Question

Ed bought $2,000 in stock shares one week before the stock price dropped $10.00. If he had waited for the price setback, he could have bought 10 more shares for the same amount. How many shares did he buy?

Answer 1

See a solution process below:

Explanation:

We can write a formula for the shares Ed bought as:

`n * p = $2000` Where:

• `n` is the number of shares Ed bought
• `p` is the the price Ed paid per share
• $2000 is the total Ed paid for all the shares

We can write an equation for what would of happened if Ed had waited one week for the price per share to drop $10 per share when he could of bought 10 more shares as:

`(n + 10)(p - $10) = $2000`

Or

`np - $10n + 10p - $100 = $2000`

Step 1" We can solve the first equation for `p`:

`(n * p)/color(red)(n) = ($2000)/color(red)(n)`

`(color(red)(cancel(color(black)(n))) * p)/cancel(color(red)(n)) = ($2000)/n`

`p = ($2000)/n`

Step 2: Substitute `($2000)/n` for `p` in the second equation and solve for `n`:

`np - $10n + 10p - $100 = $2000` becomes:

`(n * ($2000)/n) - $10n + (10 * ($2000)/n) - $100 = $2000`

`(color(red)(cancel(color(black)(n))) * ($2000)/color(red)(cancel(color(black)(n)))) - $10n + ($20000)/n - $100 = $2000`

`$2000 - $10n + ($20000)/n - $100 = $2000`

`$2000 - $100 - $10n + ($20000)/n = $2000`

`$2000 - $100 - $10n + ($20000)/n = $2000`

`$1900 - $10n + ($20000)/n = $2000`

`$1900 - color(red)($1900) - $10n + ($20000)/n = $2000 - color(red)($1900)`

`0 - $10n + ($20000)/n = $100`

`-$10n + ($20000)/n = $100`

`$10(-n + (2000)/n) = $100`

`($10(-n + (2000)/n))/color(red)($10) = ($100)/color(red)($10)`

`(color(red)(cancel(color(black)($10)))(-n + (2000)/n))/cancel(color(red)($10)) = 10`

`-n + (2000)/n = 10`

`color(red)(n)(-n + (2000)/n) = color(red)(n) xx 10`

`(color(red)(n) xx -n) + (color(red)(n) xx (2000)/n) = 10n`

`-n^2 + 2000 = 10n`

`-n^2 - color(red)(10n) + 2000 = 10n - color(red)(10n)`

`-n^2 - 10n + 2000 = 0`

`color(red)(-1)(-n^2 - 10n + 2000) = color(red)(-1) xx 0`

`n^2 + 10n - 2000 = 0`

`(n + 50)(n - 40) = 0`

Solution 1:

`n + 50 = 0`

`n + 50 - color(red)(50) = 0 - color(red)(50)`

`n + 0 = -50`

`n = -50`

Solution 2:

`n - 40 = 0`

`n - 40 + color(red)(40) = 0 + color(red)(40)`

`n - 0 = 40`

`n = 40`

The solution is: Ed bought 40 shares.

Solution 1 is an extraneous solution because Ed could not have bought a negative 50 (or -50) shares.