Question

a number between 10 and 100 is five times the sum of its digits. if 9 be added to it the digits are reversed, find the number....

Answer 1

hai friend

here is your answer

the no. between 10 and 100 which is five times the sum of digits and if 9 is added to it the digits are reversed

the number is forty five (45)

because

sum of the digits of forty five(45) is equal to

four(4) plus(+) five(5) which is equal to nine (9)

nine (9) into(*) five(5) is equal to forty five (45)

so, the equation one is true

next equation

forty five(45) plus (+) nine (9) is equal to

fifty four (54) which is the reverse of forty five(45)

so, the second equation is also true

hope it helps you

Answer 2

let the number be 10 x + y

sum of its numbers =  x+ y

according to the sum

10 x + y = 5(x + y)

10 x + y = 5 x + 5 y

10 x - 5 x = 5 y - y

5 x - 4 y = 0

if 9 is added the digits are reversed

⇒ 10 x + y + 9 = 10 y + x

    10 x - x + 9 = 10 y - y

9 x - 9 y = - 9

x - y =  - 1

simultaneous equation 5 x - 4 y = 0

                                       x - y = -1

1(5 x - 4 y = 0)

5(x - y = -1)

5 x - 4 y = 0 →a

5 x - 5 y = -5→b

On subtracting b from a

5 x - 4 y = 0

- 5 x + 5 y = 5

y = 5

 

On substituting y = 5 in x - y = - 1 

we get x = 4

The number is 10 x + y = 10(4) + 5 = 45.