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A man cheats while buying as well as while selling. While
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VeritasKarishma
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A man cheats while buying as well as while selling. While [#permalink] New post  01 Jul 2012, 21:02
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Responding to a pm:

A man cheats while buying as well as while selling. While buying he takes 10% more than what he pays for and while selling he gives 20% less than what he claims to. Find the profit percent, if he sells at 9.09% below the cost price of the claimed weight.

Options:

A. 19.81%

B. 20%

C. 37.5%

D. 25%

E. 37.5%

In weight questions, try to go one step at a time.
Say the man buys 100 pounds for $100. But he cheats and takes 110 pounds. This means HIS cost price is $10/11 per pound.
While giving 100 pounds, he actually gives only 80 pounds and charges 9.09% less i.e. 1/11 less than the cost price of 100 pounds which is $100. So he sells at 10/11 * 100 = 1000/11
HIS cost price for 80 pounds = 10/11 * 80 = 800/11
HIS selling price for 80 pounds = 1000/11

Profit = (1000/11 - 800/11)/800/11 * 100 = 25%
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Re: Incorrect Weights [#permalink] New post  01 Jul 2012, 21:02
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There is a one step calculation method too. It requires more thought but is faster.
The man takes 10% more than what he pays for. So if he claims to take 100 pounds, he pays $100 but he actually takes 110 pounds for which he will take from the customer $110. Hence, in effect, there is a 10% mark up.
While selling, he sells 20% less. This means, he claims to sell 100 pounds and gets $100 but actually sells only 80 pounds and should have got only $80 for it. So this is again a mark up of $20 on $80 which is 25%.
But he also sells at 9.09% less i.e. gives a discount of 1/11.

(1 + m1%)(1 + m2%)(1 - d%) = (1 + p%)
11/10 * 5/4 * 10/11 = (1 + p%)
profit % = 25%

To understand this formula, see: http://www.veritasprep.com/blog/2011/02 ... nt-profit/
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Re: Incorrect Weights [#permalink] New post  01 Jul 2012, 21:35
second approach seems to be easy after reading your blog post.
i am still having a hard time while trying to understnad this line
'While giving 100 pounds, he actually gives only 80 pounds and charges 9.09% less i.e. 1/11 less than the cost price of 100 pounds which is $100. '
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Re: Incorrect Weights [#permalink] New post  01 Jul 2012, 21:49
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321kumarsushant wrote:
second approach seems to be easy after reading your blog post.
i am still having a hard time while trying to understnad this line
'While giving 100 pounds, he actually gives only 80 pounds and charges 9.09% less i.e. 1/11 less than the cost price of 100 pounds which is $100. '


The man claims he is giving you 100 pounds i.e. selling you 100 pounds of product. He is supposedly giving a discount of 9.09% i.e. charging 1/11 less than the cost price of 100 pounds which is $100. So he is charging only 10/11 of $100 = $1000/11 from you for giving you "100 pounds". But actually, by cheating, he gave you only 80 pounds. So he charged you $1000/11 for 80 pounds.
Think of a fruit vendor you go to to buy some peaches. He says the cost price of the peaches is $1 per pound. But he is willing to give you a discount of 9.09%. But instead of 100 pounds of peaches, he weighs only 80 pounds and gives you less. Though he charges you for 100 pounds. How much does he charge?
9.09% less than 100 pounds i.e. 1/11 less than 100 pounds i.e. 10/11 * 100 pounds = 1000/11

Check out this post for conversion of percentages to fractions: http://www.veritasprep.com/blog/2011/02 ... rcentages/
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Re: A man cheats while buying as well as while selling. While [#permalink] New post  24 Jul 2014, 10:18
VeritasPrepKarishma wrote:
There is a one step calculation method too. It requires more thought but is faster.
The man takes 10% more than what he pays for. So if he claims to take 100 pounds, he pays $100 but he actually takes 110 pounds for which he will take from the customer $110. Hence, in effect, there is a 10% mark up.
While selling, he sells 20% less. This means, he claims to sell 100 pounds and gets $100 but actually sells only 80 pounds and should have got only $80 for it. So this is again a mark up of $20 on $80 which is 25%.
But he also sells at 9.09% less i.e. gives a discount of 1/11.
(1 + m1%)(1 + m2%)(1 - d%) = (1 + p%)
11/10 * 5/4 * 10/11 = (1 + p%)
profit % = 25%

To understand this formula, see: http://www.veritasprep.com/blog/2011/02 ... nt-profit/

Hi

How you got 1/11???highlighted part???
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Re: A man cheats while buying as well as while selling. While [#permalink] New post  24 Jul 2014, 19:53
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GGMAT760 wrote:
VeritasPrepKarishma wrote:
There is a one step calculation method too. It requires more thought but is faster.
The man takes 10% more than what he pays for. So if he claims to take 100 pounds, he pays $100 but he actually takes 110 pounds for which he will take from the customer $110. Hence, in effect, there is a 10% mark up.
While selling, he sells 20% less. This means, he claims to sell 100 pounds and gets $100 but actually sells only 80 pounds and should have got only $80 for it. So this is again a mark up of $20 on $80 which is 25%.
But he also sells at 9.09% less i.e. gives a discount of 1/11.
(1 + m1%)(1 + m2%)(1 - d%) = (1 + p%)
11/10 * 5/4 * 10/11 = (1 + p%)
profit % = 25%

To understand this formula, see: http://www.veritasprep.com/blog/2011/02 ... nt-profit/

Hi

How you got 1/11???highlighted part???


This is given in the question:
"if he sells at 9.09% below the cost price of the claimed weight."

9.09% written in fraction form is 9.09/100 = 1/11

You should know the basic percentage to fraction conversions given here: http://www.veritasprep.com/blog/2011/02 ... rcentages/
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Re: A man cheats while buying as well as while selling. While [#permalink] New post  25 Jul 2014, 01:09
VeritasPrepKarishma wrote:
GGMAT760 wrote:
VeritasPrepKarishma wrote:
There is a one step calculation method too. It requires more thought but is faster.
The man takes 10% more than what he pays for. So if he claims to take 100 pounds, he pays $100 but he actually takes 110 pounds for which he will take from the customer $110. Hence, in effect, there is a 10% mark up.
While selling, he sells 20% less. This means, he claims to sell 100 pounds and gets $100 but actually sells only 80 pounds and should have got only $80 for it. So this is again a mark up of $20 on $80 which is 25%.
But he also sells at 9.09% less i.e. gives a discount of 1/11.
(1 + m1%)(1 + m2%)(1 - d%) = (1 + p%)
11/10 * 5/4 * 10/11 = (1 + p%)
profit % = 25%

To understand this formula, see: http://www.veritasprep.com/blog/2011/02 ... nt-profit/

Hi

How you got 1/11???highlighted part???


This is given in the question:
"if he sells at 9.09% below the cost price of the claimed weight."

9.09% written in fraction form is 9.09/100 = 1/11

You should know the basic percentage to fraction conversions given here: http://www.veritasprep.com/blog/2011/02 ... rcentages/


Needless to say its a good question

9.09% = 1/11 .... This approximation makes the life easier to compute; else it was not easy to calculate

+1 Kudos to you :)
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Re: A man cheats while buying as well as while selling. While [#permalink] New post  02 Jul 2015, 05:48
Hi Bunuel,

Could you please show any easier way of solving this question ?

Thanks.
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Re: A man cheats while buying as well as while selling. While [#permalink] New post  02 Jul 2015, 05:54
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Swaroopdev wrote:
Hi Bunuel,

Could you please show any easier way of solving this question ?

Thanks.


There are perfect solutions from Karishma above. Please go through them and ask if anything remains unclear.
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Re: A man cheats while buying as well as while selling. While [#permalink] New post  08 Oct 2015, 03:27
VeritasPrepKarishma wrote:
There is a one step calculation method too. It requires more thought but is faster.
The man takes 10% more than what he pays for. So if he claims to take 100 pounds, he pays $100 but he actually takes 110 pounds for which he will take from the customer $110. Hence, in effect, there is a 10% mark up.
While selling, he sells 20% less. This means, he claims to sell 100 pounds and gets $100 but actually sells only 80 pounds and should have got only $80 for it. So this is again a mark up of $20 on $80 which is 25%.
But he also sells at 9.09% less i.e. gives a discount of 1/11.

(1 + m1%)(1 + m2%)(1 - d%) = (1 + p%)
11/10 * 5/4 * 10/11 = (1 + p%)
profit % = 25%

To understand this formula, see: http://www.veritasprep.com/blog/2011/02 ... nt-profit/

can you please clear about (1+m2)= 5/4?? i can not understand fro m where did u get this.
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Re: A man cheats while buying as well as while selling. While [#permalink] New post  08 Oct 2015, 20:43
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anik19890 wrote:
VeritasPrepKarishma wrote:
There is a one step calculation method too. It requires more thought but is faster.
The man takes 10% more than what he pays for. So if he claims to take 100 pounds, he pays $100 but he actually takes 110 pounds for which he will take from the customer $110. Hence, in effect, there is a 10% mark up.
While selling, he sells 20% less. This means, he claims to sell 100 pounds and gets $100 but actually sells only 80 pounds and should have got only $80 for it. So this is again a mark up of $20 on $80 which is 25%.
But he also sells at 9.09% less i.e. gives a discount of 1/11.

(1 + m1%)(1 + m2%)(1 - d%) = (1 + p%)
11/10 * 5/4 * 10/11 = (1 + p%)
profit % = 25%

To understand this formula, see: http://www.veritasprep.com/blog/2011/02 ... nt-profit/

can you please clear about (1+m2)= 5/4?? i can not understand fro m where did u get this.


There are two mark ups in this question -
the first mark up of 10% (he takes 110 pounds for which he will get $110 though he pays only $100 for it - that is a 10% mark up) - this gives you 1 + 10/100 = 11/10
the second mark up of 25% (he gives only 80 pounds for which he had paid $80 but charges $100 for it - the markup of $20 on $80 cost price is a 25% mark up) - this gives you (1 + 25/100) = 5/4
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Re: A man cheats while buying as well as while selling. While [#permalink] New post  27 Oct 2016, 06:54
Hi Karishma,

Appreciate your thoughts on my approach.

Let Actual cp=100
Cheated cost price – 110 [ 10% more]
Selling price – 1000/11 [9.09 % less]
Actual selling price – 80 [20% less than actual]
Overall profit = 10[110-100]+120/11 [1000/11-80] =230/11

Profit -230/11, cp-100 => profit % = 20.9 approx 21%

what's wrong in my approach.
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Re: A man cheats while buying as well as while selling. While [#permalink] New post  27 Oct 2016, 21:44
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SriramK wrote:
Hi Karishma,

Appreciate your thoughts on my approach.

Let Actual cp=100
Cheated cost price – 110 [ 10% more]
Selling price – 1000/11 [9.09 % less]
Actual selling price – 80 [20% less than actual]
Overall profit = 10[110-100]+120/11 [1000/11-80] =230/11

Profit -230/11, cp-100 => profit % = 20.9 approx 21%

what's wrong in my approach.


The question says that he takes 10% more than what he pays for. This means that if he pays $100 for 100 gms, but takes 110 gms instead.
So the cost price is actually 100/110 per gram = $90.9 per 100 gms
Cheated cost price is not 110. It is 90.9

Similarly, think about selling price too.
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Re: A man cheats while buying as well as while selling. While [#permalink] New post  13 Nov 2016, 12:45
[quote="VeritasPrepKarishma"]Responding to a pm:

A man cheats while buying as well as while selling. While buying he takes 10% more than what he pays for and while selling he gives 20% less than what he claims to. Find the profit percent, if he sells at 9.09% below the cost price of the claimed weight.

Options:

A. 19.81%

B. 20%

C. 37.5%

D. 25%

E. 37.5%

he makes a profit on buying and a profit from selling and he gives a fake discount

profit on buying

if 1 buys 1.1 then 1000/11 buys 1

100%-1000/11% = 100/11% = 9.09% approx


profit on selling ( form cheating) by decreasing weight by 20%

if 1 buys 0.8 thus 1.25 buys 1

125%-100% = 25%

fake discount = -9.09% on the deceiving cost of 1

sum the above 3 we get 25%
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Re: A man cheats while buying as well as while selling. While [#permalink] New post  21 Jan 2017, 18:12
VeritasPrepKarishma wrote:
Responding to a pm:

A man cheats while buying as well as while selling. While buying he takes 10% more than what he pays for and while selling he gives 20% less than what he claims to. Find the profit percent, if he sells at 9.09% below the cost price of the claimed weight.

Options:

A. 19.81%

B. 20%

C. 37.5%

D. 25%

E. 37.5%

In weight questions, try to go one step at a time.
Say the man buys 100 pounds for $100. But he cheats and takes 110 pounds. This means HIS cost price is $10/11 per pound.
While giving 100 pounds, he actually gives only 80 pounds and charges 9.09% less i.e. 1/11 less than the cost price of 100 pounds which is $100. So he sells at 10/11 * 100 = 1000/11
HIS cost price for 80 pounds = 10/11 * 80 = 800/11
HIS selling price for 80 pounds = 1000/11

Profit = (1000/11 - 800/11)/800/11 * 100 = 25%

Option C and E are the same. You should edit them.
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Re: A man cheats while buying as well as while selling. While [#permalink] New post  21 Jan 2017, 22:40
It helps to know the fractional equivalents. One can then proceed by taking convenient numbers. So, say the man paid for 1000 units and got 1100 units. His CP=10/11. Now he claims to sell 1100 units but only sells 880 units (20% less) and at a price 1/11th less (9.09%). So his SP is 1000, while he's incurring a cost of 880*10/11=800. So his overall profit= (1000-800)/800=1/4=25%.
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Re: A man cheats while buying as well as while selling. While [#permalink] New post  17 Feb 2017, 15:15
th23autolink_encode_start_318yq1msW3VybD1odHRwcyYjNTg7Ly9nbWF0Y2x1YiYjNDY7Y29tJiM1ODs0NDMvZm9ydW0vbWVtYmVybGlzdCYjNDY7cGhwP21vZGU9dmlld3Byb2ZpbGUmYW1wO3VuPVZlcml0YXNQcmVwS2FyaXNobWE6MzE4eXExbXNdIHRoMjNhdXRvbGlua19iYmNvZGVfc3RhcnRfYl8zMTh5cTFtcyBWZXJpdGFzUHJlcEthcmlzaG1hIHRoMjNhdXRvbGlua19iYmNvZGVfZW5kX2JfMzE4eXExbXMgWy91cmw6MzE4eXExbXNdth23autolink_encode_end_318yq1ms wrote:
Responding to a pm:

A man cheats while buying as well as while selling. While buying he takes 10% more than what he pays for and while selling he gives 20% less than what he claims to. Find the profit percent, if he sells at 9.09% below the cost price of the claimed weight.

Options:

A. 19.81%

B. 20%

C. 37.5%

D. 25%

E. 37.5%

In weight questions, try to go one step at a time.
Say the man buys 100 pounds for $100. But he cheats and takes 110 pounds. This means HIS cost price is $10/11 per pound.
While giving 100 pounds, he actually gives only 80 pounds and charges 9.09% less i.e. 1/11 less than the cost price of 100 pounds which is $100. So he sells at 10/11 * 100 = 1000/11
HIS cost price for 80 pounds = 10/11 * 80 = 800/11
HIS selling price for 80 pounds = 1000/11

Profit = (1000/11 - 800/11)/800/11 * 100 = 25%


Hi VeritasPrepKarishma ,

As I have understood , you got the man's REAL cost price i.e 10/11 per pound .

His Cost price for 80 kgs = 80 * 10/11
But his selling price for 80ks = 100 * 10/11 ( because he says its not 80 its hundred) .



And then you have calculated the difference . Where does the 9.09% factor in ?


shouldnt his selling price be = 100*10/11 - 9.09% [ 100*10/11] . which is not 100*10/11 . therefore changing the answer.

I spent quite a lot of time to understand this, any help will be appreciated.
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A man cheats while buying as well as while selling. While [#permalink] New post  26 Feb 2017, 20:13
See it this way

When the merchant buys material

what he pays for - 100 gm
what he actually gets - 110 gm


when merchant sells
what he charges for - 100 gm
what he gives - 80 gm of material

So in a single transaction he buys 110 gm for a certain amount and sells 80 for the same amount.
The difference 110-80 is what accounts for his profit. This is without considering the discount he gives


Now if i consider the discount also
discount is 9.09%

suppose cost per unit is X

for 91x he sold 80 gm
for 100x he will sell (80/91) * 100 = 88gm

So he bought 110 gm for certain amount and sold 88 gm for the same amount so his net gain is 110-88 =22gm
so for 88 gm sold he made 22gm profit

profit = 22/88 * 100 = 25%
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Re: A man cheats while buying as well as while selling. While [#permalink] New post  01 Mar 2017, 18:02
VeritasPrepKarishma wrote:
There is a one step calculation method too. It requires more thought but is faster.
The man takes 10% more than what he pays for. So if he claims to take 100 pounds, he pays $100 but he actually takes 110 pounds for which he will take from the customer $110. Hence, in effect, there is a 10% mark up.
While selling, he sells 20% less. This means, he claims to sell 100 pounds and gets $100 but actually sells only 80 pounds and should have got only $80 for it. So this is again a mark up of $20 on $80 which is 25%.
But he also sells at 9.09% less i.e. gives a discount of 1/11.

(1 + m1%)(1 + m2%)(1 - d%) = (1 + p%)
11/10 * 5/4 * 10/11 = (1 + p%)
profit % = 25%

To understand this formula, see: http://www.veritasprep.com/blog/2011/02 ... nt-profit/

Lovely way to solve... Thanks for the formula
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Re: A man cheats while buying as well as while selling. While [#permalink] New post  26 May 2017, 00:31
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VeritasPrepKarishma wrote:
There is a one step calculation method too. It requires more thought but is faster.
The man takes 10% more than what he pays for. So if he claims to take 100 pounds, he pays $100 but he actually takes 110 pounds for which he will take from the customer $110. Hence, in effect, there is a 10% mark up.
While selling, he sells 20% less. This means, he claims to sell 100 pounds and gets $100 but actually sells only 80 pounds and should have got only $80 for it. So this is again a mark up of $20 on $80 which is 25%.
But he also sells at 9.09% less i.e. gives a discount of 1/11.

(1 + m1%)(1 + m2%)(1 - d%) = (1 + p%)
11/10 * 5/4 * 10/11 = (1 + p%)
profit % = 25%

To understand this formula, see: http://www.veritasprep.com/blog/2011/02 ... nt-profit/


Responding to a pm:
Quote:
can u please clear me ..what if the question will ask for his Profit/loss percent"??

ie how can we know whether he is finally at profit or loss so that we have RHS part as (1+p%)


In case there is a loss, you will simply get a negative value for p.

(1 + m%)(1 - d%) = (1 + p%)

Say there is a mark up of 10% but a discount is given of 20%.

(1 + 10/100)*(1 - 20/100) = (1 + p%)

(11/10) * (4/5) = (1 + p%)

44/50 = 88/100 = 1 + p/100

p/100 = -12/100

p = -12
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