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M03-32

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Re: M03-32  [#permalink]

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New post 08 May 2017, 11:47
sc6h wrote:
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


For your step 1 you can simplify to:

1.1x + 0.9y = 1.05x + 1.05y
.05x = .15y
x = 3y

Then if you look at the answers, the only possible answer that matches x = 3y is $15k and $5k
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Re: M03-32  [#permalink]

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New post 26 May 2017, 22:07
Bunuel

Other GMAT material also states that Profit/Loss is calculated on Cost Price. This is the way GMAT is tested. But I agree with the others above that this is not correct and completely contradicts what we are taught in Business School, that Profit Margin is ALWAYS calculated on Selling Price.
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Re: M03-32  [#permalink]

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New post 31 May 2017, 09:57
Hi Bunnel,

I don't know if I got lucky but I did this.
Let the cost of first car be x and second be y.
The profit from first car be 0.1x and loss from second be 0.1y.
now it is obvious that the profit of first car is higher than the loss of second car as we are getting a profit of 1000.
Now we know the total profit will be 0.1x-0.1y=1000
=>x-y=10000 and there is only one answer which gives x-y as 10000 that is D.


Regards,
Arpan
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Re: M03-32  [#permalink]

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New post 23 Aug 2017, 02:29
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.
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M03-32  [#permalink]

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New post 27 Aug 2017, 10:31
I still find this question confusing and poorly written. "cost price" is not an understood term. There's cost and there's price. Profit % in the real world is profit $ margin as a percentage of price, not as a percentage of cost. Hence profit can never exceed 100%. What am I missing?
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Re: M03-32  [#permalink]

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New post 08 Oct 2017, 12:00
Hi, I am just curious about a solution path.

If we were to backsolve using choice E:

1.10(20,000)= 22,000

Profit: 2k

.9(10,000)=9,000

Loss: 1k

2k-1k= 1k net profit

Am I missing something?
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Re: M03-32  [#permalink]

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New post 08 Oct 2017, 20:16
georgetags10 wrote:
Hi, I am just curious about a solution path.

If we were to backsolve using choice E:

1.10(20,000)= 22,000

Profit: 2k

.9(10,000)=9,000

Loss: 1k

2k-1k= 1k net profit

Am I missing something?


This is not complete. The profit of $1,000 is not 5% of the total cost, which in case of E is $30,000.
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M03-32  [#permalink]

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New post 15 Nov 2017, 23:45
I approached it a bit differently by exploiting the answer choices.
C1= 10% Profit
C2= 10% loss.
Overall 5% Profit = $1000.
Cost of Each Car = ?

What we need to do is check whether the sum of C1 Profit and C2 Loss is equal to $1000.
Start with Choice C
C1= 11000 -> 10% Profit = 1100
C2+ 9000 -> 10% Loss = -900
Total is not equal to 1000 and we require a bigger C1 number

Exploit Choice D -
C1= 15000 -> 10% Profit = 1500
C2 = 5000 -> 10% Loss = -500
Total is equal to 1000 and its the 5% of total cost. Hence the correct choice.
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Re M03-32  [#permalink]

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New post 26 Dec 2017, 05:47
I think this is a high-quality question and I agree with explanation.
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Re: M03-32  [#permalink]

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New post 23 Mar 2018, 23:53
I tried to click on the feedback button but it didn't work. Anyway, here is my feedback.

I believe that there is either not very clear explanation or the answer is incorrect. You must be clear what is meany by the cost price of each car. Either Final sales price or original Purchase price.
Anyway, We get the 20000 = X*1.1 + Y*0.9
X and Y the purchase prices of cars.
We calculate the profit relative to the purchase price of the car meaning that X*1.1 will be the sales price of the first car, and we sold the second for a 10% loss, meaning that we sold it under the purchase price: Y*0.9

X*1.1 + Y*0.9 = 20000
X+Y=19000
From here x=14500 and y=4500 (purch. prices)
X*1.1 = 15950 (sales price of car X)
Y*0.9 = 4050 (sales price of car Y)

The percent difference should be understood like this:
((Sales Price) - (Purchase price))/(purch. price)
(Y-4500)/4500 = -0.1 -> From here ->
Y-4500=-450
Y=4050
The same for X.

Maybe I didn't understand the task. But anyway you should be more clear on the terminology, COST in most cases refer to the expenses (e.g. "Cost of goods sold" in accounting).
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Re: M03-32  [#permalink]

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New post 16 Aug 2018, 09:56
DJ1986 wrote:
With the Allegation method this question takes 2 seconds.

+10____5____-10

The distance from average is 5:15.


Do you have any videos or tutorials explaining allegation method?
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Re: M03-32  [#permalink]

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New post 27 Oct 2018, 07:04
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)


Just for posterity I think the \((+10)x + (-15)y = 0\) formula is wrong here. Let \(x\) be first car and \(y\) be second car. Then \(10x-10y=5(x+y) \implies 5x = 15y \implies x=3y.\)
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Re: M03-32 &nbs [#permalink] 27 Oct 2018, 07:04

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