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An article purchased by Mr. A for $x and was sold to Mr. B for $y. Wha 28 Apr 2020, 02:15
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Competition Mode Question



An article purchased by Mr. A for $x and was sold to Mr. B for $y. What was Mr. A's profit percentage?


(1) Mr. B's cost price was $20 more than Mr. A's cost price

(2) Had Mr. A sold the article for double the price than he actually sold it for, his profit would have been 600% more than his actual profit

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B
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An article purchased by Mr. A for $x and was sold to Mr. B for $y. Wha 28 Apr 2020, 04:23
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Buying price = x
Selling price = y

Q. Profit % of Mr. A = (y-x)/x = y/x - 1?

(1) y-x= 20
Since we don't have info on buying price x, we cannot determine how much Mr. A's profit percentage is
NOT SUFFICIENT

(2) 2y-x = 7*(y-x)
6x=5y
y/x = 6/5
--> Mr. A's profit percentage is: y/x - 1 = 1/5 = 20%
SUFFICIENT

FINAL ANSWER IS (B)

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An article purchased by Mr. A for $x and was sold to Mr. B for $y. Wha Updated on: 29 Apr 2020, 03:59
Originally posted by exc4libur on 28 Apr 2020, 05:41.
Last edited by exc4libur on 29 Apr 2020, 03:59, edited 1 time in total.
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Quote:
An article purchased by Mr. A for $x and was sold to Mr. B for $y. What was Mr. A's profit percentage?


(1) Mr. B's cost price was $20 more than Mr. A's cost price

(2) Had Mr. A sold the article for double the price than he actually sold it for, his profit would have been 600% more than his actual profit
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(1) insufic
y=20+x
if x=100, y=120, 120/100-1=0.2=20%
if x=20, y=40, 40/20-1=1=100%

(2) sufic
y-x/x*7=2y-x/x
7(y-x)=2y-x
5y=6x, y/x=6/5

ans (B)
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An article purchased by Mr. A for $x and was sold to Mr. B for $y. Wha 28 Apr 2020, 12:58
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An article purchased by Mr. A for $x and was sold to Mr. B for $y.
What was Mr. A's profit percentage?

Given: Cost price =$x ,Selling price=$y ,so profit (p) =y-x
Asked: (y-x)/x =?

(1) Mr. B's cost price was $20 more than Mr. A's cost price
Says y-x =20 ,what of x? (No info.)
(Not Sufficient)

(2) Had Mr. A sold the article for double the price than he actually sold it for, his profit would have been 600% more than his actual profit
Says 7p= 2y - x.........(2)
Stem: P =y-x.........(1)
Subtract (1) from (2) .: y =6p

Now Substitute y=6p into (1) x= 5p
Now (y-x)/x = (6p-5p)/5p =20%
.: Mr.A’s profit % is 20
(Sufficient)
B for Boy

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An article purchased by Mr. A for $x and was sold to Mr. B for $y. Wha 28 Apr 2020, 19:26
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1. As we are only given the difference i.e. $20, without the knowledge of the real values we can't answer the question
=> A can't be the answer

2. Let profit be P,
=> y - x = P
According to the statement,
2y - x = 7P
Two equations and two variables, hence the answer can be deduced

( Note - You shouldn't go beyond this step, during the exam)

=> y = 6P
=> x = 5p

Therefore, Profit % = (P/5P)*100 = 20%

Hence, Answer - Option B
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An article purchased by Mr. A for $x and was sold to Mr. B for $y. Wha 29 Apr 2020, 13:28
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Bunuel

Competition Mode Question



An article purchased by Mr. A for $x and was sold to Mr. B for $y. What was Mr. A's profit percentage?


(1) Mr. B's cost price was $20 more than Mr. A's cost price

(2) Had Mr. A sold the article for double the price than he actually sold it for, his profit would have been 600% more than his actual profit

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
The question asks the value of \([m]y-x/x\)\cdot100[/m].

When a question asks for a ratio, if one condition includes a ratio and the other condition just gives a number, the condition including the ratio is most likely to be sufficient. This tells us that B is most likely to be the answer to this question.
Thus we need to check condition 2) first which is ratio.

Condition 2)
We have \(2y - x = 7(y-x)\) or \(6x = 5y\).
Thus we have \(\frac{y}{x} = \frac{6}{5}\).

Then we have \(\frac{y-x}{x} \cdot 100 = (\frac{y}{x}-1) \cdot 100 = (\frac{6}{5}-1) \cdot 100 = 20\).

Condition 1)
We have \(y=x+20\).
Then \(\frac{y-x}{x} \cdot 200 = \frac{20}{x} \cdot 100\).
Since we don't know the value of x, condition 1) is not sufficient.

Therefore, B is the answer.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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An article purchased by Mr. A for $x and was sold to Mr. B for $y. Wha 09 Jul 2023, 03:32
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