Computer Games Plus needs to get rid of its copies of an old computer

Question

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

Source: Platinum GMAT
Kudos for a correct solution.

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

Source: Platinum GMAT

GMATinsight · Answer

Let, Price of Old Computer Game = $x
i.e. Revised Price of Old Computer Game = $(x-5)

Let, Old sale of Computers = y computers
i.e. Revised sales of Old Computers = (y+10) Computers

Previous Revenue = Price of One Computer x Total Computer Sold

i.e. Previous Revenue = x*y
New Revenue = (x-5)*(y+10)

But Reveue remains constant as $100

i.e. x*y = 100-------Equation (1)

and (x-5)*(y+10) = 100
i.e. x*y + 10x - 5y - 50 = 100
i.e. 100 + 10x - 5y = 100+50
i.e. 10x - 5y = 50
i.e. 2x - y = 10
i.e. y = 2x - 10-------Equation (2)

Solving Equation (1) and (2) together
x*y = 100
i.e. x*(2x - 10) = 100
i.e. 2x^2 - 10x - 100 = 0
i.e. x^2 - 5x - 50 = 0
i.e. x^2 - 10x + 5x - 50 = 0
i.e. (x - 10) (x + 5) = 0
i.e. x = 10 (Possible) or -5(Impossible due to being Negative)
i.e. x = 10
i.e. y = 10

i.e. After Implementing the Pricing Strategy, the sale of Computers = (y+10) = 10 + 10 = 20

Answer: Option  C

KS15 · Answer

Using the given options,
Let No of units=10 and rate per unit =$10
Revenue=$100
Now, reduce rate by $5, then in order to keep revenue unchanged, no of units sold=20
Revenue=20*$5=$100
Answer C

lipsi18 · Answer

Assume selling price = x and number of units =u
Therefore x*u=100
And (x-5)*(u+10) =100, solving above equation we get xu-5u+10x =150 (enter above value in this equation) we get, u=2x-10
Enter xu=100 in above equation we get u^2+10u-200 =0, solving for u we get u=10 
So after implementing the change units were(u+10) =20

Hence answer is c
Thanks,

vishwaprakash · Answer

Let x be original price and Y be the original units sold.
A be the new price and B be the New number of Units
XY = AB = 100
A = X-5
B = Y + 10
we have to find B

Lets go by substitution
Assume B = 10 (option A), Y = 0 Not posible

Assume B = 15 (option B) Y = 5, and hence X = 20 (as xy = 100) hence A should = 25 but 25 * 15 = 100 is not true

Assume B = 20 (option C) y = 10, hence X = 10, hence A = 5 20 * 5 = 100 True Hence Option C is correct answer

Like wise we can go for option D and E.

chetan2u · Answer

Hi 
the Q involves a bit of calculation and equations..
i will try this Q by forming equation of only one variable in terms of number of games..

initial number of games=y
price = \frac{100}{y} 
after discount , number=y+10

price =  \frac{100}{y+10}

given that discount is $5,  \frac{100}{y} - \frac{100}{y+10} =5
 \frac{100(y+10)-100y}{y(y+10)}=5 ..
1000=5y(y+10)
200=y(y+10)...
clearly y=10..
total=10+10=20 
ans

Bunuel · Answer

Platinum GMAT Official Solution:

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.

mvictor · Answer

actually, solved it using very few steps.
so (c-5)(n+10)=100. and cn=100.
we have cn+10c-5n-50=cn.
10c-5n=50.
2c-n=10.

now, let's check with the answer choices:
A) 10 - can't be, otherwise n=0. this is illogical, you cannot sell 0, and get 100$. so right away, out.

B) 15
this means that n=5. if we plug in the equation we got above, it means that c is 7.5. but 7.5 * 5 does not equal 100. so out.

C) 20
this means that n=10, and c=10. now, this one works, cn=100, and (c-5)(n+10)=100. since this one works, I did not even look for other answer choices. but let's test...

D) 30
means n=20, and c=15. cn is not 100. so out.

E) 50
means n=40, and c=25. cn is not 100, so out.

KarishmaB · Answer

Using a bit of logic and then looking at the options, we can arrive at the solution very quickly.

The revenue generated is $100 in both cases. Consider the original case. Since the price was later reduced by $5, obviously, initially the price must have been more than $5. This means, initially, the number of units sold must have been less than 20 (to get revenue of $100).
The first obvious choice to try for the original number of units would be 10 then. Say, if there were 10 units sold at $10 each to get revenue of $100, now 20 units will be sold at $5 each and we will get revenue of $100 again!

Answer (C)

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