Question

if two digit no is 5 times the sum of its digits and is also equal to 5 more than twice the product of digit find the number ​

Answer 1

Answer:

Required Number = 45

Step-by-step explanation:

Let the digit at unit's place be x

Let the digit at ten's place be y

Number = 10y + x

Given that two digit number is 5 times the sum of its digits

We have

10y + x = 5(x + y)

10y + x = 5x + 5y

(5x - x) + (5y - 10y) = 0

4x - 5y = 0

4x = 5y

x = 5y/4 ---- (1)

Also, two digit number is 5 more than twice the product of digit of number

Thus, we get

10y + x = 2(xy) + 5

10y + x = 2xy + 5

Putting x in above,we get

10y + 5y/4 = 2 × 5y/4 × y + 5

40y + 5y = 10y^2 + 20

45y = 10y^2 + 20

2y^2 - 9y + 4 = 0

Solving the quadratic equation using factorizatio

2y^2 - 8y - y + 4 = 0

2y(y - 4) - 1(y - 4) = 0

(2y - 1)(y - 4) = 0

Either 2y - 1 = 0      or      y - 4 = 0

          y = 1/2          or       y = 4

Digit can never be a fraction.

Thus, y = 4

Putting the value of y in (1), we get

x = 5/4 × 4

x = 5

Thus, required number  = 10(4) + 5 = 45