The digit at the tens place of a two-digit number is four times that is the units place. If the digits
reversed, the new number will be 54 less than the original number. Find the number
Answers
Given:
Sum of digits of the two digits number is four times that in the unit's place.
If the digits are reversed the new number will be 54 less that the original number.
Find:
The original number
Solution:
Let the digits at unit place be y.
Let the digits at ten's place be x.
Number = 10x + y
Sum of digits of the two digits number is four times that in the unit's place.
=> x + y = 4y
=> x = 4y - y
=> x = 3y ......(i).
If the digits are reversed the new number will be 54 less that the original number.
Number obtained by reversing the digits = 10y + x
Number obtained by reversing the digits = 10x + y - 54
=> 10y + x = 10x + y - 54
=> 54 = 10x + y - 10y - x
=> 54 = 9x - 9y
=> 54 = 9(x - y)
=> 54/9 = x - y
=> 6 = x - y .........(ii).
Putting the value of x from Eq (i). in Eq (ii).
=> 6 = x - y
=> 6 = 3y - y
=> 6 = 2y
=> y = 6/2
=> y = 3
Putting the value of y in Eq (ii).
=> 6 = x - y
=> 6 = x - 3
=> -x = -6 - 3
=> -x = -9
=> x = 9
So, Number = 10x + y
=> 10(9) + 3
=> 90 + 3
=> 93
Hence, the original number is 93.
I hope it will help you.
Regards.
Answer:
=> 10y + x = 10x + y - 54
=> 54 = 10x + y - 10y - x
=> 54 = 9x - 9y
=> 54 = 9(x - y)
=> 54/9 = x - y
=> 6 = x - y .........(ii).
Step-by-step explanation: