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A car dealership sold two cars: Car A at a 10% profit and Car B at a 10% loss, resulting into an overall profit of 5% from the two sales. If the dealership's profit from these two cars was $1,000, what was the initial cost of cars A and B?
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
_________________
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
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Official Solution:
A car dealership sold two cars: Car A at a 10% profit and Car B at a 10% loss, resulting into an overall profit of 5% from the two sales. If the dealership's profit from these two cars was $1,000, what was the initial cost of cars A and B?
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 the dealership had an overall profit of 5% and that equals to $1,000, we can determine that the total cost price of the two cars was $20,000 (0.05*x = 1,000 giving x = 20,000).
Among the options, only options C and D give the total cost price of the two cars as $20,000. Let's check one of them:
C. Cost prices of the two cars are given as $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 \($1,100-$900=$200\), which is not equal to the total profit of $1,000. Therefore, option C is incorrect.
So, we are left with only option D, which must be correct.
Answer: D
_________________
A car dealership sold two cars: Car A at a 10% profit and Car B at a 10% loss, resulting into an overall profit of 5% from the two sales. If the dealership's profit from these two cars was $1,000, what was the initial cost of cars A and B?
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 the dealership had an overall profit of 5% and that equals to $1,000, we can determine that the total cost price of the two cars was $20,000 (0.05*x = 1,000 giving x = 20,000).
Among the options, only options C and D give the total cost price of the two cars as $20,000. Let's check one of them:
C. Cost prices of the two cars are given as $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 \($1,100-$900=$200\), which is not equal to the total profit of $1,000. Therefore, option C is incorrect.
So, we are left with only option D, which must be correct.
Answer: D
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shasadou
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josegf1987
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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General Discussion
imperfectdark
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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)
(+10)x + (-15)y = 0 yielding that x = 3y. The only combination of prices which matches the ratio is (D)
imperfectdark
Yes, that's correct.
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Posts: 3
Bunuel
Hi Bunuel, can you elaborate more about how we go about setting up the weighted average equation? Thanks.
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.
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.
Erina89
Profit and loss are on costs price. So, $20,000 was total cost price of two cars.
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With the Allegation method this question takes 2 seconds.
+10____5____-10
The distance from average is 5:15.
+10____5____-10
The distance from average is 5:15.
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.
If it helped you then quickly hit that +Kudos button
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.
If it helped you then quickly hit that +Kudos button
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.
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.
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I think this is a high-quality question and I agree with explanation.
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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
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
Let x be the cost price of both cars
5% of x = 1000,
therefore, x = 20,000
Option C and D both total to 20,000. We just check which one it is.
10% of 15,000 = 1500
10% of 5,000 = 500
Total profit = profit - loss = 1500 - 500 = 1000.
Thats the answer : D.
5% of x = 1000,
therefore, x = 20,000
Option C and D both total to 20,000. We just check which one it is.
10% of 15,000 = 1500
10% of 5,000 = 500
Total profit = profit - loss = 1500 - 500 = 1000.
Thats the answer : D.
TheNightKing
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Bunuel
I used options.
We made an overall Profit. So let's assume that it is more likely that the higher priced model is sold at Profit.
A. $5,000 and $1,000
+500-100=400 Nope
B. $9,000 and $5,000
+900-500=400 Nope
C. $11,000 and $9,000
+1100-900=200 Nope Nope
D. $15,000 and $5,000
1500-500=1000. Okay. Lets make sure we are right.
Total Cost Price=20,000. Profit =1000 which is 5% of 20,000. Looks like we are in good shape.
Bunuel
First car \(=X\)
Second car\(= Y\)
\(1.10X+0.90Y=1.05X+1.05Y\)
\(=0.05X=0.15Y\)
\(=X=3Y\)
From options, only D satisfies this condition. So:
First car \(=$15,000\)
Second car \(= $5,000\)
The answer is \(D\)
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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I think this is a high-quality question and I agree with explanation.
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I think this is a high-quality question and I agree with explanation. This question can be more quickly solved by algebra:
first car price: x, second car price: y
10x/100 -10y/100 = 5 (x+y)/100
on simplifying , x=3y implies price of first car is a multiple of second car and multiple is 3.
quickly glance thru options, only option D works.
first car price: x, second car price: y
10x/100 -10y/100 = 5 (x+y)/100
on simplifying , x=3y implies price of first car is a multiple of second car and multiple is 3.
quickly glance thru options, only option D works.
Posts: 9
I think this is a high-quality question and I agree with explanation. I'm sorry Bunuel, I have read this question solution and the discussion probably 50 times.
I am missing something here.
I understand 0.1a - 0.1b = 1000, so a - b = 10000
What I'm struggling with is the "cost price"
5 percent of TOTAL SALES = 1000, thus TOTAL SALES = 20 000
So how are we getting cost price = 20 000?
I saw Bunuel's response to this twice but I'm still at a loss.
Sorry, and thank you.
I am missing something here.
I understand 0.1a - 0.1b = 1000, so a - b = 10000
What I'm struggling with is the "cost price"
5 percent of TOTAL SALES = 1000, thus TOTAL SALES = 20 000
So how are we getting cost price = 20 000?
I saw Bunuel's response to this twice but I'm still at a loss.
Sorry, and thank you.
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