Question

Q.The sum of the digits of a two digit number is 5. The digit obtained by
increasing the digit in ten's place by unity is one-eighth of the number. Find the
number.​

Answer 1

Let the digit in unit place be y and digit in tens place be x

the number will be  10x+y

therefore from first condition x+y=5.......(1)

from second condition x+1=1/8*10x+y

therefore 8x+8=10x+y                8=10x-8x+y                   8=2x+y....................................(2)

now   subtracting  equation  (1) from (2)

  2x+y=8-(x+y)=-5     x   =3now putting  value of x in equation (1)

3+y=5    y=5-3   y  =2 

so the number is 32

Answer 2

Solution :-

Let the digit at the ten's place be x ; and the digit at the unit's place be y.

∵ The digit by increasing the digit in ten's place by unity = x + 1.

A/Q,

$x + y = 5 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ....(i) \\ and \: x + 1 = \frac{1}{8}(10x + y) \\ \implies \: 8(x + 1) = 10x + y \\ \implies \: 8x + 8 = 10x + y \\ \implies \: 2x + y = 8 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(ii)$

On Subtracting (i) from (ii),

we get , x = 3.

Now,

On substituting this value of x in (i) ,

we get,

3x + y = 5

=> y = 5-3

=> y = 2.

Hence, the required number is 32.