Sum of the digits of a two digits number is 9. The number obtained by interchanging of the digits/ exceed the given number by 27 find the number
Answers
Answer:
36
Step-by-step explanation:
Let that number is 10a+b
sum of the digits a+b= 9 (1)
10b+a-(10a+b)=27
10(b-a)-(b-a)=27
(b-a)(10-1)= 27
b-a = 27/9
b-a = 3 (2)
by adding 1 and 2
2b = 12
b= 6
a+6=9
a=3
Answer:
hey mate
let the no. at ones place be "x"
and no. at tens place be"9-x"
original no.
tens ones
9-x x
so, original no.=1*x+10(9-x)
=x+90-10x
original no=90-9x
on reversing
tens ones
x 9-x
so, new number=1(9-x)+10*x
=9-x+10x
new no.=9+9x
A.T.Q
9+9x=90-9x+27
9x+9x=90-9+27
18x= 81+27
18x=108
x=108/18=6
so,the original no.is tens ones
9-x x
9-6 6
and the answer is 36
hope this will help you
good evening