Question

The sum of the digits of the two-digit number is 6. If the digits are reversed
the new number is 36 more than and
the original number .find the number.

Answer 1

Solution :

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

The sum of the digits of the two - digit number is 6. If the digit's are reversed the new number is 36 more than the original number.

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

The number.

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

Let the ten's digit be r

Let the one's digit be m

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

A/q

$\longrightarrow\sf{r+m=6}\\\\\longrightarrow\sf{r=6-m....................(1)}$

&

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

So;

$\longrightarrow\sf{10r+m+36=10m+r}\\\\\longrightarrow\sf{10r-r+m-10m=-36}\\\\\longrightarrow\sf{9r-9m=-36}\\\\\longrightarrow\sf{9(r-m)=-36}\\\\\longrightarrow\sf{r-m=\cancel{\dfrac{-36}{9} }}\\\\\longrightarrow\sf{r-m=-4}\\\\\longrightarrow\sf{6-m-m=-4\:\:\:\:[from(1)]}\\\\\longrightarrow\sf{6-2m=-4}\\\\\longrightarrow\sf{-2m=-4-6}\\\\\longrightarrow\sf{-2m=-10}\\\\\longrightarrow\sf{m=\cancel{\dfrac{-10}{-2} }}\\\\\longrightarrow\sf{\red{m=5}}$

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

$\longrightarrow\sf{r=6-5}\\\\\longrightarrow\sf{\red{r=1}}$

Thus;

$\underbrace{\bf{The\:original\:number=10r+m=10(1)+5=10+5=15}}}}}$

Answer 2

Given,

• Sum of digits of two-digit number = 6
• Digits reversed, new number = 36 more than original number.

To find,

• Original number

Solution,

Let the digit at unit's place be a & digit at ten's place be b.

∴ Original number = a + 10b

AccordinG to QuesTion :

$:\implies\sf a + b = 6 \\\\ :\implies\sf a = 6 - b \qquad\dots (1) \\\\\sf When\: digits \: reversed, \\\\\therefore\sf Reversed \: number = b + 10a \\\\{\underline{\sf\green{Case \: ll \: :}}}\\\\ :\implies\sf b + 10a = (a + 10b) + 36 \\\\ :\implies\sf b + 10a = a + 10b + 36 \\\\ :\implies\sf 10a - a + b - 10b = 36 \\\\ :\implies\sf 9a - 9b = 36 \\\\ :\implies\sf 9(a - b) = 36 \\\\ :\implies\sf a - b = 36/9 \\\\ :\implies\sf a - b = 4 \\\\\qquad\scriptsize\sf{\big [Putting\: Value\: from\: (1) \big ]} \\\\ :\implies\sf 6 - b - b = 4 \\\\ :\implies\sf 6 - 2b = 4 \\\\ :\implies\sf -2b = 4 - 6 \\\\ :\implies\sf -2b = -2 \\\\ :\implies\sf b = \cancel{\dfrac{-2}{-2}} \\\\ :\implies\large\sf\blue{b = 1 }$

$\quad{\underline{\textsf{\dag\: Putting this value in (1) }}} \\\\ :\implies\sf a = 6 - 1 \\\\ :\implies\large\sf\red{a = 5 }$

$\qquad\small\underline{\underline{\sf\green{\dag \: Finding\: the\: number \: : }}} \\\\ :\implies\sf Original \: number = 5 + 10(1) \\\\ :\implies\sf Original \: number = 5 + 10 \\\\ :\implies\underline{\boxed{\textsf{\textbf{Original \: number = 15 }}}}$