Given: A two-digit number is 4 more than 6 times the sum of its digits.
If 18 is subtracted from the number, the digits are reversed.
Solution:
Let the digit in the unit's place be x and the digit at the tens place be y.
Number = 10y + x
The number obtained by reversing the order of the digits is = 10x + y
ATQ:
Condition: 1
10y + x = 6(x + y) + 4
10y + x = 6x + 6y + 4
10y + x - 6x - 6y = 4
- 5x + 4y = 4
5x - 4y = - 4 ............... (1)
Condition : 2
(10y + x) - 18 = 10x + y
10y + x - 10x - y = 18
- 9x + 9y = 18
- 9(x - y) = 18
x - y = - 18/9
x - y = - 2 ..............(2)
On multiplying equation (2) by 4:
4x - 4y = -8............(3)
On Subtracting equation (3) from
equation (1), we obtain:
5x - 4y = - 4
4x - 4y = - 8
(-) (+) (+)
----------------------
x = 4
On putting x = 4 in eq (1) we obtain:-
5x - 4y = - 4
5(4) - 4y = - 4
20 - 4y = - 4
- 4y = - 4 - 20
- 4y = - 24
y = 24/4
y = 6
Now, Number = 10y + x = 10 × 6 + 4 = 60 + 4 = 64
Hence, the number is 64