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Computer Games Plus needs to get rid of its copies of an old computer game. If it lowers the cost of the old computer game by $5 dollars, it can increase sales of the old computer game by 10 units and still generate exactly $100 of revenue from the old game. How many units of the old computer game did Computer Games Plus sell after implementing the new selling strategy?

Correct Answer:

**C**Set up a set of equations for each selling strategy.

- Assign variables to pieces of the problem:

Let n = number of computer game units sold with the old strategy

Let p = price of computer game per unit with the old strategy - We know that before the new strategy, the total revenue was $100. Based on the number of copies sold at the price of the old strategy:

np = $100. - We know that with the new strategy, the total revenue was $100. Based on 10 more copies sold and a $5 reduction in the price per unit:

(n+10)(p-$5) = $100. - Expand this second equation:

np - 5n + 10p - 50 = 100. - From the first equation, we know that np = $100, so plug that in for np:

100 - 5n + 10p - 50 = 100. - From the first equation, if we divide by n, we end up with p = $100/n. Plug this value in for p:

100 - 5n + 10(100/n) - 50 = 100. - Combine like terms:

50 - 5n + 1000/n = 100. - Subtract 50 from each side:

1000/n - 5n = 50. - Divide by 5:

200/n - n = 10. - Multiply through by n:

200 - n^{2}= 10n - Subtract 10n from both sides:

200 - n^{2}- 10n = 0 - Rearrange:

-n^{2}- 10n + 200 = 0 - Multiply by -1:

n^{2}+ 10n - 200 = 0 - Factor the quadratic:

(n+20)(n-10) = 0 - Thus n= -20, or n=10. Since n cannot be negative, n in this case must be 10. However, do not be tricked into choosing A at this point. We want the number sold with the
*new*strategy, not the old. - n=10 means 10 copies were sold with the old strategy. The new strategy sold 10 more, resulting in a total of 20 copies sold.

n_{new}= n_{old}+ 10 = 10 + 10 = 20

The answer is C.