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# A dealer offers a cash discount of 20% and still makes a pro

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Manager
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A dealer offers a cash discount of 20% and still makes a pro [#permalink]  25 Nov 2010, 07:20
1
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Difficulty:

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Question Stats:

41% (04:02) correct 59% (02:26) wrong based on 168 sessions
A dealer offers a cash discount of 20% and still makes a profit of 20% when he further allows 16 articles to be sold at the cost price of 12 articles to a particular sticky bargainer. How much percent above the cost price were his articles listed?

A. 100%
B. 80%
C. 75%
D. 66+2/3%
E. 55%
[Reveal] Spoiler: OA
Manager
Joined: 08 Oct 2010
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Re: Profit/Loss question [#permalink]  25 Nov 2010, 07:37
Here is an official explanation of solution method from "Math Question Bank For GMAT Winners" at www.my-gmat.com:

Let the CP (cost price) of the article be $x, since he earns a profit of 20%, hence SP (selling price)=1.2x. Loss=[(16-12)/16]*100=25% His selling price=SP*0,75 Now--> SP*0.75=1.2x SP=1.2x/0.75=1.6x This SP is arrived after giving a discount of 20% on MP (marked price). Hence --> MP=1.6x/0.8=2x --> it means that article has been marked 100% above the cost price. Can somebody can suggest an alternative solution method with a better explanation?! I appreciate your help +KUDOS for you! Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5682 Location: Pune, India Followers: 1415 Kudos [?]: 7362 [4] , given: 186 Re: Profit/Loss question [#permalink] 25 Nov 2010, 15:50 4 This post received KUDOS Expert's post 4 This post was BOOKMARKED feruz77 wrote: A dealer offers a cash discount of 20% and still makes a profit of 20% when he further allows 16 articles to be sold at the cost price of 12 articles to a particular sticky bargainer. How much percent above the cost price were his articles listed? a) 100% b) 80% c) 75% d) 66+2/3% e) 55% I have explained the Mark up - Discount - Profit relation here: http://gmatclub.com/forum/discount-problem-104001.html#p810682 This gives us the formula (1 + m/100)(1 - d/100) = (1 + p/100) (1 + m/100)(1 - 20/100) = (1 + 20/100) m = 50 So mark up was 50% in a situation where 12 articles were sold and charged for. i.e. effective mark up turned out to be 50% though his actual mark up was higher since he gave away 16 articles for the cost of 12. Lets say the cost price of each of the 16 articles was$1. Then total cost price was $16 and effective markup was 50% higher on$16 i.e. 16 + 8 = $24. So he charged$24 from the customer. This must have been the marked price on 12 articles. So each article must have a marked price of $2 i.e. a mark up of 100% It is definitely a tricky question. If any parts of the solution are unclear, ask. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Manager
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Re: Profit/Loss question [#permalink]  26 Nov 2010, 02:13
2
KUDOS
1
This post was
BOOKMARKED
In the case of cost/selling/discount stuff problems, just keep ur brain cool and for any problem, note down the below 5 things first.
These 5 points will take u further, I swear.

1) C=The actual price = the COST PRICE. è hoe much did it cost to the seller to get that product into his shop.
2) S=selling price è what is the price that he is selling the product for. Hence if S>C è profit and if S<C è Loss.
3) X=|S-C|, difference (profit/loss) he made by removing the $30, for example, sticker and pasting$40 sticker. After all it is his shop and the product (the ball) is in his shop (court)
4) D=the cheating, but cleverly calculated, discount after he made the changes as specified in point3.
5) GOLDEN POINTS: THE DISCOUNT IS ALWAYS ON “S” – THE SELLING PRICE and THE profit/loss IS ALWAYS ON “C” – THE ACTUAL/COST PRICE

Coming to the original qtn:
Qtn= How much percent above the cost price were his articles listed – this is point 3 above, i.e. asking for X but in terms of percentage.
Hence,the simplified question is à what percent of C is X è X/C * 100 = S-C/C *100 è (S/C – 1) * 100

Given.
Discount of 20% = 1-1/5 = 4/5 times S (as per point5)
Profit of 20% = 1+1/5 = 6/5 TIMES C (as per point5)

Discount of 20% MADE Profit of 20% ( a clever seller, isn’t he)
Hence
New SSelling price = (4/5)S

Not only this. Further more.
he further allows 16 articles to be sold at the cost price of 12 articles to a particular sticky bargainer
 He further reduced the selling price by 12/16
 New S = (12/16)(4/5)S

(12/16)(4/5)S give the profit of (6/5)C
 (12/16)(4/5)S = (6/5)C
 S/C = 2
As the simplified question is à what percent of C is X è X/C * 100 = S-C/C *100 è (S/C – 1) * 100

Answer = (2-1) * 100 = 100%
Or u from S=2C, we can simply say the clever seller has removed the $30 (i.e C), for an example, sticker and pasted the$60 (i.e S=2C) sticker.
 100% over the cost price is listed

Regards,
Murali.
Manager
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Kudos [?]: 144 [5] , given: 73

Re: Profit/Loss question [#permalink]  11 Aug 2013, 07:41
5
KUDOS
Tricky Question - But the key is to, as always, understand the question inside out

Given

Cash Discount - 20%
Profit - 20%
Items Sold - 16
Price Sold at = List Price of 12

Assume List price = $10 Total Invoice =$120
-20% Cash Discount = $96 Let cost price of 16 items be x So total cost = 16*x Given the shopkeeper had a profit of 20% 16 * x * 120/100 = 96 or x =$5

Which means his products were listed at $10 which is a 100% markup over$5

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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]  18 Feb 2014, 08:17
1
KUDOS
feruz77 wrote:
A dealer offers a cash discount of 20% and still makes a profit of 20% when he further allows 16 articles to be sold at the cost price of 12 articles to a particular sticky bargainer. How much percent above the cost price were his articles listed?

A. 100%
B. 80%
C. 75%
D. 66+2/3%
E. 55%

Wow! this one one tough one. Let's see

Let's call 'P' the price and 'C' the cost

We have 0.8P = 1.2C

Now we also know that we had an additional (16-12)/16=4/16=3/4 discount on the price

Therefore (4/5)(3/4)P = 6/5 C

P/C = 2

So they won 100%.

Hope this is clear
Cheers
J
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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]  18 Feb 2014, 21:38
How much time taken to solve this problem?? Its taking above 2 minutes for me. Any suggestions please?
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SVP
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Kudos [?]: 315 [0], given: 355

Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]  27 May 2014, 16:10
PareshGmat wrote:
How much time taken to solve this problem?? Its taking above 2 minutes for me. Any suggestions please?

Yeah around 1.20 for me, but 2 minutes is good for average student
Senior Manager
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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]  23 Sep 2014, 01:07
VeritasPrepKarishma wrote:
feruz77 wrote:
A dealer offers a cash discount of 20% and still makes a profit of 20% when he further allows 16 articles to be sold at the cost price of 12 articles to a particular sticky bargainer. How much percent above the cost price were his articles listed?
a) 100%
b) 80%
c) 75%
d) 66+2/3%
e) 55%

I have explained the Mark up - Discount - Profit relation here: http://gmatclub.com/forum/discount-problem-104001.html#p810682

This gives us the formula (1 + m/100)(1 - d/100) = (1 + p/100)
(1 + m/100)(1 - 20/100) = (1 + 20/100)
m = 50
So mark up was 50% in a situation where 12 articles were sold and charged for. i.e. effective mark up turned out to be 50% though his actual mark up was higher since he gave away 16 articles for the cost of 12.

Lets say the cost price of each of the 16 articles was $1. Then total cost price was$16 and effective markup was 50% higher on $16 i.e. 16 + 8 =$24. So he charged $24 from the customer. This must have been the marked price on 12 articles. So each article must have a marked price of$2 i.e. a mark up of 100%

It is definitely a tricky question. If any parts of the solution are unclear, ask.

Hi Karishma,
I got this far:
VeritasPrepKarishma wrote:
This gives us the formula (1 + m/100)(1 - d/100) = (1 + p/100)
(1 + m/100)(1 - 20/100) = (1 + 20/100)

Since he gave away 16 for the price of 12, I multiplied the left side by 3/4 and got the right result.
My question is how you proceeded from that part:
VeritasPrepKarishma wrote:
Lets say the cost price of each of the 16 articles was $1. Then total cost price was$16 and effective markup was 50% higher on $16 i.e. 16 + 8 =$24. So he charged $24 from the customer. This must have been the marked price on 12 articles. So each article must have a marked price of$2 i.e. a mark up of 100%

Once you got the 24$, why did you calculate how much of a markup it was on the 12 items? I didn't get the transition between the 12 and 16. Can you help with that? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5682 Location: Pune, India Followers: 1415 Kudos [?]: 7362 [0], given: 186 A dealer offers a cash discount of 20% and still makes a pro [#permalink] 23 Sep 2014, 20:41 Expert's post ronr34 wrote: VeritasPrepKarishma wrote: feruz77 wrote: A dealer offers a cash discount of 20% and still makes a profit of 20% when he further allows 16 articles to be sold at the cost price of 12 articles to a particular sticky bargainer. How much percent above the cost price were his articles listed? a) 100% b) 80% c) 75% d) 66+2/3% e) 55% I have explained the Mark up - Discount - Profit relation here: http://gmatclub.com/forum/discount-problem-104001.html#p810682 This gives us the formula (1 + m/100)(1 - d/100) = (1 + p/100) (1 + m/100)(1 - 20/100) = (1 + 20/100) m = 50 So mark up was 50% in a situation where 12 articles were sold and charged for. i.e. effective mark up turned out to be 50% though his actual mark up was higher since he gave away 16 articles for the cost of 12. Lets say the cost price of each of the 16 articles was$1. Then total cost price was $16 and effective markup was 50% higher on$16 i.e. 16 + 8 = $24. So he charged$24 from the customer. This must have been the marked price on 12 articles. So each article must have a marked price of $2 i.e. a mark up of 100% It is definitely a tricky question. If any parts of the solution are unclear, ask. Hi Karishma, I got this far: VeritasPrepKarishma wrote: This gives us the formula (1 + m/100)(1 - d/100) = (1 + p/100) (1 + m/100)(1 - 20/100) = (1 + 20/100) Since he gave away 16 for the price of 12, I multiplied the left side by 3/4 and got the right result. My question is how you proceeded from that part: VeritasPrepKarishma wrote: Lets say the cost price of each of the 16 articles was$1. Then total cost price was $16 and effective markup was 50% higher on$16 i.e. 16 + 8 = $24. So he charged$24 from the customer. This must have been the marked price on 12 articles. So each article must have a marked price of $2 i.e. a mark up of 100% Once you got the 24$, why did you calculate how much of a markup it was on the 12 items? I didn't get the transition between the 12 and 16.
Can you help with that?

Here is the logic I used:

Say, the cost price of each article is $1. We get that the effective mark up is 50%. So the buyer buys products worth$16 which were effectively marked up at $16 + 50% of 16 =$24. So the buyer saw the marked price as $24. But this was the marked price on 12 articles which have a cost price of$12. So actual markup is 100%.

I am not sure what logic you used to multiply the left hand side by 3/4.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 28 Aug 2013 Posts: 101 Location: India Concentration: Operations, Marketing Schools: Insead '14, ISB '15 GMAT Date: 08-28-2014 GPA: 3.86 WE: Supply Chain Management (Manufacturing) Followers: 0 Kudos [?]: 27 [1] , given: 23 Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink] 25 Sep 2014, 00:18 1 This post received KUDOS feruz77 wrote: A dealer offers a cash discount of 20% and still makes a profit of 20% when he further allows 16 articles to be sold at the cost price of 12 articles to a particular sticky bargainer. How much percent above the cost price were his articles listed? A. 100% B. 80% C. 75% D. 66+2/3% E. 55% I find assuming CP (Sticker) as 100 was easier . Total cost of 12 articles = 1200 He offer 20 % discount on it , therefore Selling price = 1200 - 1200*.2 = 960 SP of 16 articles comes out to be 960 , therefore 1 article cost : 60 Now, On selling articles at 60, shopekeeper still manage to make 20% profit , therefore his actual CP will be calculated by 1.2*Actual CP = 60 Actual CP = 50, Difference between value on sticker & actual CP is 100-50 = 50, therefore % increase = 100% _________________ G-prep1 540 --> Kaplan 580-->Veritas 640-->MGMAT 590 -->MGMAT 2 640 --> MGMAT 3 640 ---> MGMAT 4 650 -->MGMAT 5 680 -- >GMAT prep 1 570 Give your best shot...rest leave upto Mahadev, he is the extractor of all negativity in the world !! Manager Joined: 13 Sep 2014 Posts: 71 WE: Engineering (Consulting) Followers: 0 Kudos [?]: 14 [0], given: 42 Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink] 25 Sep 2014, 22:08 VeritasPrepKarishma Question says "...16 articles to be sold at the cost price of 12 articles..." cost price is actually - rather should be - marked price? as per OA and discussions I understood; if C.P. is 1$ then marked price is 2$now if 16 articles are sold at 12$ (C.P. of 12$) the shopkeeper is actually at loss of 25% Hope my doubt is clear. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 5682 Location: Pune, India Followers: 1415 Kudos [?]: 7362 [0], given: 186 Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink] 27 Sep 2014, 03:35 Expert's post sanket1991 wrote: VeritasPrepKarishma Question says "...16 articles to be sold at the cost price of 12 articles..." cost price is actually - rather should be - marked price? as per OA and discussions I understood; if C.P. is 1$ then marked price is 2$now if 16 articles are sold at 12$ (C.P. of 12$) the shopkeeper is actually at loss of 25% Hope my doubt is clear. The questions implies that 16 articles are sold at the price of 12. Cost price here does not mean the price at which the 12 articles were bought. That would be a certain loss. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: A dealer offers a cash discount of 20% and still makes a pro   [#permalink] 27 Sep 2014, 03:35
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