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A car dealership sold two cars: the first car at a 10% profit and the second car at a 10% loss, which gave them an overall profit of 5% from these two sales. If the dealership's total profit was $1000, what was the cost price of each car?

A. $5,000 and $1,000
B. $9,000 and $5,000
C. $11,000 and $9,000
D. $15,000 and $5,000
E. $20,000 and $10,000
[Reveal] Spoiler: OA

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Official Solution:

A car dealership sold two cars: the first car at a 10% profit and the second car at a 10% loss, which gave them an overall profit of 5% from these two sales. If the dealership's total profit was $1000, what was the cost price of each car?

A. $5,000 and $1,000
B. $9,000 and $5,000
C. $11,000 and $9,000
D. $15,000 and $5,000
E. $20,000 and $10,000


Since 5% profit equals to $1,000 of profit, then total cost price of two cars was $20,000 (\(0.05*x=1,000\) giving \(x=20,000\)).

Now, options C and D both fit. Let's check one of them:

C. $11,000 and $9,000. The profit from the first car would be \(0.1*\$11,000=\$1,100\) and the loss from the second car would be \(0.1*\$9,000=\$900\). So, the overall profit would be \($1,100-$900=$200 \ne $1,000\). So, the answer must be D.


Answer: D
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Is it valid to set this up as a weighted average? If the overall % gain was 5% then we could solve this as:

(+10)x + (-15)y = 0 yielding that x = 3y. The only combination of prices which matches the ratio is (D)

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New post 07 Oct 2014, 03:23
imperfectdark wrote:
Is it valid to set this up as a weighted average? If the overall % gain was 5% then we could solve this as:

(+10)x + (-15)y = 0 yielding that x = 3y. The only combination of prices which matches the ratio is (D)


Yes, that's correct.
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Bunuel wrote:
imperfectdark wrote:
Is it valid to set this up as a weighted average? If the overall % gain was 5% then we could solve this as:

(+10)x + (-15)y = 0 yielding that x = 3y. The only combination of prices which matches the ratio is (D)


Yes, that's correct.


Hi Bunuel, can you elaborate more about how we go about setting up the weighted average equation? Thanks.

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josegf1987 wrote:
Bunuel wrote:
imperfectdark wrote:
Is it valid to set this up as a weighted average? If the overall % gain was 5% then we could solve this as:

(+10)x + (-15)y = 0 yielding that x = 3y. The only combination of prices which matches the ratio is (D)


Yes, that's correct.


Hi Bunuel, can you elaborate more about how we go about setting up the weighted average equation? Thanks.


Hi

this shortcut looks the following way: please the pic.

1. Draw a number line and place values (5 is midpoint, Y is 15 points from the midpoint because it is a loss)
2. The closer to average the greater the weight (X is greater that Y)
3. Configure a proportion where X gets the corresponding value (greater value because weight is greater)

4. X relates to Y just the same way as 15 relates to 5, because X weighs more than Y.

5. Only answer D matches our fraction 15:5
>> !!!

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New post 08 Sep 2015, 00:24
hi

an algebraic solution + back solving:

1. We are given that we derived profit from sale of car 1 that translates into 10% of cost1 or 0.1*cost1. We are given that we encountered loss on a standalone sale of the other car: (10% of cost2) - eventually (-0.1 * cost2)

2. We are also given that the toal profits sums up to: 1000 = 0.1*c1 + (-0.1*c2)

3. Now let us test the values the answers provide to us. I usually start with C: 0.1 of 11 000 = 1 100 (profit) and -0.1 of 9 000 is -900 (loss). The sum is 1 100 - 900 = 200. INCORRECT.

D: 10% of 15 000 is 1 500 (gain). -10% of 5 000 is -500 (loss). This sums up to 1500 - 500 = 1000. CORRECT
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Hello Bunuel,

Thanks for a very good question.

Don't you feel question needs rewording for better clarity:

......Overall profit of 5% from these two sales. If the dealership's total profit was $1000....


While reading through the question I interpreted it at 5% of profits from two sales and Total profits were $1000.

Hence amount of profit from two sales 50 (5% of 1000).

Please suggest.

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New post 27 Jan 2016, 23:28
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.

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New post 27 Jan 2016, 23:34
Erina89 wrote:
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.


Kindly check the discussion above. Also, can you please be a little bit more specific what exactly is unclear in the question/solution?
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New post 27 Jan 2016, 23:50
Hi Bunuel,

sorry I just clicked on the feedback button of the question. I think I probably misunderstood the question stem. Are the 20,000 the total sales price of the cars? And don't I have to calculate the profit/loss of each car like that: 1,1x=15,000 =>~13,636 => 15,000 - 13,636 is the actual profit of this car?

Thanks for your Support.

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New post 28 Jan 2016, 02:18
Erina89 wrote:
Hi Bunuel,

sorry I just clicked on the feedback button of the question. I think I probably misunderstood the question stem. Are the 20,000 the total sales price of the cars? And don't I have to calculate the profit/loss of each car like that: 1,1x=15,000 =>~13,636 => 15,000 - 13,636 is the actual profit of this car?

Thanks for your Support.


Profit and loss are on costs price. So, $20,000 was total cost price of two cars.
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New post 25 Feb 2016, 14:07
With the Allegation method this question takes 2 seconds.

+10____5____-10

The distance from average is 5:15.

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New post 01 Aug 2016, 04:14
My approach was

0.1*x-0.1*y=0.5*(x+y)

.05x=.15y

x=3y

which one of the options proves the equation above? Just D

15=3*5

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New post 01 Aug 2016, 06:06
Bunuel wrote:
Official Solution:


Since 5% profit equals to $1,000 of profit, then total cost price of two cars was $20,000 (\(0.05*x=1,000\) giving \(x=20,000\)).


Hi Bunuel,

Can I suggest an edition to your answer explanation?

The passage above would be more clear if you had used (\(0.05*(x+y)=1,000\) giving \(x+y=20,000\)) then the following "C and D are the only possible answers" would be also more clear

since, x + y = 20
then, 15 + 5 = 20 and 11 + 9 = 20

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New post 09 Aug 2016, 23:24
Bunuel wrote:
Erina89 wrote:
Hi Bunuel,

sorry I just clicked on the feedback button of the question. I think I probably misunderstood the question stem. Are the 20,000 the total sales price of the cars? And don't I have to calculate the profit/loss of each car like that: 1,1x=15,000 =>~13,636 => 15,000 - 13,636 is the actual profit of this car?

Thanks for your Support.


Profit and loss are on costs price. So, $20,000 was total cost price of two cars.


Bunuel, I agree with Erina, that the question stem is quiet confusing. There should be an indication that the profit/loss is on the cost price, and not the sale price, or the question should be somehow restated. The reasoning is that profit, which in this case, I guess, means profit margin, is always on the sale price. I haven't met profit on the cost price, since it doesn't make sense.

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I am doing mathematical approach.

1. 1.1x + 0.9y = 1.05(x+y)
2. 0.05(x+y) = 1,000

if you solve the two equations above, you will get x=15,000 y=5,000

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New post 18 Mar 2017, 22:18
Hello friends, sharing with you the pure algebraic approach.

Let A & B be two cars.

As per the given info, we can deduce two equations.

As per the profit, which is 10% of A - 10% of B = 1000 that is 0.1A - 0.1B = 1000,
Multiplying the above equation we get that A - B = 10,000.

5% of total cost of A & B = 1000. So, 100% of the cost is 20,000.
OR
5/100 = 1000/x, then x=20,000.
OR
0.5(A+B) = 1000, A+B = 20,000.

Now we have two equations A+B = 20,000 & A-B = 10,000
Solving those we get that A = 15,000
That is one of the options so no need to solve for B.

CORRECT ANSWER D.

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New post 20 Mar 2017, 21:25
Hi
I was so confused with the question. $20,000 should be the sale price of both cars (cause including 5% profit). So, the question should be finding the sale price of each car (not sale price). Please help make it clear. :roll:
Thanks

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New post 21 Mar 2017, 04:05
ChiDuong wrote:
Hi
I was so confused with the question. $20,000 should be the sale price of both cars (cause including 5% profit). So, the question should be finding the sale price of each car (not sale price). Please help make it clear. :roll:
Thanks


Profit and loss are on costs price always. So, $20,000 was total cost price of two cars.
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