Problem Solving

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?
A) 10
B) 15
C) 20
D) 30
E) 50

please help me with this one. Thanks!
[spoiler]OA: C[/spoiler]

I just did it by substituting the answer options
lowering the cost by 5 dollars increase the number of units by 10..
20 satisfies the condition.
it says it still made 100 dollars revenue
so initially company wanted to sell 10 units each priced 10 dollars
after lowering the price..number of units is 10. price of each game unit is 5 dollars

nhai2003 wrote: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?
A) 10
B) 15
C) 20
D) 30
E) 50

please help me with this one. Thanks!

As noted, backsolving (working backwards from the choices) is a great approach on this type of question. In general, if the answer choices are numbers and you have a word problem with a simple question, backsolving is a great choice.

Of course, we could also solve algebraically:

x = original number of copies sold
p = original price

xp = (x + 10)(p - 5)

and

xp = 100

So:

100 = xp + 10p - 5x - 50

100 = 100 + 10p - 5x - 50

50 + 5x = 10p

10 + x = 2p

5 + x/2 = p

and subbing in:

x(x/2 + 5) = 100

(1/2)x^2 + 5x - 100 = 0

x^2 + 10x - 200 = 0

(x + 20)(x - 10) = 0

x = -20 or + 10

Negative solutions make no sense, so x=10.

Going back to the question: under the NEW plan, how many are sold:

so, we want (x + 10) = 20... choose (C).

After seeing the algebra, I'm sure you're on board with backsolving as a better approach!

Thanks Stuart!