Someone bought pens for 240 in total. If the same money had bought one more pen, each pen would have cost 1 less on average. How many pens were bought?

Step-by-step solution

1. 1

   Choose the unknown that makes the prices easy to write. Let $x$ be the number of pens actually bought. Then the actual average price is

   $\frac{240}{x}$

   Choosing the count rather than the price as the unknown is what keeps the algebra to a single variable.

2. 2

   Express the hypothetical price. With one extra pen for the same money, the count is $x + 1$ and the price becomes

   $\frac{240}{x+1}$

   More pens for the same total means a lower unit price, so this second expression is the smaller of the two.

3. 3

   Turn the comparison into an equation. The hypothetical price is $1$ less than the actual price, so the actual price minus the hypothetical price equals $1$:

   $\frac{240}{x} - \frac{240}{x+1} = 1$

   Getting this subtraction the right way round matters — reversing it gives $-1$ and no positive solution.

4. 4

   Combine the fractions and clear denominators.

   $240\left(\frac{1}{x} - \frac{1}{x+1}\right) = 240 \cdot \frac{(x+1) - x}{x(x+1)} = \frac{240}{x(x+1)} = 1$

   so

   $x(x+1) = 240 \ \Longrightarrow \ x^2 + x - 240 = 0$

5. 5

   Factor and solve the quadratic. Two numbers multiplying to $-240$ and adding to $1$ are $16$ and $-15$:

   $x^2 + 16x - 15x - 240 = x(x+16) - 15(x+16) = (x+16)(x-15) = 0$

   $x = -16 \quad \text{or} \quad x = 15$

6. 6

   Reject the impossible root and check. A count of pens cannot be negative, so $x = -16$ is discarded and $x = 15$. Checking against the story: $240/15 = 16$ per pen, and with one more pen $240/16 = 15$ per pen — exactly $1$ less. So 15 pens were bought.