = 11$

$\sf \longrightarrow \: x + 5 = 11$

$\sf \longrightarrow \: x = 11 - 5$

$\sf \longrightarrow \: x = \boxed{ \red{ \sf \: 6}}$

Now Original Number :

$\star \sf \: Original \: Number = 10(6 + 5) = \huge \boxed{ \purple{ \sf \: 65}}$