Question

solve the following​

Answer 1

$\underline{\textbf{\huge{Answer}}}$


Sum of Digits=16

Let Tens Place Digit=x

Let Ones Place Digit=y

A/Q

x+y=16

x=16-y. (Equation 1)

Now Original Number=10x+y

Reverse Number=10y+x

Reverse Number is 18 less than Original Number.

A/Q

10x+y-18=10y+x

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

10x+y-10y-x=18

9x-9y=18

Common=9

9(x-y)=18

x-y=18/9

x-y=2

We know that x=16-y

Then

x-y=2

16-y-y=2

16-2y=2

16-2=2y

14=2y

y=14/2

y=7

We get y=7

We know

x=16-y

x=16-7

x=9

Original Number=10x+y

=10×9+7

=97

$\underline{\textbf{\huge{Explanation.}}}$

Why Original number=10x+y

Since Digit at Tens Place=x

At ones Place=y

How 2 Digit Number 98 can be written

If Tens Place Digit=9

Ones Place Digit=8

Can We write as 9+8=17

No it will be Written as =10×9+8=98

So,

$\boxed{\mathbf{\huge{Original\: Number=97}}}$