Question

the sum of the digits of a two-digit number is 7 then the number obtained by reversing the digit is 2 more than twice the original number find the number​

Answer 1

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Answer 2

The  original number is 07

Step-by-step explanation:

Given as :

Statement I

The sum of the digits of a two-digit number is 7

The original number = 10 x + y

Let The first digit = y

Let The second digit = x

So, 10 × x + 1 × y = 7

i.e  10 x + y = 7                       ...........1

Statement II

The number obtained by reversing the digit is 2 more than twice the original number .

The number obtained by reversing the digit = 10 × y + 1 × x

i.e The number obtained by reversing the digit = 10 y + x

So,

( 10 y + x ) = 2 + 2 ( 10 x + y )

i.e 10 y +  x = 2 + 20 x + 2 y

Or, 10 y - 2 y +  x - 20 x = 2

Or, 8 y - 19 x = 2                    ....2

Solving eq 1 and eq 2

( 8 y - 19 x ) - 8 (  10 x + y ) = 2 - 8 × 7

Or, (8 y - 8 y) + (- 19 x - 80 x) = 2 - 56

Or, 0 - 99 x = 54

∴   y = $\dfrac{54}{-99}$

i.e   y = $\dfrac{-6}{11}$

Put the value of y in eq 1

10 x + y = 7

i.e  10 x -  $\dfrac{6}{11}$ = 7

Or,  10 x = 7 + $\dfrac{6}{11}$

∴     10 x = $\dfrac{83}{11}$

Now, The original number = 10 x + y

i.e     The original number = $\dfrac{83}{11}$   + ( $\dfrac{-6}{11}$ )

Or,    The original number = $\dfrac{83-6}{11}$

or,   The original number = $\dfrac{77}{11}$

∴  The original number = 7

Hence, The  original number is 07  Answer