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[TheProphet]

Here's the problem:
"A dealer purchased a number of equally priced articles for a total of cost of £120. She sold all but one of them, making a profit of £1.50 on each article with a total revenue of £135. How many articles did she purchase?"

Here's my solution so far:
number of articles bought = a
price bought at = p
ap = 120
135 = (p + 1.5)(a - 1)
135 = ap - p + 1.5a - 1.5
136.5 = ap + 1.5a - p
136.5 = 120 + 1.5a - p
16.5 = 1.5a - p
11 = a - p

Anything I do after this ends with some ridiculous answer, so what am I doing wrong?

[kaijyuu]

The word problem is a bit vague as to whether the profit counts the unsold article or not. The answer is either 10 or 11 depending on that.

[Darvi]

16.5 = 1.5a - p
11 = a - p

This. It bothers me.

[TheProphet]

The word problem is a bit vague as to whether the profit counts the unsold article or not. The answer is either 10 or 11 depending on that.

I don't think the profit counts the unsold article. How did you get those answers?

[alway]

16.5 = 1.5a - p
11 = a - p

This. It bothers me.

[Simmura McCrea]

16.5 = 1.5a - p
11 = a - p

This. It bothers me.

Possibly because it's wrong. That second line should read 11 = a - (p/1.5)

[Karlito]

ap = 120

...

136.5 = ap + 1.5a - p
136.5 = 120 + 1.5a - p

This bothers me more. If you're going to substitute in for a variable, substitute in for all instances of that variable.

[TheProphet]

16.5 = 1.5a - p
11 = a - p

This. It bothers me.

Possibly because it's wrong. That second line should read 11 = a - (p/1.5)

You know that feeling you get when you couldn't see something right in front of you? I think I just overdosed on it.
Thanks everyone.

[Darvi]

ap = 120

...

136.5 = ap + 1.5a - p
136.5 = 120 + 1.5a - p

This bothers me more. If you're going to substitute in for a variable, substitute in for all instances of that variable.

...there is only one instance of ap so I don't see what you mean.

[zombie urist]

Spoiler: My Solution (click to show/hide)

number of articles bought = a
price bought at = p
ap = 120 #eq 1
135 = (p + 1.5)(a - 1)
135 = ap - p + 1.5a - 1.5 #multiply out
135 = 120 + 1.5a - p - 1.5 #substitute eq 1
16.5 = 1.5a - p #isolate variables
1.5a - p - 16.5=0 #move everything to LHS
1.5a - 120/a -16.5 = 0 #substittute eq 1 again
1.5a^2 - 16.5 a - 120 = 0 #multiply everything by a
a = -5, 16 #quadratic equation
a = 16, p = 7.5 #negative a doesn't make sense for this problem

By the way theres a new maths help thread somewhere.

Edit: http://www.bay12forums.com/smf/index.php?topic=116800.0