A car dealer received a shipment of c new cars from the car company at

Total cost \(= c*p=cp\)

total selling price for 1st lot \(= d*f=df\)

Total selling price for 2nd lot \(= (c-d)*m=cm-dm\)

Hence Total selling price \(= df+cm-dm\)

Hence Net Profit \(= df+cm-dm-cp=d(f-m)+c(m-p)\)

Option C

Agree this method is quite cumbersome as already suggested, but its an alternate method

Let the car dealer have ordered c=10 cars each priced at p=100
Hence, the cost price that the cars cost the car dealer was 1000(10*100)

It has been given that the dealer was able to sell d=4 cars with a profit of 25 making f=125
The selling price of the 4 cars is 4*(100+25) = 500
The remaining cars(6 in number) were sold at the a loss of 25 dollars each(m=75)
The selling price of these cars was 6*75 = 450, making the total selling price 450+500 = 950

The selling of the cars leads to a loss of 50, profit is -50

Plugging the values into the equation, we get
A. df – cp – mp = 1250 - 1000 - 2500 (need not calculate, can't be -50)
B. df + f(d – m) – cp = 1250 + 25(-21) - 1000 (need not calculate, can't be -50)
C. d(f – m) + c(m – p) = 4(125-75) + 10(75-100) = 200 + 10(-25) = 200 - 250 = -50

Hence, Option C {d(f – m) + c(m – p)} is the correct answer

Hi,
The highlighted part is not correct:

According to your solution, it should be

d(f – m) + c(m – p) = 4(25-75) + 10(75-100) = - 200 + 10(-25) = -200 - 250 = -450....it does not match the answer you provided above (-50). However, your math is totally correct in example you provided.

I tried many other sets of number which could match choice C.

I wonder where it is wrong in plugging number.

Hi niks18,

The choices has something incorrect.

If you have d cars and f profit for each, so the revenue is not d * f. It must be d (f+P)...Do you agree???

If I have 1 car with price 100 and profit 20.......So revenue would be 1 (20+100)=120 and i is NOT 1 * 20 =20

What do you think???

Total cost for c Cars = \(c∗p\)

The total selling price for d cars = \(d∗f\)

The total selling price remaining cars = \((c−d)∗m\)

Hence Total selling price for all cars = \(df+cm−dm\)

Hence Net Profit = \(df+cm−dm−cp\)

Hence, Option C

Make sense. Thanks for your reply.

Make sense. Thanks for your reply.[/quote]


Hi Mo2men

I think I was also making the same mistake.
If we change the value of f, we should get Option C as the answer.

Have corrected the solution accordingly

The total cost of the c cars was cp dollars.

The revenue made on d cars was df dollars, and the revenue made on the remaining (c - d) cars was (c - d)m = cm - dm dollars. Thus, the total revenue for the c cars was df + cm - dm dollars.

Therefore, the total profit made on selling the c cars was:

(df + cm - dm) - cp = d(f - m) + c(m - p) dollars

Answer: C

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