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A man cheats while buying as well as while selling. While
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01 Jul 2012, 22:02
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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
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01 Jul 2012, 22:02
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 ... ntprofit/
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Re: Incorrect Weights
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01 Jul 2012, 22:35
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
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01 Jul 2012, 22:49
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
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24 Jul 2014, 11:18
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 ... ntprofit/Hi How you got 1/11???highlighted part???



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Re: A man cheats while buying as well as while selling. While
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24 Jul 2014, 20:53
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 ... ntprofit/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
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25 Jul 2014, 02:09
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 ... ntprofit/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
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02 Jul 2015, 06:48
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
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02 Jul 2015, 06:54
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
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08 Oct 2015, 04:27
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 ... ntprofit/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
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08 Oct 2015, 21:43
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 ... ntprofit/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
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27 Oct 2016, 07:54
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[110100]+120/11 [1000/1180] =230/11
Profit 230/11, cp100 => 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
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27 Oct 2016, 22:44
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[110100]+120/11 [1000/1180] =230/11
Profit 230/11, cp100 => 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
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13 Nov 2016, 13:45
[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
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21 Jan 2017, 19:12
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
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21 Jan 2017, 23:40
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= (1000800)/800=1/4=25%.
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Re: A man cheats while buying as well as while selling. While
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17 Feb 2017, 16:15
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
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26 Feb 2017, 21:13
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 11080 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 11088 =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
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01 Mar 2017, 19:02
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 ... ntprofit/Lovely way to solve... Thanks for the formula



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Re: A man cheats while buying as well as while selling. While
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26 May 2017, 01:31
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 ... ntprofit/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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Re: A man cheats while buying as well as while selling. While
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