Question

Q..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

Answer:

x + y = 11

x = 11 - y

10y + x = 10x + y + 27

9y -9x = 27

y- x = 3

y - 11 + y = 3

2y = 14

y = 7

x = 4

The number is 47.

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Answer 2

S O L U T I O N :

Let the ten's digit be r & let the one's digit be m respectively.

Where as,

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

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

A/q

$\underbrace{\bf{1^{st}\:Case\::}}$

$\mapsto\tt{r+ m=11}$

$\mapsto\tt{r=11-m.............(1)}$

$\underbrace{\bf{2^{nd}\:Case\::}}$

$\mapsto\tt{10m + r = 10r+ m + 27}$

$\mapsto\tt{10m - m + r -10r =27}$

$\mapsto\tt{9m -9r =27}$

$\mapsto\tt{9(m-r) =27}$

$\mapsto\tt{m-r =\cancel{27/9}}$

$\mapsto\tt{m-r =3}$

$\mapsto\tt{m-(11-m) =3\:\:\:[from(1)]}$

$\mapsto\tt{m-11+m =3}$

$\mapsto\tt{2m =3+11}$

$\mapsto\tt{2m =14}$

$\mapsto\tt{m =\cancel{14/2}}$

$\mapsto\bf{m = 7}$

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

$\mapsto\tt{r = 11-7}$

$\mapsto\bf{r = 4}$

Thus,

The number = 10r + m

The number = 10(4) + 7

The number = 40 + 7

The number = 47 .