Question

6. A 2-digit number is 5 more than 3 times the sum of
its digits. If 45 is added to the number, the digits are
reversed. Find the number.​

Answer 1

AnswEr :

Let the Number be (10x + y) where y is One's Digit and x is Ten's Digit.

☢ According to the Question :

↠ Number = 5 + 3(Sum of Digits)

↠ (10x + y) = 5 + 3(x + y)

↠ 10x + y = 5 + 3x + 3y

↠ 10x – 3x + y – 3y = 5

↠ 7x – 2y = 5 ⠀— eq.( I )

________________________

☢ According to the Question Again :

↠ Original No. + 45 = Reverse No.

↠ (10x + y) + 45 = (10y + x)

↠ 10x + y + 45 = 10y + x

↠ 10x – x + y – 10y + 45 = 0

↠ 9x – 9y + 45 = 0

↠ 9x – 9y = – 45

↠ 9(x – y) = – 45

• Dividing both term by 9

↠ x – y = – 5

↠ x = (y – 5) ⠀— eq.( II )

━━━━━━━━━━━━━━━━━━━━━━━━

★ Putting the value of x in eq. ( I ) :

⇢ 7x – 2y = 5

⇢ 7(y – 5) – 2y = 5

⇢ 7y – 35 – 2y = 5

⇢ 5y = 5 + 35

⇢ 5y = 40

• Dividing both term by 5

⇢ y = 8

________________________

★ Putting the value of y in eq. ( II ) :

⇢ x = (y – 5)

⇢ x = (8 – 5)

⇢ x = 3

◗ Number = (10x + y) = (10 × 3 + 8) = 38

∴ Therefore, Orginal Number will be 38.

Answer 2

AnswEr :

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

A 2-digit number is 5 more than 3 times the sum of it's digits . If 45 is added to the number, the digits are reversed.

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

The number.

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

Let the ten's place be R

Let the one's place be M

$\bf{\underbrace{\bf{The\:original\:number\:=\:10R+M}}}}$

$\bf{\underbrace{\bf{The\:reversed\:number\:=\:10M+R}}}}$

$\bf{\red{\large{\underline{\underline{\tt{A.T.Q.\::}}}}}}$

$\Rightarrow\tt{10R+M=3(R+M)+5}\\\\\\\Rightarrow\tt{10R+M=3R+3M+5}\\\\\\\Rightarrow\tt{10R-3R+M-3M=5}\\\\\\\Rightarrow\tt{\green{7R-2M=5.........................(1)}}$

&

$\Rightarrow\tt{10R+M+45=10M+R}\\\\\\\Rightarrow\tt{10R+M-10M-R=-45}\\\\\\\Rightarrow\tt{9R-9M=-45}\\\\\\\Rightarrow\tt{9(R-M)=-45}\\\\\\\Rightarrow\tt{R-M=\cancel{\dfrac{-45}{9} }}\\\\\\\Rightarrow\tt{R-M=-5}\\\\\\\Rightarrow\tt{\green{R=-5+M.....................(2)}}$

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

$\hookrightarrow\tt{7(-5+M)-2M=5}\\\\\\\hookrightarrow\tt{-35+7M-2M=5}\\\\\\\hookrightarrow\tt{-35+5M=5}\\\\\\\hookrightarrow\tt{5M=5+35}\\\\\\\hookrightarrow\tt{5M=40}\\\\\\\hookrightarrow\tt{M=\cancel{\dfrac{40}{5} }}\\\\\\\hookrightarrow\tt{\red{M=8}}$

Putting the value of M in equation (2), we get;

$\hookrightarrow\tt{R=-5+8}\\\\\\\hookrightarrow\tt{\red{R=3}}$

∴ The original number is 10R+M = 10(3) + 8

The original number is 30 + 8 = 38 .