Question

in a two digit number, the ten's digit is three times the unit's digit, when the numbee is decreased by 54, the digits are reversed. find the number.​

Answer 1

Answer :-

Let the digit of unit's place be x

and digit of ten's place be y

then,

• x = 2y ......(i)

and number = 10y + x

now, number obtained by reversing the digits = 10x + y

and it is given that the digits interchange their places of 27 is added to the number.

➪ number + 27 = new number

$\bold{: \implies 10y + x + 27 = 10x + y } \: \: \: \\ \\ \bold{: \implies 10y - y + x - 10x = - 27 } \\ \\ \bold{: \implies 9y - 9x = - 27 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies \cancel9(y - x) = \cancel9( - 3)} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies y - x = - 3 } .........(ii)\: \: \: \: \: \: \: \: \: \: \: \:$

now, put x = 2y in eq.(ii) from (i)

$\bold{: \implies y - 2y= - 3 } \\ \\ \bold{: \implies \cancel - y = \cancel- 3} \: \: \: \: \: \: \: \\ \\ \bold{: \implies y = 3} \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

now, substitute the value of y in eq.(i)

$\bold{: \implies x = 2 \times 3} \\ \\ \bold{: \implies x = 6 } \: \: \: \: \: \: \: \:$

so, digit of unit's place is 6 and digit of ten's place is 3

then, the number is

• 10 × 3 + 6 = 36

hence, the number is 36