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Paul bought a number of shares of stock for a total of $3000.
Three months later the stock had increased in value by $5 per share and he sold all
but 50 shares and regained his original investment of $3000. How many shares did he sell?
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Let n be the number of shares Paul sold.
Then the original number of shares he bought was (n+50).
The buying price per share was dollars.
The selling price per share was dollars.
The difference of these prices was 5 dollars, according to the problem,
which gives this "price" equation
- = 5 dollars. (1)
To solve this equation, multiply both sides by n*(n+50), then simplify step by step
3000*(n+50) - 3000*n = 5n*(n+50)
3000n + 150000 - 3000n = 5n^2 + 250n
5n^2 + 250n - 150000 = 0
n^2 + 50n - 30000 = 0
(n+200)*(n-150) = 0
The equation has two roots, -200 and 150.
Of these two roots, we discard the negative one and accept the positive value n= 150.
CHECK. To check, substitute n= 150 into equation (1). You will get
- = 20 - 15 = 5, which is precisely correct.
ANSWER. Paul sold 150 shares.
Solved.