Question

Q15. The sum of the digits of a two digit number is 10. If the number formed by reversing the digits is
greater than the original number by 18, find the original numb
​

Answer 1

$\huge{\underline{\underline{\mathcal{SOLUTION}}}}$ :

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

Then, Number = 10y + x

ATQ,

x + y = 10 --> ( i )

10x +y - ( 10y + x ) = 18 --> ( ii )

10x - x + y - 10y = 18

9x - 9y = 18

x - y = 2 --> ( iii )

Adding i and iii,

2x = 12

x = 6

Putting x =6 in i,

6 + y = 10

y = 4

Number = 10y + x = 10 × 4 + 6 = 46

$\boxed{\boxed{\tt{46}}}$

Answer 2

ones digit be x and tens digit be y

original number = x+10y

reversed no = 10x+y

a/q

x+y=10......(i)

again, 10x+y=10y+x+18

9x-9y=18

x-y=18/9

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

adding eqn(i) and eqn(ii)

x+y=10

x-y=2

------------

2x=12

•x=6

putting value of x in (1)

6+y=10

•y=4

hence no is x+10y= 6+10*4=40+6=46

•hence no is 46