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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]
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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!
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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]
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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

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Murali.
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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]
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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%.

A is the correct answer

Hope this is clear
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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]
How much time taken to solve this problem?? Its taking above 2 minutes for me. Any suggestions please?
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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]
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
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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]
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: https://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?
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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: https://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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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%

:lol:
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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]
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$)
the shopkeeper is actually at loss of 25%
Hope my doubt is clear.
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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]
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sanket1991 wrote:
VeritasPrepKarishma
Question says "...16 articles to be sold at the cost price of 12 articles..."
cost price is actually - rather should be - marked price?
as per OA and discussions I understood; if C.P. is 1$ 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.
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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]
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Karishma ,
I dont understand this part " 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%". Can you please explain ?
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Monisha2402 wrote:
Karishma ,
I dont understand this part " 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%". Can you please explain ?


Here is another way to think about this if you understood the formula:

Let's say the cost price of each article was $1. So cost of 16 articles would be $16.
Say, there is no mark up and the articles are sold at $1 too. If 16 articles are given away at the price of 12 articles, the selling price is $12.
This means a discount of 4/16 * 100 = 25%

So there are two independent discounts involved: a cash discount of 20% and another discount of 25%. Use successive percentages for that.

(1 + m/100) * (1 - 20/100) * (1 - 25/100) = (1 + 20/100)

(1 + m/100) * (1 - 1/5) * ( 1 - 1/4) = ( 1 + 1/5)

(1 + m/100) = (6/5) * (5/4) * ( 4/3)

m = 100
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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]
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: https://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.


The only confusion in the wording here is "16 articles at the COST PRICE of 12 articles"...if you consider this..it would be a loss...but if its "Selling price" instead of this..kinds of makes sense to me then..please explain..
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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]
lastshot 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 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%

:lol:


Can someone please explain the formula to calculate actual CP :

actual CP will be calculated by 1.2*Actual CP = 60


Regards
Kshitij
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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]
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%


What I don't understand is how is the dealer making a profit of 20% when SP of 16 = CP of 12? Isn't this a loss? Is there something missing in the question?
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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]
Let the listed price of 1 article be x, Thus the listed price of 16 articles is 16x.

The shopkeeper sold 16 articles at the price of 12 articles = 12x
He would have further given a cash discount of 20%, making the final price= 9.6x

We are told that 9.6x = 120% of the cost of 16 articles (Say P).

Thus P= (9.6x*100)/120 = 8x

8x is the cost. Since the articles were listed at 16x, the shopkeeper has listed at 100% above the original rate.
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Re: A dealer offers a cash discount of 20% and still makes a pro [#permalink]
VeritasKarishma 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: https://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,

If we just see that the new set of information is nothing but another discount of 75% provided to the customer, making it a problem of successive discount of 20% and 75% respectively, I think we can solve it faster.

Do you think this approach is right?

Regards,
Deepak Patel
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