Bunuel

Other GMAT material also states that Profit/Loss is calculated on Cost Price. This is the way GMAT is tested. But I agree with the others above that this is not correct and completely contradicts what we are taught in Business School, that Profit Margin is ALWAYS calculated on Selling Price.

1) a,b - prices of cars:
10% of a - 10% of b = 1,000
a) 500-100 nope
b) 900-500 nope
c) 1100-900 nope
d) 1500-500 =1000
e) 2000-1000 = 1000 .
So we have d and e left.
2) 0,05 (a+b) = 1000:
d) 0,05 (20k) = 1k OK
e) 0,05 (30k) = 1.5 Nope.
hence d.

Hi, I am just curious about a solution path.

If we were to backsolve using choice E:

1.10(20,000)= 22,000

Profit: 2k

.9(10,000)=9,000

Loss: 1k

2k-1k= 1k net profit

Am I missing something?

I approached it a bit differently by exploiting the answer choices.
C1= 10% Profit
C2= 10% loss.
Overall 5% Profit = $1000.
Cost of Each Car = ?

What we need to do is check whether the sum of C1 Profit and C2 Loss is equal to $1000.
Start with Choice C
C1= 11000 -> 10% Profit = 1100
C2+ 9000 -> 10% Loss = -900
Total is not equal to 1000 and we require a bigger C1 number

Exploit Choice D -
C1= 15000 -> 10% Profit = 1500
C2 = 5000 -> 10% Loss = -500
Total is equal to 1000 and its the 5% of total cost. Hence the correct choice.

I tried to click on the feedback button but it didn't work. Anyway, here is my feedback.

I believe that there is either not very clear explanation or the answer is incorrect. You must be clear what is meany by the cost price of each car. Either Final sales price or original Purchase price.
Anyway, We get the 20000 = X*1.1 + Y*0.9
X and Y the purchase prices of cars.
We calculate the profit relative to the purchase price of the car meaning that X*1.1 will be the sales price of the first car, and we sold the second for a 10% loss, meaning that we sold it under the purchase price: Y*0.9

X*1.1 + Y*0.9 = 20000
X+Y=19000
From here x=14500 and y=4500 (purch. prices)
X*1.1 = 15950 (sales price of car X)
Y*0.9 = 4050 (sales price of car Y)

The percent difference should be understood like this:
((Sales Price) - (Purchase price))/(purch. price)
(Y-4500)/4500 = -0.1 -> From here ->
Y-4500=-450
Y=4050
The same for X.

Maybe I didn't understand the task. But anyway you should be more clear on the terminology, COST in most cases refer to the expenses (e.g. "Cost of goods sold" in accounting).

Just for posterity I think the \((+10)x + (-15)y = 0\) formula is wrong here. Let \(x\) be first car and \(y\) be second car. Then \(10x-10y=5(x+y) \implies 5x = 15y \implies x=3y.\)

I really like HimJ 's approach to this. It's much simpler than the official explanation or any weighted average type as its the type of logic I'd apply in business.

For clarification:
We can derive two equations

The first from profit
Let A and B represent the cost price

0.1A - 0.1B = 1000 -> 10% profit on cost price of A less 10% loss on cost price of B
A - B = 10,000

0.05 (A+B) = 1,000 -> we are told that the overall profit of $1000 represents 5%
A+B = 20,000 -> from simplifying

Combining the equations
2A = 30,000
A = 15,000

Substitute back in, B = 5000
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