GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 19 Mar 2019, 10:08

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# M03-32

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 53706

### Show Tags

16 Sep 2014, 00:21
1
16
00:00

Difficulty:

55% (hard)

Question Stats:

70% (01:26) correct 30% (01:46) wrong based on 141 sessions

A car dealership sold two cars: the first car at a 10% profit and the second car at a 10% loss, which gave them an overall profit of 5% from these two sales. If the dealership's total profit was $1000, what was the cost price of each car? 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 _________________ Math Expert Joined: 02 Sep 2009 Posts: 53706 M03-32 [#permalink] ### Show Tags 16 Sep 2014, 00:21 Official Solution: A car dealership sold two cars: the first car at a 10% profit and the second car at a 10% loss, which gave them an overall profit of 5% from these two sales. If the dealership's total profit was$1000, what was the cost price of each car?

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 5% profit equals to $1,000 of profit, then total cost price of two cars was$20,000 ($$0.05*x=1,000$$ giving $$x=20,000$$).

Now, options C and D both fit. Let's check one of them:

C. $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 $$1,100-900=200 \ne 1,000$$. So, the answer must be D.

_________________
Intern
Joined: 06 Jan 2013
Posts: 11
Location: United States
Concentration: Finance, Economics
GMAT Date: 11-15-2013
GPA: 3.5
WE: Education (Education)

### Show Tags

07 Oct 2014, 02:45
3
1
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)
Math Expert
Joined: 02 Sep 2009
Posts: 53706

### Show Tags

07 Oct 2014, 03:23
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)

Yes, that's correct.
_________________
Intern
Joined: 20 Apr 2015
Posts: 4

### Show Tags

02 Jun 2015, 17:15
1
Bunuel wrote:
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)

Yes, that's correct.

Hi Bunuel, can you elaborate more about how we go about setting up the weighted average equation? Thanks.
Senior Manager
Joined: 12 Aug 2015
Posts: 283
Concentration: General Management, Operations
GMAT 1: 640 Q40 V37
GMAT 2: 650 Q43 V36
GMAT 3: 600 Q47 V27
GPA: 3.3
WE: Management Consulting (Consulting)

### Show Tags

Updated on: 08 Sep 2015, 08:58
2
2
josegf1987 wrote:
Bunuel wrote:
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)

Yes, that's correct.

Hi Bunuel, can you elaborate more about how we go about setting up the weighted average equation? Thanks.

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
>> !!!

You do not have the required permissions to view the files attached to this post.

_________________

KUDO me plenty

Originally posted by shasadou on 08 Sep 2015, 00:10.
Last edited by shasadou on 08 Sep 2015, 08:58, edited 2 times in total.
Senior Manager
Joined: 12 Aug 2015
Posts: 283
Concentration: General Management, Operations
GMAT 1: 640 Q40 V37
GMAT 2: 650 Q43 V36
GMAT 3: 600 Q47 V27
GPA: 3.3
WE: Management Consulting (Consulting)

### Show Tags

08 Sep 2015, 00:24
hi

an algebraic solution + back solving:

1. We are given that we derived profit from sale of car 1 that translates into 10% of cost1 or 0.1*cost1. We are given that we encountered loss on a standalone sale of the other car: (10% of cost2) - eventually (-0.1 * cost2)

2. We are also given that the toal profits sums up to: 1000 = 0.1*c1 + (-0.1*c2)

3. Now let us test the values the answers provide to us. I usually start with C: 0.1 of 11 000 = 1 100 (profit) and -0.1 of 9 000 is -900 (loss). The sum is 1 100 - 900 = 200. INCORRECT.

D: 10% of 15 000 is 1 500 (gain). -10% of 5 000 is -500 (loss). This sums up to 1500 - 500 = 1000. CORRECT
_________________

KUDO me plenty

Verbal Forum Moderator
Joined: 15 Apr 2013
Posts: 181
Location: India
Concentration: General Management, Marketing
GMAT Date: 11-23-2015
GPA: 3.6
WE: Science (Other)

### Show Tags

25 Oct 2015, 01:47
1
Hello Bunuel,

Thanks for a very good question.

Don't you feel question needs rewording for better clarity:

......Overall profit of 5% from these two sales. If the dealership's total profit was $1000.... While reading through the question I interpreted it at 5% of profits from two sales and Total profits were$1000.

Hence amount of profit from two sales 50 (5% of 1000).

Intern
Joined: 04 Sep 2015
Posts: 6

### Show Tags

27 Jan 2016, 23:28
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.
Math Expert
Joined: 02 Sep 2009
Posts: 53706

### Show Tags

27 Jan 2016, 23:34
Erina89 wrote:
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate.

Kindly check the discussion above. Also, can you please be a little bit more specific what exactly is unclear in the question/solution?
_________________
Intern
Joined: 04 Sep 2015
Posts: 6

### Show Tags

27 Jan 2016, 23:50
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?

Math Expert
Joined: 02 Sep 2009
Posts: 53706

### Show Tags

28 Jan 2016, 02:18
Erina89 wrote:
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?

Profit and loss are on costs price. So, $20,000 was total cost price of two cars. _________________ Manager Joined: 05 Jul 2015 Posts: 100 Concentration: Real Estate, International Business GMAT 1: 600 Q33 V40 GPA: 3.3 Re: M03-32 [#permalink] ### Show Tags 25 Feb 2016, 14:07 With the Allegation method this question takes 2 seconds. +10____5____-10 The distance from average is 5:15. Manager Joined: 23 Jun 2009 Posts: 178 Location: Brazil GMAT 1: 470 Q30 V20 GMAT 2: 620 Q42 V33 M03-32 [#permalink] ### Show Tags Updated on: 01 Aug 2016, 06:13 My approach was 0.1*x-0.1*y=0.5*(x+y) .05x=.15y x=3y which one of the options proves the equation above? Just D 15=3*5 Originally posted by felippemed on 01 Aug 2016, 04:14. Last edited by felippemed on 01 Aug 2016, 06:13, edited 1 time in total. Manager Joined: 23 Jun 2009 Posts: 178 Location: Brazil GMAT 1: 470 Q30 V20 GMAT 2: 620 Q42 V33 Re: M03-32 [#permalink] ### Show Tags 01 Aug 2016, 06:06 Bunuel wrote: Official Solution: Since 5% profit equals to$1,000 of profit, then total cost price of two cars was $20,000 ($$0.05*x=1,000$$ giving $$x=20,000$$). Hi Bunuel, Can I suggest an edition to your answer explanation? The passage above would be more clear if you had used ($$0.05*(x+y)=1,000$$ giving $$x+y=20,000$$) then the following "C and D are the only possible answers" would be also more clear since, x + y = 20 then, 15 + 5 = 20 and 11 + 9 = 20 Current Student Joined: 08 Jan 2015 Posts: 76 Re: M03-32 [#permalink] ### Show Tags 09 Aug 2016, 23:24 Bunuel wrote: Erina89 wrote: 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. Profit and loss are on costs price. So,$20,000 was total cost price of two cars.

Bunuel, I agree with Erina, that the question stem is quiet confusing. There should be an indication that the profit/loss is on the cost price, and not the sale price, or the question should be somehow restated. The reasoning is that profit, which in this case, I guess, means profit margin, is always on the sale price. I haven't met profit on the cost price, since it doesn't make sense.
Intern
Joined: 16 Dec 2016
Posts: 2

### Show Tags

22 Dec 2016, 16:20
1
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
Intern
Joined: 15 Dec 2016
Posts: 9
Location: India
GMAT 1: 600 Q44 V28
GPA: 3.5
WE: Marketing (Real Estate)

### Show Tags

18 Mar 2017, 22:18
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.

If it helped you then quickly hit that +Kudos button
_________________

GEARING UP FOR THE GMAT RETAKE.

Intern
Joined: 23 Oct 2013
Posts: 4
WE: Consulting (Law)

### Show Tags

20 Mar 2017, 21:25
Hi
I was so confused with the question. $20,000 should be the sale price of both cars (cause including 5% profit). So, the question should be finding the sale price of each car (not sale price). Please help make it clear. Thanks Math Expert Joined: 02 Sep 2009 Posts: 53706 Re: M03-32 [#permalink] ### Show Tags 21 Mar 2017, 04:05 ChiDuong wrote: Hi I was so confused with the question.$20,000 should be the sale price of both cars (cause including 5% profit). So, the question should be finding the sale price of each car (not sale price). Please help make it clear.
Thanks

Profit and loss are on costs price always. So, \$20,000 was total cost price of two cars.
_________________
Re: M03-32   [#permalink] 21 Mar 2017, 04:05

Go to page    1   2    Next  [ 33 posts ]

Display posts from previous: Sort by

# M03-32

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.