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M03-32

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Joined: 01 Nov 2016
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Concentration: Technology, Operations
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.
Re M03-32   [#permalink] 26 Dec 2017, 05:47

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