Question

a number consist of two digits whose sum is 9. the number formed by reversing the digits exceeds twice the original number is 18. find the original number.

Answer 1

Heya user ,
Here is your answer !!

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Let the ten's digit be x
Let the unit's digit be y .
So , x + y = 9
So , y = 9 - x .

The actual number is = 10x + y .

Acc. to the question ,
( 10y + x ) - 2 ( 10x + y ) = 18 .
=> 10y + x - 20x - 2y = 18.
=> 8y - 19x = 18 .
=> 8 ( 9 - x ) - 19x = 18 [ Substituting the value of y ] .
=> 72 - 8x - 19x = 18
=> 72 - 18 - 27x = 0
=> 54 = 27x
=> x = 2 .

So ,
y = 9 - 2
= 7 .

So , the original number is ( 2*10 + 7 ) = 27 . [ Answer ] .

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Hope it helps !!