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Ms. Adams sold two properties, X and Y, for $30,000 each. She sold pro
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22 Jun 2015, 04:23
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Ms. Adams sold two properties, X and Y, for $30,000 each. She sold property X for 20 % more than she paid for it and sold property Y for 20% less than she paid for it. If expenses are disregarded , what was her total net gain or loss, if any, on the two properties ? (A) Loss of $1,250 (B) Loss of $2,500 (C) Gain of $1,250 (D) Gain of $2,500 (E) Neither a net gain nor a net loss
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Ms. Adams sold two properties, X and Y, for $30,000 each. She sold pro
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Updated on: 22 Jun 2015, 04:46
CharmiShah wrote: Ms. Adams sold two properties, X and Y, for $30,000 each. She sold property X for 20 % more than she paid for it and sold property Y for 20% less than she paid for it. If expenses are disregarded , what was her total net gain or loss, if any, on the two properties ?
(A) Loss of $1,250 (B) Loss of $2,500 (C) Gain of $1,250 (D) Gain of $2,500 (E) Neither a net gain nor a net loss There is a property to solve such questions with Common Selling Price and Common %gain and loss. such cases always result in a loss and... Total %Loss = (Common Gain% or Loss%/10)^2 Hence Here Loss% = (20/10)^2 = 4% which means he recovered only 96% of his investment which amount to a total revenue = 30000 + 30000 = 60000 i.e. 96% of cost = 60000 therefore, 4% of cost ( Loss) = $2500 Answer: Option
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Originally posted by GMATinsight on 22 Jun 2015, 04:36.
Last edited by GMATinsight on 22 Jun 2015, 04:46, edited 1 time in total.




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Re: Ms. Adams sold two properties, X and Y, for $30,000 each. She sold pro
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22 Jun 2015, 04:44
CharmiShah wrote: Ms. Adams sold two properties, X and Y, for $30,000 each. She sold property X for 20 % more than she paid for it and sold property Y for 20% less than she paid for it. If expenses are disregarded , what was her total net gain or loss, if any, on the two properties ?
(A) Loss of $1,250 (B) Loss of $2,500 (C) Gain of $1,250 (D) Gain of $2,500 (E) Neither a net gain nor a net loss ALTERNATECOST PRICE = SELLING PRICE  PROFIT i.e. COST PRICE of X = 30,000  (20/100)* COST i.e. 1.2 Cost of X = 30000 i.e. Cost of X = 30000/1.2 = 25000 Similarly, COST PRICE = SELLING PRICE + LOSS i.e. Cost of Y = 30,000 + (20/100)* COST i.e. 0.8 Cost of Y = 30000 i.e. Cost of Y = 30000/0.8 = 37500 Total Cost of X and Y = 25000 + 37500 = 62500 Total Revenue = 62500  60000 = 2500 Answer: Option
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Re: Ms. Adams sold two properties, X and Y, for $30,000 each. She sold pro
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22 Jun 2015, 05:04
COST PRICE = SELLING PRICE  PROFIT
i.e. COST PRICE of X = 30,000  (20/100)* COST i.e. 1.2 Cost of X = 30000 > I got that because of 20% you did 1.2 but why 1.2 Cost of X = 30000 ..? i.e. Cost of X = 30000/1.2 = 25000 > why division ?
Similarly, COST PRICE = SELLING PRICE + LOSS
i.e. Cost of Y = 30,000 + (20/100)* COST i.e. 0.8 Cost of Y = 30000 > Same as above query that 20% loss is 0.80 but what it has to do with X that I didnt get. i.e. Cost of Y = 30000/0.8= 37500> why division ?
Total Cost of X and Y = 25000 + 37500 = 62500 Total Revenue = 62500  60000 = 2500
I could not get this solution. Could you please explain .



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Re: Ms. Adams sold two properties, X and Y, for $30,000 each. She sold pro
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22 Jun 2015, 05:08
CharmiShah wrote: COST PRICE = SELLING PRICE  PROFIT
i.e. COST PRICE of X = 30,000  (20/100)* COST i.e. 1.2 Cost of X = 30000 > I got that because of 20% you did 1.2 but why 1.2 Cost of X = 30000 ..? i.e. Cost of X = 30000/1.2 = 25000 > why division ?
Similarly, COST PRICE = SELLING PRICE + LOSS
i.e. Cost of Y = 30,000 + (20/100)* COST i.e. 0.8 Cost of Y = 30000 > Same as above query that 20% loss is 0.80 but what it has to do with X that I didnt get. i.e. Cost of Y = 30000/0.8= 37500> why division ?
Total Cost of X and Y = 25000 + 37500 = 62500 Total Revenue = 62500  60000 = 2500
I could not get this solution. Could you please explain . These are simple arithmatic calculations COST of X = 30,000  (20/100)* COST of X i.e. COST of X = 30,000  0.2* COST of X i.e. COST of X+ 0.2* COST of X = 30,000 i.e. 1.2* COST of X = 30,000 i.e. COST of X = 30,000 / 1.2 = 25000 Similarly, COST of X = 30,000 + (20/100)* COST of X i.e. COST of X = 30,000 + 0.2* COST of X i.e. COST of X 0.2* COST of X = 30,000 i.e. 0.8* COST of X = 30,000 i.e. COST of X = 30,000 / 0.8 = 37500 I hope it helps!
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Re: Ms. Adams sold two properties, X and Y, for $30,000 each. She sold pro
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22 Jun 2015, 05:20
yes now I got it. many thanks.



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Re: Ms. Adams sold two properties, X and Y, for $30,000 each. She sold pro
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22 Jun 2015, 05:31
CharmiShah wrote: yes now I got it. many thanks. Tradition of extending " Thanks" on GMAT CLUB is by pressing the button of +1KUDOS
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Re: Ms. Adams sold two properties, X and Y, for $30,000 each. She sold pro
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22 Jun 2015, 19:17
CharmiShah wrote: Ms. Adams sold two properties, X and Y, for $30,000 each. She sold property X for 20 % more than she paid for it and sold property Y for 20% less than she paid for it. If expenses are disregarded , what was her total net gain or loss, if any, on the two properties ?
(A) Loss of $1,250 (B) Loss of $2,500 (C) Gain of $1,250 (D) Gain of $2,500 (E) Neither a net gain nor a net loss Check out this post: http://www.veritasprep.com/blog/2014/04 ... nanother/It discusses how and why we get this formula: When two items are sold at the same selling price, one at a profit of x% and the other at a loss of x%, there is an overall loss. The loss% = (x^2/100)% So the loss % here is (20^2/100)% = 4% Now just approximate. The Cost price will be higher than the selling price and overall selling price is 2*30,000 = $60,000 4% of selling price (60,000) is 2400 so 4% of cost price will be slightly higher. Answer must be (B)
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Re: Ms. Adams sold two properties, X and Y, for $30,000 each. She sold pro
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14 Sep 2017, 14:27
The underlying concept here is percent change.
The problem give us the selling price, so we need to find the original price of each property before their respective increase and decrease in price.
Step 1The selling price of the first property is $30,000. this is the price AFTER an increase of 20%, so the original price is x times 1.20 since the price INCREASED 20%. That leaves us with x*1.20=30,000>x=25,000. Thus, the price increased $5,000
Step 2The selling price of the Property y is 30,000. This is the price AFTER a decrease of 20%, so the original price is y times .80 since 1.00.20=.80. That leaves us with y*.80=30,000y=37,500. Thus the price decreased 75,00
Finally, we need to find out whether or not Ms. Adams made money. So we vet the two prices. 50007500=2500. So overall, she lost 2500.
This problem is a lot easier to manage if you make a table using the formula for percent change...



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Re: Ms. Adams sold two properties, X and Y, for $30,000 each. She sold pro
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20 Sep 2017, 14:48
CharmiShah wrote: Ms. Adams sold two properties, X and Y, for $30,000 each. She sold property X for 20 % more than she paid for it and sold property Y for 20% less than she paid for it. If expenses are disregarded , what was her total net gain or loss, if any, on the two properties ?
(A) Loss of $1,250 (B) Loss of $2,500 (C) Gain of $1,250 (D) Gain of $2,500 (E) Neither a net gain nor a net loss We can let x = cost of property X. Thus, the expression (30,000  x) represents the amount of profit she made on property X. Since she sold property X for 20% profit, the profit earned from the sale of property X is: 30,000  x = 0.2x 30,000 = 1.2x 25,000 = x So, she made a profit of 30,000  25,000 = 5,000 dollars on property X. We can let y = cost of property Y. Thus, the expression (30,000  y) is negative and represents the loss she experienced from the sale of property Y. Since she sold property Y at a 20% loss, the loss from the sale of property Y is: 30,000  y = 0.2y 30,000 = 0.8y 37,500 = y So, she incurred a loss of 37,500  30,000 = 7,500 dollars on property Y. Since she made 5,000 dollars on property X and lost 7,500 dollars on property Y, she had a net loss of 2,500 dollars. Answer: B
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Re: Ms. Adams sold two properties, X and Y, for $30,000 each. She sold pro
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Re: Ms. Adams sold two properties, X and Y, for $30,000 each. She sold pro
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