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Math Expert V
Joined: 02 Sep 2009
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A car dealer received a shipment of c new cars from the car company at  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 54% (02:35) correct 46% (02:33) wrong based on 49 sessions

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A car dealer received a shipment of c new cars from the car company at a base price of p dollars per car. If the dealer was able to sell d cars at a profit for f dollars each, but had to sell the rest of the cars at a loss for m dollars each, which of the following represents the total profit generated on the sales of the automobiles?

A. df – cp – mp
B. df + f(d – m) – cp
C. d(f – m) + c(m – p)
D. cp – df – f(c – d)
E. df – m(c – d) – cm

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A car dealer received a shipment of c new cars from the car company at  [#permalink]

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Bunuel wrote:
A car dealer received a shipment of c new cars from the car company at a base price of p dollars per car. If the dealer was able to sell d cars at a profit for f dollars each, but had to sell the rest of the cars at a loss for m dollars each, which of the following represents the total profit generated on the sales of the automobiles?

A. df – cp – mp
B. df + f(d – m) – cp
C. d(f – m) + c(m – p)
D. cp – df – f(c – d)
E. df – m(c – d) – cm

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
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Re: A car dealer received a shipment of c new cars from the car company at  [#permalink]

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Bunuel wrote:
A car dealer received a shipment of c new cars from the car company at a base price of p dollars per car. If the dealer was able to sell d cars at a profit for f dollars each, but had to sell the rest of the cars at a loss for m dollars each, which of the following represents the total profit generated on the sales of the automobiles?

A. df – cp – mp
B. df + f(d – m) – cp
C. d(f – m) + c(m – p)
D. cp – df – f(c – d)
E. df – m(c – d) – cm

Too many variables so plugging in will be too confusing.

Focus on what you need.

Total profit = Revenue - Cost
Revenue = df + (c - d)m = df + cm - dm
Cost = cp

Total profit = df + cm - dm - cp

Option (C) matches.
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Senior PS Moderator V
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A car dealer received a shipment of c new cars from the car company at  [#permalink]

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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
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A car dealer received a shipment of c new cars from the car company at  [#permalink]

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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.
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A car dealer received a shipment of c new cars from the car company at  [#permalink]

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niks18 wrote:
Bunuel wrote:
A car dealer received a shipment of c new cars from the car company at a base price of p dollars per car. If the dealer was able to sell d cars at a profit for f dollars each, but had to sell the rest of the cars at a loss for m dollars each, which of the following represents the total profit generated on the sales of the automobiles?

A. df – cp – mp
B. df + f(d – m) – cp
C. d(f – m) + c(m – p)
D. cp – df – f(c – d)
E. df – m(c – d) – cm

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

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???
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Posts: 7953
Re: A car dealer received a shipment of c new cars from the car company at  [#permalink]

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1
Mo2men wrote:
niks18 wrote:
Bunuel wrote:
A car dealer received a shipment of c new cars from the car company at a base price of p dollars per car. If the dealer was able to sell d cars at a profit for f dollars each, but had to sell the rest of the cars at a loss for m dollars each, which of the following represents the total profit generated on the sales of the automobiles?

A. df – cp – mp
B. df + f(d – m) – cp
C. d(f – m) + c(m – p)
D. cp – df – f(c – d)
E. df – m(c – d) – cm

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

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???

Hi..
The choices are ok..

If the statement read... d cars were sold at a profit OF f dollars meant profit was fd.

But if statement reads ... d cars were sold at a profit FOR f dollars MEANS revenue is fd and f is more than the base price
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Re: A car dealer received a shipment of c new cars from the car company at  [#permalink]

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Bunuel wrote:
A car dealer received a shipment of c new cars from the car company at a base price of p dollars per car. If the dealer was able to sell d cars at a profit for f dollars each, but had to sell the rest of the cars at a loss for m dollars each, which of the following represents the total profit generated on the sales of the automobiles?

A. df – cp – mp
B. df + f(d – m) – cp
C. d(f – m) + c(m – p)
D. cp – df – f(c – d)
E. df – m(c – d) – cm

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
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Posts: 2341
Re: A car dealer received a shipment of c new cars from the car company at  [#permalink]

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chetan2u wrote:
Hi..
The choices are ok..

If the statement read... d cars were sold at a profit OF f dollars meant profit was fd.

But if statement reads ... d cars were sold at a profit FOR f dollars MEANS revenue is fd and f is more than the base price

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Posts: 3333
Location: India
GPA: 3.12
Re: A car dealer received a shipment of c new cars from the car company at  [#permalink]

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Mo2men wrote:
chetan2u wrote:
Hi..
The choices are ok..

If the statement read... d cars were sold at a profit OF f dollars meant profit was fd.

But if statement reads ... d cars were sold at a profit FOR f dollars MEANS revenue is fd and f is more than the base price

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
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Re: A car dealer received a shipment of c new cars from the car company at  [#permalink]

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Bunuel wrote:
A car dealer received a shipment of c new cars from the car company at a base price of p dollars per car. If the dealer was able to sell d cars at a profit for f dollars each, but had to sell the rest of the cars at a loss for m dollars each, which of the following represents the total profit generated on the sales of the automobiles?

A. df – cp – mp
B. df + f(d – m) – cp
C. d(f – m) + c(m – p)
D. cp – df – f(c – d)
E. df – m(c – d) – cm

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

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