# Question b0d8e

Oct 21, 2015

$200$

#### Explanation:

Your goal here is to wirte two equations with two unknowns, the price per item, $P$, and the number of items, $x$.

Let's say that at the initial price of $P$ you were able to buy a total of $x$ items for $45000$. This means that you can write

$P \cdot x = 45000 \text{ " " } \textcolor{p u r p \le}{\left(1\right)}$

Now the price per item is reduced by 10%. This is equivalent to saying that the new price, ${P}^{'}$, is equal to 90% of the old price.

SInce now $45000$ can buy you 25 items more, you can say that

${P}^{'} \cdot \left(x + 25\right) = 45000$

This is equivalednt to

${\overbrace{\frac{90}{100} \cdot P}}^{\textcolor{b l u e}{= {P}^{'}}} \cdot \left(x + 25\right) = 45000 \text{ " " } \textcolor{p u r p \le}{\left(2\right)}$

Use equation $\textcolor{p u r p \le}{\left(1\right)}$ to find

$P = \frac{45000}{x}$

and use this value in equation $\textcolor{p u r p \le}{\left(2\right)}$ to get

$\frac{90}{100} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{45000}}}}{x} \cdot \left(x + 25\right) = \textcolor{red}{\cancel{\textcolor{b l a c k}{45000}}}$

Solve this equation for $x$ to get

$\frac{90}{100} \cdot \frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}} + \frac{90}{100} \cdot \frac{1}{x} \cdot 25 = 1$

This is equivalent to

$90 x + 2250 = 100 x$

$10 x = 2250 \implies x = \frac{2250}{10} = 225$

The initial number of items is thus equal to $225$. This means that the initial price per item was

$\textcolor{p u r p \le}{\left(1\right)} \implies P = \frac{45000}{225} = \textcolor{g r e e n}{200}$

So, initially you had $225$ items at $200$ each. The price decreased by 10%# to

$\frac{90}{100} \cdot 200 = 180$

which means that you can now buy $250$ items, $25$ more thn you had initially

$180 \cdot \left(225 + 25\right) = 45000$