The sum of a two digit number and the number obtained by reversing the order of its digits is 99. The digits differ by 3, find the number.
Answers
Answered by
17
Let 10X+Y be the number...
reversing the digits 10Y+X
10X+Y+10Y+X=11X+11Y
11X+11Y=99
X-Y=3
X=3+Y
11(X+Y)=99
11(3+Y+Y)=99
11(2Y+3)=99
2Y+3=9
2Y=6
Y=3
As X=3+Y
X=3+3=6
Hence the number is 63 or 36...
It is because :-
63+36=99
36+63=99
Note:- In the exam u have to mention that the number is 63 (or) 36... {MENTION BOTH THE NUMBERS}
Hope this helps!
cheers! (:
reversing the digits 10Y+X
10X+Y+10Y+X=11X+11Y
11X+11Y=99
X-Y=3
X=3+Y
11(X+Y)=99
11(3+Y+Y)=99
11(2Y+3)=99
2Y+3=9
2Y=6
Y=3
As X=3+Y
X=3+3=6
Hence the number is 63 or 36...
It is because :-
63+36=99
36+63=99
Note:- In the exam u have to mention that the number is 63 (or) 36... {MENTION BOTH THE NUMBERS}
Hope this helps!
cheers! (:
harsh878:
Why you have take 10 and 11
10 because it is tens digit place..
both sum 10X+X so its 11
Ok thanks sir
Similar questions