Question

The sum of digits of a two digit number is
11. If the digit at ten's place is increased by
5 and the digit at unit's place is decreased by
5, the digits of the number are found to be
reversed. Find the original number.
​

Answer 1

Answer:

original no. 10

Step-by-step explanation:

let, x and y be unit place and tens place digits resp.

original no = 10y+x

now, 10y+x=11......(1)

ATQ

tens place digit=10y+5

unit place digit=x-5

so,new sum after revert

(x-5)*10 + (10y+5)= 11

10x-50+10y+5=11

10x+10y=56.....(2)

eq1-eq2

9x=45

x=5

and y=1/2

hence, required no.= 10