Divide 50 in two parts so that the sum of reciprocals is (1/12), the numbers are?
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Hi ,
Let x , 50 - x are two parts .
Sum of reciprocals = 1/12
1/x + 1/( 50 - x ) = 1/12
( 50 - x + x )/[ x( 50 - x ) ] = 1/12
50 × 12 = 50x - x²
x² - 50x + 600 = 0
x² - 20x - 30x + 600 = 0
x( x - 20 ) - 30( x - 20 ) = 0
( x - 20 ) ( x - 30 ) = 0
x - 20 = 0 or x - 30 = 0
x = 20 or x = 30
Therefore ,
Required numbers are ,
( 20 , 30 ) or ( 30 , 20 )
I hope this helps you.
: )
Let x , 50 - x are two parts .
Sum of reciprocals = 1/12
1/x + 1/( 50 - x ) = 1/12
( 50 - x + x )/[ x( 50 - x ) ] = 1/12
50 × 12 = 50x - x²
x² - 50x + 600 = 0
x² - 20x - 30x + 600 = 0
x( x - 20 ) - 30( x - 20 ) = 0
( x - 20 ) ( x - 30 ) = 0
x - 20 = 0 or x - 30 = 0
x = 20 or x = 30
Therefore ,
Required numbers are ,
( 20 , 30 ) or ( 30 , 20 )
I hope this helps you.
: )
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