Question

The sum of the digits of a 2-digit number is 11. The number obtained by
interchanging the digits exceeds the original number by 27. Find the number. ​

Answer 1

GivEn:

• The sum of the digits of a 2-digit number is 11.

• The number obtained by interchanging the digits exceeds the original number by 27.

To find:

• Find the number.

SoluTion:

Lets the digit of the number in one's place be x and the number in ten's place be y.

Therefore, the number is xy.

As per given Question,

The sum of the digits of a 2-digit number is 11.

$\implies$ x + y = 11

$\implies$ y = 11 - x⠀⠀⠀⠀⠀(1)

If one's and ten's place of the number is x and y respectively.

Therefore the number is,

$\implies$ 10y + x

Now, Put the value of x from eq(1) -

$\implies$ 10(11 - x) + x

$\implies$ 110 - 10x + x

$\implies$ 110 - 9x

★ After interchanging,

$\implies$ 10x + y

$\implies$ 10x + (11 - x)

$\implies$ 11 + 9x

━━━━━━━━━━━━━━━

The number obtained by interchanging the digits exceeds the original number by 27.

$\implies$ 11 + 9x = 110 - 9x + 27

$\implies$ 11 + 9x = 137 - 9x

$\implies$ 9x + 9x = 137 - 11

$\implies$ 18 = 126

$\implies\sf x = \cancel{ \dfrac{126}{18}}$

$\implies$ x = 7

★ Now, Put the value of x in eq(1) -

$\implies$ y = 11 - 7

$\implies$ y = 4

━━━━━━━━━━━━━━━

★ Put ths value of x and y in,

$\implies$ 10y + x

$\implies$ 10 × 4 + 7

$\implies$ 40 + 7

$\implies$ 47

$\therefore$ Hence, the required number is 47.

Answer 2

Solution :

Let the ten's place number be r & one's place number be m respectively;

$\boxed{\bf{Original\:number=10r+m}}}\\\boxed{\bf{Reversed\:number=10m+r}}}$

A/q

$\longrightarrow\sf{r+m=11}\\\\\longrightarrow\sf{r=11-m....................(1)}$

&

$\longrightarrow\sf{10m+r=10r+m+27}\\\\\longrightarrow\sf{10m-m+r-10r=27}\\\\\longrightarrow\sf{9m -9r=27}\\\\\longrightarrow\sf{9(m-r)=27}\\\\\longrightarrow\sf{m-r=\cancel{27/9}}\\\\\longrightarrow\sf{m-r=3}\\\\\longrightarrow\sf{m-(11-m) = 3\:\:[from(1)]}\\\\\longrightarrow\sf{m-11+m=3}\\\\\longrightarrow\sf{2m-11=3}\\\\\longrightarrow\sf{2m=3+11}\\\\\longrightarrow\sf{2m=14}\\\\\longrightarrow\sf{m=\cancel{14/2}}\\\\\longrightarrow\bf{m=7}$

∴Putting the value of m in equation (1),we get;

$\longrightarrow\sf{r=11-7}\\\\\longrightarrow\bf{r=4}$

Thus;

$\boxed{\sf{The\:number=10r+m = [10(4) + 7]=[40+7]=\boxed{\bf{47}}}}$