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GMAT Practice Question (One of Hundreds)
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.