# Fun Vacation to Maine [closed]

Some college friends decided to take a vacation in Maine together. They each had to chip in \$90 for the weeks' lodgings. They tried to convince three more people to go with them, to reduce the cost per person to \$22.50, but they were unsuccessful. How many friends actually want?

• interesting puzzle - I have to ask is the text formatting with italics etc. deliberate? - if so then we will treat it as a clue - if not do you need help formatting your question.
– tom
Oct 30 '18 at 21:52
• looks like this is a textbook-style math question. maybe it belongs over at math.stackexchange.com? Oct 30 '18 at 21:54

I believe that:

$$9$$ people went.

Math:

$$90x=y$$ ($$x$$ people times $$\90$$) = cost of trip
$$(x+3) \times 67.5 = y$$
$$67.5x + 202.5 = 90x$$
$$22.5x = 202.5$$
$$x = 9$$

• looks good, you beat me to it... :-)
– tom
Oct 30 '18 at 21:54
• I think you've got this inverted. The OP says "reduce the cost ... to $22.50," not "reduce the cost ... by$22.50." That makes the answer ten minus the answer you came up with. Oct 31 '18 at 3:40
• It originally said "reduce the cost by". (If you check the edit, you can kind of see it) Oct 31 '18 at 16:10

I believe the number of friends who went to Maine is:

1 (can we really still say 'friends' went to maine??)

The maths:

Let C be the total cost of the vacation
Let X be the number of people attending the vacation

From the question, we have the following equations:
C/X = 90
C/X+3 = 22.5

We can solve this set of simultaneous equations as follows:
C = 90X
90X/X+3 = 22.5
90X = 22.5X + 67.5
(90-22.5)X = 67.5
X = 67.5/67.5
X = 1