Question

The sum of a two digit number and another formed by reversing the digit is 99. Five added to the number gives 4 less than 6 times the sum of its digit . find the number

Answer 1

Answer:

Let the tens place digit be x

units place digit be y.

Therefore, original number = 10x + y

Number obtained on reversing the digits = 10y + x

Equation 1........... 10x + y + 10y + x = 99

                                11x + 11y = 99

Dividing throughout by 11,

   X + y = 9 ...........  (1)

Equation 2..........5 + 10x + y = 6(x+y) -4

5 + 10x +y = 6x + 6y - 4

4x - 5y = - 9 ............ (2)

NOW SOLVE EQUATIONS (1) AND (2) BY ANY METHOD AND GET THE ANSWER FOR x AND y.

THEN SUBSTITUTE IN (10x +y) TO GET THE ORIGINAL NUMBER.

HOPE THIS HELPS!!
(^_^)

Answer 2

Solution:

Suppose, the tens place digit be p

&, units place digit be q.

Therefore, original number = 10p + q

Number obtained on reversing the digits is
= 10q + p

Equation 1...........

= 10p + q + 10q + p = 99

= 11p + 11q = 99

Dividing throughout by 11,
 
 p + q = 9 ...........  (1)

Equation 2..........

= 5 + 10p + q = 6(p+q) -4

= 5 + 10p +q = 6p + 6q - 4

= 4p - 5q = - 9 ............ (2)

NOW SOLVE EQUATIONS (1) AND (2) BY ANY METHOD AND GET THE ANSWER FOR p AND q.

THEN SUBSTITUTE IN (10p +q) TO GET THE ORIGINAL NUMBER.

HOPE THIS HELPS!!