Question

Sum of the digit is Two-digit is 9.Interchange the digit resulting new no is greater than the original no by 27.What is the two digit no?​

Answer 1

Let Tens place Digit=x

Let Ones Place Digit=y

Sum of Digit=9

x+y=9

x=9-y. (Equation 1)

Original Number=10x+y

Reverse Number=10y+x

Reverse is Greater than Original by 27

A/Q

10x+y+27=10y+x

10y+x-(10x+y)=27

10y+x-10x-y=27

9y-9x=27

9(Y-X)=27. (common=9)

Y-X=27/9

Y-X=3

We get Y-X=3

x=9-y

y-(9-y)=3

y-9+y=3

2y=3+9

2y=12

y=12/2

y=6

We know

x+y=9

x+6=9

x=9-6

x=3

So,

x=3

y=6

Number=10x+y

=10×3+6

=36

Now Understand why we take Number=10x+y.

Let Understand

Let Ones Place Digit=3

Let Tens Place Digit=9

How can It We Original number. 93

By adding=9+3=12 no wrong.

If We Multiply 10 to tens Digit number=9 then add 3 it will be original number

=93

Just like 10x+y

$\boxed{\mathbf{\huge{Number=36}}}$

Answer 2

HELLO DEAR FRIEND

$\large{let \: the \: digits \: be \: x \: and \: y}$

$\large{so \: number \:will \: 10x + y }$

$\large{atq,}$

$\large{x + y = 9 \:...(1)}$

$\large{also, }$

$\large{10x + y =10y + x + 27}$

$\large{10x - x + y - 10y = 27}$

$\large{9x - 9y = 27}$

$\large{ x - y = 3 \: ...(2)}$

$\large{(1) + (2)}$

$\large{x + y + x - y = 9 + 3}$

$\large{2x = 12}$

$\large{ x = \frac{ \large{12}}{ \large{2}} }$

$so , \: \large{x = 6}$

$\large{from \: equation \: (1), }$

$\large{x + y = 9}$

$= > \large{6 + y = 9}$

$or, \large{y = 9 - 6}$

$so, \large{y = 3}$

$\huge \boxed{x = 6}$

$\huge \boxed{y = 3}$

NUMBER = 10X + Y

= 10(6) + 3

= 60 + 3

$\huge \boxed{Number = 63}$ans

HOPE IT HELPS YOU DEAR FRIEND

THANKS