Question

At a convention of monsters, 2/5 have no horns, 1/7 have one horn, 1/3 have two horns, and the remaining 26 have three or more horns. How many monsters are attending the convention

Answer 1

I hope this helps.....

Answer 2

Let x be the number of monsters:
a) number of 0 horn = 2x/5
b) number of 1 horn = x/7
c) number of 2 horns = x/3

The 26 remaining monsters are:

(x) - (2x/5 + x/7 + x/3) = 26, solve for x,

x - 2x/5 - x/7 - x/3 = 26, reduce to same denominator:(105)

x - 92x/105 = 2730/105

and x = 210 monsters