Question

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

Answer 1

let the digits are x and y so..
If i write a no. xy then= 10x + y
e.g. 32= 10×3+2
so.. reversing of xy is yx and
yx=10y+x

Now..
A/Q : x-y=3 eq1
10x+ y+ 10y+x = 99 eq 2
so from eq 2
10x+y+10y+x= 99
11x+11y=99
x+y=9 ...... eq 3

Now
from eq 1
we have
x-y=3
x=3+y
substituing value of x in eq 3
x+y=9
3+y+y=9
2y=6
y=3
Substituting value of y in eq1
x-y=3
x-3=3
x=6
so the no. is 36 or 63
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Answer 2

★Hey Mate ★


here is your answer !!


let the number be tenth digit be X


let the. unit digit be Y



hence , the number is =>


10x + y



according to question !

it's. reversing digit is =>

10 y + x


now adding we get




10x + y + 10 y + x = 99

=> 11 ( x + y ) = 99

=> x + y = 9 --------------------(1)



now , according to question !!


if the digit is differ by 3

then


x -y =3 -----------------------(2)

or y -x = 3



from equation 1 and 2 we get


x + y = 9

x - y = 3

( + ) (+ ) = ( +)
============


2x = 12

x = 6

then y = > 6 - y = 3

y = 3


hence , the two digit number is


10 x + y = 10 × 6 + 3

=>. 63


now case 2


when.

y - x = 3,



x + y = 9

y - x = 3
=========
2 y = 12

y => 6


hence x = 3


hence the number is


10 x + y


=> 10 × 3 + 6

=> 36


hence , two Answer will be possible.


that is :- 63 , 36



hope it helps you dear !!!


$thanks$
!!