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

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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  B
Joined: 30 Apr 2017
Posts: 11

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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.
Intern  B
Joined: 25 Nov 2016
Posts: 4
Location: India
GMAT 1: 600 Q47 V26 WE: Web Development (Computer Software)

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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
Intern  B
Joined: 17 May 2016
Posts: 13
Location: United States
Schools: AGSM '18

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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.
Intern  B
Joined: 20 Aug 2017
Posts: 38
Location: United States (FL)
Schools: Stanford '20 (D)
GMAT 1: 750 Q49 V42 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  B Joined: 26 Sep 2017 Posts: 3 Re: M03-32 [#permalink] Show Tags 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? Math Expert V Joined: 02 Sep 2009 Posts: 58410 Re: M03-32 [#permalink] Show Tags 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. _________________ Intern  B Joined: 27 Dec 2016 Posts: 12 M03-32 [#permalink] Show Tags 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  S Joined: 20 Sep 2012 Posts: 6 Location: Russian Federation Concentration: Finance, General Management GMAT 1: 710 Q49 V38 GPA: 3.5 Re M03-32 [#permalink] Show Tags I think this is a high-quality question and I agree with explanation. Intern  Joined: 23 Mar 2018 Posts: 1 Re: M03-32 [#permalink] Show Tags 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). Manager  S Joined: 21 Apr 2018 Posts: 73 Location: India GMAT 1: 710 Q50 V35 GMAT 2: 750 Q49 V42 Re: M03-32 [#permalink] Show Tags 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? Intern  B Joined: 24 Apr 2016 Posts: 30 Re: M03-32 [#permalink] Show Tags 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 $$10x-10y=5(x+y) \implies 5x = 15y \implies x=3y.$$ Intern  B Joined: 28 Aug 2016 Posts: 11 Re: M03-32 [#permalink] Show Tags 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) (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). 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.. VP  P Joined: 14 Feb 2017 Posts: 1213 Location: Australia Concentration: Technology, Strategy Schools: LBS '22 GMAT 1: 560 Q41 V26 GMAT 2: 550 Q43 V23 GMAT 3: 650 Q47 V33 GMAT 4: 650 Q44 V36 WE: Management Consulting (Consulting) Re: M03-32 [#permalink] Show Tags 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
_________________
Goal: Q49, V41

+1 Kudos if I have helped you Re: M03-32   [#permalink] 02 Aug 2019, 17:08

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