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Manager
Joined: 01 Nov 2016
Posts: 58
Concentration: Technology, Operations

Re: M0332
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08 May 2017, 12:47
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



Intern
Joined: 30 Apr 2017
Posts: 11

Re: M0332
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26 May 2017, 23:07
BunuelOther 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.



Intern
Joined: 25 Nov 2016
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Location: India
WE: Web Development (Computer Software)

Re: M0332
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31 May 2017, 10:57
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 0.1x0.1y=1000 =>xy=10000 and there is only one answer which gives xy as 10000 that is D.
Regards, Arpan



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Joined: 17 May 2016
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Location: United States

Re: M0332
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23 Aug 2017, 03:29
1) a,b  prices of cars: 10% of a  10% of b = 1,000 a) 500100 nope b) 900500 nope c) 1100900 nope d) 1500500 =1000 e) 20001000 = 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.



Intern
Joined: 20 Aug 2017
Posts: 38
Location: United States (FL)
GPA: 3.4

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?



Intern
Joined: 26 Sep 2017
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Re: M0332
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08 Oct 2017, 13:00
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
2k1k= 1k net profit
Am I missing something?



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Joined: 02 Sep 2009
Posts: 58410

Re: M0332
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08 Oct 2017, 21:16
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
2k1k= 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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Intern
Joined: 27 Dec 2016
Posts: 12

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.



Intern
Joined: 20 Sep 2012
Posts: 6
Location: Russian Federation
Concentration: Finance, General Management
GPA: 3.5

Re M0332
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26 Dec 2017, 06:47
I think this is a highquality question and I agree with explanation.



Intern
Joined: 23 Mar 2018
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Re: M0332
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24 Mar 2018, 00:53
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) (Y4500)/4500 = 0.1 > From here > Y4500=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).



Manager
Joined: 21 Apr 2018
Posts: 73
Location: India
GMAT 1: 710 Q50 V35 GMAT 2: 750 Q49 V42

Re: M0332
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16 Aug 2018, 10:56
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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Joined: 24 Apr 2016
Posts: 30

Re: M0332
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27 Oct 2018, 08:04
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 \(10x10y=5(x+y) \implies 5x = 15y \implies x=3y.\)



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Joined: 28 Aug 2016
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Re: M0332
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16 Jan 2019, 15:42
lolikons wrote: 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) (Y4500)/4500 = 0.1 > From here > Y4500=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). I have to agree, i also found the question confusing. Usually, profit is % OF Sales or % On cost. In this problem, assuming that x and y are costs, and 1.1x and 0.9 y are sales prices, i would build equations which will not lead me to any of the answer choices (unless i am missing something): 1.05 (x+y) = 1.1 x + 0.9y = 1000  profit as % change on cost price 0.05 (1.1x + 0.9y) = 1000  profit as % of final sales price in my understanding we can't really take 0.05 of initial cost and equate it to profit, we should take 1.05. of initial cost and equate it to profit..



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Re: M0332
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02 Aug 2019, 17:08
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
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