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

Let's assume that the total monsters are "M"

Now, according to the question

M-[(2M/5)+(M/7)+(M/3)]=26

M-[(42+15+35)/105]M =26

M[1-(92/105)]=26

M(13/105) = 26

M = (26x105)/13

M = 210

So there are 210 monsters in total.

Answer 2

let the total number of monsters be x

$\frac{2}{5} x + \frac{1}{7} x + \frac{1}{3} x + 26 = x \\ \frac{42x + 15x + 35x + 2730}{105} = x \\ 92x + 2730 = x \times 105 \\ 105x - 92x = 2730 \\ 13x = 2730 \\ x = \frac{2730}{13} \\ x = 210 \\ no \: horns = \frac{2}{5}x = \frac{2}{5} \times 210 = 84 \\ one \: horn = \frac{1}{7}x = \frac{1}{7} \times 210 = 30 \\ two \: horns\: = \frac{1}{3} x = \frac{1}{3} \times 210 = 70$

no horns = 84

one horns = 30

two horns = 70

more than one horns = 26

total monsters = 210