Question

sum of digits of two digit number is 11 the given number is less than number obtained by interchanging digits by 9 find number​

Answer 1

AnswEr:

$\mathsf{Let\:us\: Consider\:that\:the\: Digit\; Ten's\:place\;be\:x}$

$\sf{And,\:Ones\: place\:y}$

$\sf{Two\: Digit\; Number\:is:-}$

$:\implies\sf\: 10x + y$

$\sf{Sum\:of\; Digits\:two\: Number\;is\;11\:\red{[Given]}}$

$\bigstar\bold{\underline{\sf{According\:to\: Question\: Now,}}}$

$:\implies\sf\: x + y = 11$ ---------[Equation 1]

$:\implies\sf\: xy\:is \:less\:than\; yx\:by\:9$

$:\implies\sf\: xy + 9 = yx$

$:\implies\sf\: 10x + y = 10y + x - 9$

$:\implies\sf\: 9x - 9y =-9$

$:\implies\sf\: x - y = -1$ ----------[Equation 2]

$\rule{120}2$

$\:\:\:\:\;\:\:\:\:\:\:\dag\:\footnotesize\bold{\underline{\underline{\sf{\purple{[From \: Equations\: (1) \:\&(2)]}}}}}$

$:\implies\sf\: x \: \cancel{+ y} = 11$

$:\implies\sf\: x \: \cancel {- y} = -1$

$:\implies\sf\: 2x = 10$

$:\implies\sf\: x = \dfrac{10}{2}$

$:\implies\small{\underline{\boxed{\sf{\red{x\:=\:5}}}}}$

$\sf{Substituting\:the\:Value\:of\:x\; in\: Equation\:(2)}$

$:\implies\sf\: x - y = - 1$

$:\implies\sf\: 5 - y = - 1$

$:\implies\sf\: y = 5 + 1$

$:\implies\small{\underline{\boxed{\sf{\red{y\:=\:6}}}}}$

$\sf{Now,\; Finding\: Number:-}$

$:\implies\sf\: 10 x + y$

$:\implies\sf\: 10(5)+ 6$

$:\implies\sf\: 50 + 6$

$:\implies\large\boxed{\sf{\purple{56}}}$

$\bold{\underline{\sf{Hence,\:The\; \: Original \: Number\: is\: 56.}}}$

Answer 2

Solution :

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

Sum of digits of two number is 11, the given number is less than number obtained by interchange digits by 9.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The number.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

Let the ten's digit number be r

Let the one's digit number be m

$\underline{\sf{The\:Original\:number=10r+m}}}}\\\underline{\sf{The\:Reversed\:number=10m+r}}}}$

So;

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

A/q

$\mapsto\sf{10r+m=10m+r-9}\\\\\mapsto\sf{10r-r+m-10m=-9}\\\\\mapsto\sf{9r-9m=-9}\\\\\mapsto\sf{9(r-m)=-9}\\\\\mapsto\sf{r-m=\cancel{\dfrac{-9}{9} }}\\\\\mapsto\sf{r-m=-1}\\\\\mapsto\sf{11-m-m=-1\:\:\:\:\:\:[from(1)]}\\\\\mapsto\sf{11-2m=-1}\\\\\mapsto\sf{-2m=-1-11}\\\\\mapsto\sf{-2m=-12}\\\\\mapsto\sf{m=\cancel{\dfrac{-12}{-2} }}\\\\\mapsto\sf{\red{m=6}}$

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

$\mapsto\sf{r=11-6}\\\\\mapsto\sf{\red{r=5}}$

Thus;

The original number = 10r + m

The original number = 10(5) + 6

The original number = 50 + 6 = 56