Question

A number consists of two digits. The digits at tens place is two times the digit at units place. The number formed by reversing thedigit is 27 less than the original number. Find the original number.

Answer 1

Let the ten's digit and the one's digit be x and y respectively.

Given

The ten's digit is two times the unit's digit.

$\implies x = 2y\\\implies x - 2y =0 --(1)$

The number formed by reversing the digit is 27 less than the original number.

$\implies 10y+x=10x+y-27\\\implies 10y+x-10x-y=-27\\\implies -9x+9y=-27\\\implies x-y=3--(2)$

Solving (1) and (2), we get,

$x-2y=0\\\underline{x-y=3}\\\underline{\underline{-y=-3}}\\\implies y = 3\\\implies x = 6$

$\boxed{\boxed{\bold{Therefore, \ the \ required \ number \ is \ 63}}}}}}$

                                                         

Answer 2

Concept

In this type of question assume unit's place digit x and ten's digit y

So the two-digit number will of the form of $10y+x$ .

Given

We have given that digits at tens place is two times the digit at units place and the number formed by reversing the digit is 27 less than the original number.

Find

We are asked to find the original number

Solution

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

Two-digit number be $=10y+x$ .

On interchanging the digit number will be $=10x+y$ .

According to the first statement which is "digits at tens place is two times the digit at units place " equation will be

 $y=2x$    ....(1)

According to the second statement which is "The number formed by reversing the digit is 27 less than the original number " equation will be

Again, new number = original number - 27

           $10x+y=10y+x-27\\9x-9y=-27$

Dividing by 9 , we  get

 $x-y=-3$       ....(2)

Putting the value of y = 2x in equation (2) .

$x-2x=-3\\-x=-3\\x=3$

$y=2(3)=6$

The two-digit number will be $=10(6)+3$

                                                  $=60+3=63$

Therefore, the value of the original number is 63.

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