Question

the sum of digits of a 2 digit number is 11. if the number obtained by reversing the digits is 9 less than the original number, find the number

Answer 1

let the ones place digit be y and the tens place digit be x.
Therefore, the number is 10x + y
After reversing the number, we get = 10y + x
ATQ,
sum of digits = 11
x + y = 11 ............(1)

Also,
10y + x = 10x + y - 9
10y - y + x - 10x = -9
9y - 9x = -9
y - x = -1
x - y = 1..................(2)

now add equation (1) and equation (2)
we get

2x = 12
x = 6

put this in equation (1)
6 + y = 11
y = 5

the number is = 10 x + y = 10 × 6 + 5 = 60 + 5 = 65

Answer 2

hi!
here's your answer!

Let the digit at tens place be x
and at the units place be y.

By the given condition ,

x + y = 11..........................(1)

Number =10x + y
Reversed number=10y + x

By the second condition .

10y + x = 10x + y - 9

10y - y + x - 10x = -9

9y - 9x = -9

(Divide throughout by 9)

y - x = -1

x - y = 1..........................(2)

[By adding (1) & (2)]

x + y = 11 ................. (1)
x - y = 1 ..................(2)
_________
2x = 12
x=12/2
x = 6.

Substitute x=6 in equation no.(1)

x + y =11
6 + y =11
y=11-6
y=5

so,
x = 6 ; y =5

Number:
10x + y
=10(6) +5
=60+5
=65.

Therefore,
The required number is 65.

hope this helps ! ! !#