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⭐IN A TWO DIGIT NATURAL NUMBER, THE DIGITS DIFFER BY 1.THE PRODUCT OF THE NUMBER AND THE NUMBER OBTAINED BY REVERSING THE DIGITS IS 252.FIND THE NUMBER.⭐
CLASS 10 CHAPTER:QUADRATIC EQUATIONS
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Answers
According to the questions,
digits of the required number are differ by 1.
Let,
Unit's place digit be x
Ten's place digit be ( x + 1 )
Number = 10( x + 1 ) + x
Given, product of the number and the number obtained by reversing the digits is 252.
Original number = 10( x + 1 ) + x
Number when digit are reversed = 10x + ( x + 1 )
Product = [ [10( x + 1 ) + x ] [ 10x + ( x + 1 ) ]
= > 252 = [ 10x + 10 + x ] [ 10x + x + 1 ]
= > 252 = [ 11x + 10 ] [ 11x + 1 ]
= > 252 = 121x^2 + 11x + 110x + 10
= > 252 = 121x^2 + 121x + 10
= > 121x^2 + 121x + 10 - 252 = 0
= > 121x^2 + 121x - 242 = 0
= > 121( x^2 + x - 2 ) = 0
= > x^2 + x - 2 = 0
= > x^2 + ( 2 - 1 ) x - 2 = 0
= > x^2 + 2x - x - 2 = 0
= > x( x + 2 ) - ( x + 2 ) = 0
= > ( x + 2 ) ( x - 1 ) = 0
= > x = - 2 or x = 1
It is given that the required number is a natural number, so x can't be equal to - 2. Therefore, x = 1 .
Then,
Unit's place of the number = x
Unit's place of the number = 1
Ten's place of the number = x + 1
Ten's place of the number = 1 + 1
Ten's place of the number = 2
Therefore, required natural number = xy = 21
Required number = 21 .