Question

the sum of the digit of a two digit no.is 7. if the radius are reversed , the new no. increased by 3 equals 4 times the original no.Find the original no.​

Answer 1

Correct Question :

The sum of the digit of a two digit no. is 7. Kf the digits are reversed, the new no. increased by 3 equals to 4 times the original no. Find the Original Number.

AnswEr :

Let the digit at unit's place be y and the digit at ten's place be x. Then,

⋆ Original Number will be : (10x + y)

⋆ Reversed Number will be : (10y + x)

⋆ Sum of Digit : (x + y) = 7⠀⠀⠀—eq.( I )

• According to the Question Now :

⇒ New Number + 3 = 4 times Old Number

⇒ (10y + x) + 3 = 4(10x + y)

⇒ 10y + x + 3 = 40x + 4y

⇒ 10y - 4y + x - 40x + 3 = 0

⇒ 6y - 39x + 3 = 0

⇒ 3(2y - 13x + 1) = 0

• Dividing Both term by 3

⇒ 2y - 13x + 1 = 0

⇒ 1 = 13x - 2y

⇒ 13x - 2y = 3⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀—eq.( II )

_________________________________

• Multiplying eq.( I ) by 2 and adding in ( II ):

⇝ 2x + 2y = 14

⇝ 13x - 2y = 3

_________________

⇝ 2x + 13x = 14 + 3

⇝ 15x = 15

• Dividing Both term by 15

⇝ x = 1

• Putting the Value of x in eq.( I ) :

⇝ x + y = 7

⇝ 1 + y = 7

⇝ y = 7 - 1

⇝ y = 6

_________________________________

• Original Number Formed will be :

↠ (10x + y)

↠ {(10 × 1) + 6}

↠ (10 + 6)

↠ 16

∴ Therefore, Original Number is 16.

━━━━━━━━━━━━━━━━━━━━━━━━

• Verification :

⇒ Reversed No. + 3 = (4 × Original No.)

⇒ (10y + x) + 3 = 4 × (10x + y)

⇒ (10 × 6 + 1) + 3 = 4 × (10 × 1 + 6)

⇒ (60 + 1) + 3 = 4 × (10 + 6)

⇒ 61 + 3 = 4 × 16

⇒ 64 = 64 ⠀⠀⠀⠀⠀⠀Hence, Verified!

Answer 2

$\huge\sf\red{\underline{Appropriate\:Question}}$

The sum of the digit of a two digit no.is 7. if the digits are reversed, the new no. increased by 3 equals 4 times the original no. Find the original no.

$\bold{\huge{\boxed{\purple{\tt{Answer-}}}}}$

Original number = 16

$\bold{\huge{\boxed{\pink{\mathfrak{Explanation-}}}}}$

Let the tens digit be x and unit digit be y.

Number = 10x + y

After reversing digits,

Number = 10y + x

________________

$\bold{\large{\underline{\boxed{\tt{\green{As\:per\:the\:question}}}}}}$

Case 1).

The sum of the digit of a two digit no.is 7.

$\bold{\blue{\boxed{\large{\sf{x + y = 7\:--->(1)}}}}}$

Case 2).

If the digits are reversed, the new no. increased by 3 equals 4 times the original no.

$\leadsto$ $\tt{ ( 10y + x ) + 3 = 4 ( 10x + y ) }$

$\leadsto$ $\tt{10y + x + 3 = 40x + 4y}$

$\leadsto$ $\tt{10y - 4y + x - 40x = -3}$

$\leadsto$ $\tt{6y - 39x = -3}$

Taking 3 as common,

$\leadsto$ $\tt{2y - 13x = -1 }$

$\leadsto$ $\bold{\large{\boxed{\red{\sf{13x - 2y = 1 \:----> (2)}}}}}$

__________________

Multiply equation (1) by 2,

$\bold{\pink{\boxed{\large{\sf{2x + 2y = 14\:--->(3)}}}}}$

Adding equation (2) and (3),

$\leadsto$ $\tt{2x + 2y + 13x - 2y = 14 + 1}$

$\leadsto$ $\tt{15x = 15}$

$\leadsto$ $\tt{x\:=\:{\dfrac{15}{15}}}$

$\leadsto$ $\bold{\large{\boxed{\red{\rm{x\:=\:1}}}}}$

Now, put the value of x in equation (3)

$\leadsto$ $\tt{2(1) + 2y = 14}$

$\leadsto$ $\tt{2 + 2y = 14}$

$\leadsto$ $\tt{2y = 14 - 2}$

$\leadsto$ $\tt{2y = 12}$

$\leadsto$ $\leadsto$ $\tt{y\:=\:{\dfrac{12}{2}}}$

$\leadsto$ $\bold{\large{\boxed{\purple{\rm{y\:=\:6}}}}}$

___________________

$\bold{\large{\boxed{\boxed{\tt{\red{Ten's\:digit\:=\:x\:=\:1}}}}}}$

$\bold{\large{\boxed{\boxed{\tt{\red{Unit's\:digit\:=\:y\:=\:6}}}}}}$

$\bold{\large{\boxed{\boxed{\tt{\red{\therefore{Original\:Number\:=\:10x\:+\:y=\:\:10\:\times\:1\:+\:6\:=\:10\:+\:6\:=\:16}}}}}}}$