Question

the sum of the digits of a two digit number is 10 the digit in ones place is 9 times the digit at tens place find the number​

Answer 1

Answer:

Let the number be (10x + y).

Given:

Case - 1:

5 times the sum of the digits of the number is 9 less than the number formed by reversing digits.

· 5(x + y) = (10y + x) - 9

→ 5x + 5y = 10y + x - 9

→ 5x - x = 10y - 5y - 9

→ 4x = 5y - 9

➡X= (5y - 9)/4 -- equation (1).

Case - 2:

And, 4 times the value of the digits at ones place is equal to half the place value of digit at tens place (10x/2).

4(y) = 1/2* (10x)

→ 4y = 5x

Putting the value of x from equation - (1) we get,

→ 4y = 5 [ (5y - 9)/4]

On cross multiplication we get,

16y = 25y - 45

→ 16y - 25y = - 45

→-9y = - 45

→y = (-45)/(- 9)

→y=5

Substitute the value of y in equation (1)

→ x = (5y - 9)/4

→ x = [5(5) - 9 ]/4

→x = 25-9/4

→x = 16/4

→ X= 4

→ Number = 10(4) + 5 = 40 + 5 = 45.

Hence the required number is 45.