Sum of the digits of two-digit number is 9. The number obtained by interchanging the digits exceeds the given number by 27. Find the given number.

Question :- Sum of the digits of two-digit number is 9. The number obtained by interchanging the digits exceeds the given number by 27. Find the given number.

 

Solution :-

Let the unit’s digit = x

Then ten's digit = (9 –x)

 

Number formed by these digits

        = 10 × ten's digit + unit's digit

        = 10 (9 – x) + x

        = 90 – 10x + x

        = 90 – 9x

 

When the digits are interchanged, unit’s digit becomes 9 – x and ten’s digit becomes x.

Ten’s digit	Unit’s didit	Number
9 – x	x	10 (9 – x) + x
x	9 – x	10x + (9 –x)

 

Number formed on interchanging the digits

         = 10x + (9 – x)

         = 9x + 9

 

It is given that new number exceeds the given number by 27

i.e., new number – given number = 27

i.e., (9x + 9) – (90 – 9x) = 27

ð  ⇒ 9x + 9 – 90 + 9x = 27

ð  ⇒ 18x – 81 = 27

ð  ⇒ 18x = 27 + 81

ð  ⇒ 18x = 108

ð  ⇒ x = 108 / 18

ð  ⇒ x = 6

 

Required number = (90 – 9x)

                                 = 90 – 9 x 6

                                 = 90 – 54

                                 = 36

 

Check :-

Sum of the digits = 3 + 6 = 9

Number obtained on interchanging the digits = 63

We have, 63 – 36 = 27