Number Problem 12 Solution

This is the solution to a number algebra word problem that was asked by Zakeus who said , "The numerator and denominator of a fraction are together equal to 100. Increase the numerator by 18 and decrease the denominator by 16 and the fraction is doubled. What is the fraction?"


Ok you want to set up 2 equations the 1st one is easy.   x + y = 100 so y = 100 - x

The other would be  (x + 18)/(y - 16) = (x/y)*2

(x + 18) y = 2x (y - 16)  Now substitute 100-x for y....

(x + 18)(100 - x) = 2x(100 - x - 16)

100 x + 1800 - x^2 -18x = 2x(84-x)

82x + 1800 -x^2 = 168x - 2x^2  Now combine like terms...

x^2 - 86x + 1800 = 0    

(x - 50)(x - 36) = 0   Now 50/50 is 1 and that's not the fraction we are looking for  ( y = 100 - 50 = 50)

So substituting 36 in for x we have y = 100 - 36 = 64     So the fraction is 36/64  Now to verify this is correct add 18 to the numerator and subtract 16 from the denominator and we get ..

54/48  ...       Ok now to verify if this is double the other equation lets reduce the equations.     36/64  4 is the gcd so we get 9/16     and for 54/48 , 3 is the gcd and we get 18/16   and   9/16 * 2 does equal 18/16, so we have the correct answer....