The sum of two digit number and the number obtained by reversing the order of its digit is 99.if the digits differ by 3,find the number?

Asked by FEHMEEDAH NAZNEEN | 30th Aug, 2010, 09:38: PM

Expert Answer:

Let the ten's place be x and units place be y.
so the original number becomes 10x+y
so
 the number obtained by reversing the digits is 10y+x
ATQ
(10x+y)+(10y+x)=99
11x+11y=99
x+y=11....(i)
also
ATQ,
digits differ by 3
so
either x-y=3 or x-y=-3
lets take
x-y=3..(ii)
so , solving  (i) and (ii)
2x=14
so
x=7
and y=4
so original number is 74
next
 lets take x-y=-3..(iii)
so,
solving (i) and (iii)
2x=8
so x=4
and
y=7
so the number is 47

Answered by  | 31st Aug, 2010, 08:59: AM

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