The sum of the digits of a two digits number is 12.if the digits are revesed the new number is 12 less than twice the original number
Answers
Let ten's digit be "M" and one's digit be "N".
The sum of digits of two digit number is 12.
According to question,
=> M + N = 12
=> M = 12 - N _____ (eq 1)
If digits are reversed the new number is 12 less than twice the original number.
Original number = 10M + N
Reversed number = 10N + M
According to question,
=> 10N + M = 2(10M + N) - 12
=> 10N + M = 20M + 2N - 12
=> 10N - 2N + M - 20M = - 12
=> 8N - 19M = - 12
=> 8N - 19(12 - N) = - 12
=> 8N - 228 + 19N = - 12
=> 27N = - 12 + 228
=> 27N = 216
=> N = 8
Put value of N in (eq 1)
=> M = 12 - 8
=> M = 4
Original number = 10M + N
=> 10(4) + 8
=> 48
ANSWER:-
Given:
The sum of the digits of a two digits number is 12. If the digits are reversed the new number is 12 less than twice the original number.
Solution:
Let the unit's digit be x &
Let the tens digit be y.
Sum of the digits of a two number is given to be 12.
Therefore,
x + y= 12
x= 12 -y...........(1)
Original number is 10x + y
Number obtained on reversing the digits= 10y+x
According to the question:
(10y + x) = 2(10x + y)-12
=) 10y + x = 20x + 2y-12
=) 10y-2y+ x -20x= -12
=) 8y -19x = -12
=) 8y -19(12-y)= -12 [using (1)]
=) 8y - 228+19y= -12
=) 27y -228= -12
=) 27y = -12 +228
=) 27y = 216
=) y= 216/27
=) y= 8
Now,
Putting the value of y in eq. (1);
x = 12 - 8
=) x= 4
Therefore,
The number is required is original number;
10x +y
=)10(4) + 8
=) 40 + 8