Question

11. The digit at the ten's place of a two-digit number is four times the digit at one's place. If the sum
of this number and the number formed by reversing the digits is 55, find the numbers.​

Answer 1

Answer:

$\bigstar{\bold{Original\:number=41}}$

$\bigstar{\bold{Reversed\:number=14}}$

Step-by-step explanation:

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

• Digits at ten's place = 4 times the digit at one's place
• Sum of the number and reversed number = 55

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• The numbers

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Let the one's digit be x

→ Let the ten's digit be y

→ Hence,

  The number = 10y + x

→ By given,

  y = 4x ----(1)

→ Now reversing the number the number becomes,

  Reversed number = 10x + y

→ By given we know that sum of original number and reversed number is 55

→ Hence,

  10y + x + 10x + y = 55

→ Substitute the value of y from equation 1

  10 × 4x + x + 10x + 4x = 55

  40x + 15x = 55

  55x = 55

      x= 55/55

      x = 1

→ Hence the one's digit is 1

→ Substitute the value of x in equation 1

  y = 4x

  y = 4 × 1

  y = 4

→ Hence the ten's digit is 4

→ Therefore the number is,

   Original number = 10y + x

   Original number = 10 × 4 + 1

   Original number = 41

$\boxed{\bold{Original\:number=41}}$

→ Reversed number = 10x + y

  Reversed number = 10 × 1 + 4

  Reversed number = 14

$\boxed{\bold{Reversed\:number=14}}$

$\Large{\underline{\underline{\bf{Verification:}}}}$

→ By given,

  y = 4x

  4 = 4 × 1

  4 = 4

→  10y + x + 10x + y = 55

   10 × 4 + 1 + 10 × 1 + 4 = 55

   40 + 1 + 10 + 4 = 55

   50 + 5  = 55

   55 = 55

→ Hence verified.