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During a sale, a clothing store sold each shirt at a price [#permalink]

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30 Aug 2009, 07:01

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D

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79% (01:57) correct
21% (01:22) wrong based on 134 sessions

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During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of $25. Did the store sell more sweaters than shirts during the sale? (1) The average (arithmetic mean) of the prices of all the shirts and sweaters that the store sold during the sale was $21. (2) The total of the prices of all the shirts and sweaters that the store sold during the sale was $420.

In the xy-plane, does the line with equation y=3x+2 contain the point(r,s)? (1) (3r+2-s)(4r+9-s)=0 (2) (4r-6-s)(3r+2-s)=0

Re: GMAT Practice Test answer explanations - DS [#permalink]

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30 Aug 2009, 08:11

2) In the xy plane, does the line with the equation y = 3x + 2 contain the point (r,s)? 1) (3r + 2 - s)(4r + 9 - s) = 0 2) (4r + 6 - s)(3r + 2 - s) = 0

The line y = 3x + 2 contains the point means it passes through (r,s)

=> it should satisfy the equation.

=> s = 3r +2 or, 3r + 2 - s = 0 --- equation 1st...

1.) (3r + 2 - s)(4r + 9 - s) = 0

It is possible when (3r + 2 - s) = 0 or (4r + 9 - s) = 0 if (3r + 2 - s) = 0, then , (r,s) satisfies the equation ( from 1st) but if, (4r + 9 - s) = 0 then, we can not be sure that (r,s) satisfies the equation ( from 1st) Hence, insufficient..

2.) Same as 1

Combining both,

both are possible only when (3r + 2 - s) = 0 ..Hence, (r,s) lies on the line..

Re: GMAT Practice Test answer explanations - DS [#permalink]

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30 Aug 2009, 09:08

First question.

Without any equations and thinking logically, it can be shown that the stmt 1 is sufficient. The average price is 21. Price per shirt =15, price per sweater = 25. If the equal quantity of shirts and sweater was sold, then the average would be 20. Since average is greater than 20, the store sold more items with higher price - sweaters.

Re: GMAT Practice Test answer explanations - DS [#permalink]

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22 Nov 2009, 18:54

gmate2010 wrote:

2) In the xy plane, does the line with the equation y = 3x + 2 contain the point (r,s)? 1) (3r + 2 - s)(4r + 9 - s) = 0 2) (4r + 6 - s)(3r + 2 - s) = 0

The line y = 3x + 2 contains the point means it passes through (r,s)

=> it should satisfy the equation.

=> s = 3r +2 or, 3r + 2 - s = 0 --- equation 1st...

1.) (3r + 2 - s)(4r + 9 - s) = 0

It is possible when (3r + 2 - s) = 0 or (4r + 9 - s) = 0 if (3r + 2 - s) = 0, then , (r,s) satisfies the equation ( from 1st) but if, (4r + 9 - s) = 0 then, we can not be sure that (r,s) satisfies the equation ( from 1st) Hence, insufficient..

2.) Same as 1

Combining both,

both are possible only when (3r + 2 - s) = 0 ..Hence, (r,s) lies on the line..

Hence, sufficient.. C is the answer..

I have the question

the question stem itself gives us the info that 3r+2-s = 0

so from statement 1, applying question stem, the equation statisfies irrespective of of what (4r + 9 - s) is.

Re: GMAT Practice Test answer explanations - DS [#permalink]

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14 Dec 2009, 20:17

ISBtarget wrote:

gmate2010 wrote:

2) In the xy plane, does the line with the equation y = 3x + 2 contain the point (r,s)? 1) (3r + 2 - s)(4r + 9 - s) = 0 2) (4r + 6 - s)(3r + 2 - s) = 0

The line y = 3x + 2 contains the point means it passes through (r,s)

=> it should satisfy the equation.

=> s = 3r +2 or, 3r + 2 - s = 0 --- equation 1st...

1.) (3r + 2 - s)(4r + 9 - s) = 0

It is possible when (3r + 2 - s) = 0 or (4r + 9 - s) = 0 if (3r + 2 - s) = 0, then , (r,s) satisfies the equation ( from 1st) but if, (4r + 9 - s) = 0 then, we can not be sure that (r,s) satisfies the equation ( from 1st) Hence, insufficient..

2.) Same as 1

Combining both,

both are possible only when (3r + 2 - s) = 0 ..Hence, (r,s) lies on the line..

Hence, sufficient.. C is the answer..

I have the question

the question stem itself gives us the info that 3r+2-s = 0

so from statement 1, applying question stem, the equation statisfies irrespective of of what (4r + 9 - s) is.

same for statement 2 as well, so i chose D

I think from statement one 3r+2-s may be 0 or second part may be zero...similarly for statement 2, combining both will prove that 3r+2-s is zero

Re: GMAT Practice Test answer explanations - DS [#permalink]

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28 Mar 2012, 07:53

from statement 2,

shirts x sweaters y

15x +20y = 420

Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation? If there is a common multiple for

Re: GMAT Practice Test answer explanations - DS [#permalink]

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28 Mar 2012, 09:04

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dchow23 wrote:

from statement 2,

shirts x sweaters y

15x +20y = 420

Can we say that since 60 is a common multiple between the 15 and 20, there will be more than one answer that can satisfy the equation? If there is a common multiple for

Not sure that I understand your question correctly but generally for such kind of question it's always better to check whether an equation whose solution must be integers only has one or more set of variables satisfying it.

During a sale, a clothing store sold each shirt at a price of $15 and each sweater at a price of $25. Did the store sell more sweaters than shirts during the sale?

(1) The average (arithmetic mean) of the prices of all the shirts and sweaters that the store sold during the sale was $21 --> since the average sale price of $21 is closer to $25 than to $15 then store sold more sweaters than shirts (sweaters have more weight in the weighted average and thus pulled it closer to $21). Sufficient.

(2) The total of the prices of all the shirts and sweaters that the store sold during the sale was $420 --> 15x + 25y =420 --> 3x+5y=84. If x=8 and y=12 then x<y but if x=13 and y=5 then x>y. Not sufficient.

Re: During a sale, a clothing store sold each shirt at a price [#permalink]

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28 Mar 2012, 09:04

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In the XY plane, does the line with equation y=3x+2 contain the point (r,s)?

Line with equation \(y=3x+2\) contains the point \((r,s)\) means that when substituting \(r\) ans \(s\) in line equation: \(s=3r+2\) (or \(3r+2-s=0\)) holds true.

So basically we are asked to determine whether \(3r+2-s=0\) is true or not.

(1) \((3r+2-s)(4r+9-s)=0\) --> either \(3r+2-s=0\) OR \(4r+9-s=0\) OR both. Not sufficient.

(2) \((4r-6-s)(3r+2-s)=0\) --> either \(3r+2-s=0\) OR \(4r-6-s=0\) OR both. Not sufficient.

(1)+(2) Both \(4r+9-s=0\) and \(4r-6-s=0\) can not be true (simultaneously), as \(4r-s\) can not equal to both -9 and 6, hence \(3r+2-s=0\) must be true. Sufficient.

Re: During a sale, a clothing store sold each shirt at a price [#permalink]

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09 Sep 2013, 07:58

Dear Bunuel,

For following statement, you provided two different sets of values for x,y. I tried to identify these values by simply plugging multiples of one variable, but this approach is taking more than 2min, is there any method or trick related to lcm or any thing which we can use to quickly identify all possible values for the equation? I am really struggling with this hit and trial approach, what we should look for while selecting these values.

(2) The total of the prices of all the shirts and sweaters that the store sold during the sale was $420 --> 15x + 25y =420 --> 3x+5y=84. If x=8 and y=12 then x<y but if x=13 and y=5 then x>y. Not sufficient.

Thanks _________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

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Re: During a sale, a clothing store sold each shirt at a price
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09 Sep 2013, 07:58

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