pushpitkc wrote:
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 f=25
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 = 100 - 1000 - 2500 (need not calculate, can't be -50)
B. df + f(d – m) – cp = 100 + 25(-21) - 1000 (need not calculate, can't be -50)
C. d(f – m) + c(m – p) = 4(25-75) + 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.