Question

a number of 2 digit exceeds 4 times the sum of its digit by 6 and it is increased by 9 on reversing the digits find the number

Answer 1

let the unit digit of the number be x and ten's digit be y.

the number = 10y+x

10 y+x = 4(x+y) + 6

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

- 3x +6y = 6

- x + 2y = 2...............(1)

the number by reversing its digit = 10 x + y

10y+x = 10x+y + 9

-9x + 9y = 9

-x + y = 1.........(2)

subtracting eq (2) from eq(1):

y = 1 and x = 0

thus the number is 10.

Answer 2

Let the digits of the number be x and y.
Therefore the number is 10x + y.

First case:

10x + y = 4(x +y) + 6
or, 10x + y = 4x + 4y + 6
or, 10x - 4x + y - 4y = 6
or, 6x - 3y = 6
or, 3(2x -y) = 6
or, 2x - y = 6/3
or, 2x - y = 2 ...(1)

Second case:

10y + x = 10x + y + 9
or, 10y - y + x - 10x = 9
or, 9y - 9x = 9
or, 9(y-x) = 9
or, y - x = 9/9
or, y - x = 1 ...(2)

Adding equations (1) and (2),

2x - y = 2
- x + y = 1
--------------------
x = 3


Now we find the value of y.

y-x = 1
y-3 = 1
y = 1+3 = 4

Therefore the number
= 10x + y
= 10 • 3 + 4
= 30 + 4
= 34

Ans: The required number is 34.

Hope that helps :)