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For three consecutive odd integers, the product of seven a

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emmak
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For three consecutive odd integers, the product of seven a [#permalink] New post   22 Mar 2013, 01:03
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For three consecutive odd integers, the product of seven and the median of the integers is thirty-five greater than the sum of the other two integers. What is the average of Set A if Set A contains the original three consecutive odd integers as well as the products of those integers and 2?

A. 9.0
B. 10.5
C. 11.0
D. 12.5
E. 13.0

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Re: For three consecutive odd integers, the product of seven a [#permalink] New post   22 Mar 2013, 02:20
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emmak wrote:
For three consecutive odd integers, the product of seven and the median of the integers e is thirty-five greater than the sum of the other two integers. What is the average of Set A if Set A contains the original three consecutive odd integers as well as the products of those integers and 2?

9.0

10.5

11.0

12.5

13.0

This questions indicates the poor quality of some of VERITAS questions


Let the three consecutive odd numbers be 2n-1, 2n+1, 2n+3

median is 2n+1

the product of seven and the median of the integers e is thirty-five greater than the sum of the other two integers

(2n+1) * 7 = 2n-1 + 2n+3 + 35
14n + 7 = 4n + 37
10n = 30
n =3

numbers are 5,7,9
the average of Set A if Set A contains the original three consecutive odd integers as well as the products of those integers and 2
Set will contain 5,7,9 and ,5*2, 7*2 and 9*2

Avg = (5+7+9)*(1+2)/6 = 21/2 = 10.5

So, asnwer will be B

hope it helps!
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Re: For three consecutive odd integers, the product of seven a [#permalink] New post   22 Mar 2013, 03:49
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emmak wrote:
For three consecutive odd integers, the product of seven and the median of the integers is thirty-five greater than the sum of the other two integers. What is the average of Set A if Set A contains the original three consecutive odd integers as well as the products of those integers and 2?

A. 9.0
B. 10.5
C. 11.0
D. 12.5
E. 13.0


Set A={x-2, x, x+2}, where x is an odd integer.

The product of seven and the median of the integers is thirty-five greater than the sum of the other two integers --> 7x=(x-2)+(x+2)+35 --> x=7.

So, set A is {5, 7, 9}.

Now, the new set would be {5, 7, 9, 2*5, 2*7, 2*9} --> (average)=(5+7+9+2*5+2*7+2*9)/6=10.5

Answer: B.

I don't think it's a poor quality question.
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Re: For three consecutive odd integers, the product of seven a [#permalink] New post   25 Mar 2013, 05:08
Hi,

I was able to arrive up to finding these 3 consecutive odd integers as 5,7,9
but then, i misunderstood the question stem and arrived at Set A as,
A = {2,5,7,9,315}

Quote :
"Set A contains the original three consecutive odd integers as well as the products of those integers and 2?"

It says, product of those integers so I did 5x7x9 = 315 and 2

I suppose, question stem can further be improved to make it clear...or my understadning is weak..!!!
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Re: For three consecutive odd integers, the product of seven a [#permalink] New post   29 Mar 2013, 10:11
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Bunuel wrote:

I don't think it's a poor quality question.


Disagree, the last sentence is very confusing. Although I found the sequence, I took the average of 5,7,9, 5*7*9 & 2. To my suprise none of the answer choices were close to my answer :shock: :roll:
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Re: For three consecutive odd integers, the product of seven a [#permalink] New post   26 Apr 2013, 16:39
jainpiyushjain wrote:
Bunuel wrote:

I don't think it's a poor quality question.


Disagree, the last sentence is very confusing. Although I found the sequence, I took the average of 5,7,9, 5*7*9 & 2. To my suprise none of the answer choices were close to my answer :shock: :roll:



Yeah I agree that too. I did the same and spent nearly 7 mins while cracking my head around this. My answer was nowhere close to the options .
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Re: For three consecutive odd integers, the product of seven a [#permalink] New post   26 Apr 2013, 20:50
emmak wrote:
For three consecutive odd integers, the product of seven and the median of the integers is thirty-five greater than the sum of the other two integers. What is the average of Set A if Set A contains the original three consecutive odd integers as well as the products of those integers and 2?

A. 9.0
B. 10.5
C. 11.0
D. 12.5
E. 13.0


Let us assume that the initial set was [x-2,x,x+2] and we need to find out the average of the set [x-2,x,x+2, 2(x-2),2x,2(x+2)].
Required ans = (x+2 + x + x-2 + 2(x-2) +2x + 2(x+2))/6 => 9x/6

i.e. if we somehow get x we are good to go.

From the question we know that 7x = x+2 + x-2 + 35 => x=7

So avg = 9*7/6 = 10.5

OA should be B
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Re: For three consecutive odd integers, the product of seven a [#permalink] New post   09 Mar 2014, 21:42
I also had no idea how to interpret that last sentence. Possibly an inaccurate translation/transcription, as most Veritas questions I have come across are of a high quality and well written.

That being said, I am really hoping the actual GMAT does not contain ambiguous statements, or even potentially ambiguous statements, where success may hinge on interpretation/semantics.
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Re: For three consecutive odd integers, the product of seven a [#permalink] New post   11 Mar 2014, 09:26
same here m not geting the last sentence..
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Re: For three consecutive odd integers, the product of seven a [#permalink] New post   11 Mar 2014, 09:58
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sanjoo wrote:
same here m not geting the last sentence..


Check here: for-three-consecutive-odd-integers-the-product-of-seven-a-149610.html#p1200936
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Re: For three consecutive odd integers, the product of seven a [#permalink] New post   13 Mar 2014, 03:25
The last part of the question 'What is the average of Set A if Set A contains the original three consecutive odd integers as well as the products of those integers and 2?' is quite confusing. It should have been phrased as ' the product of each odd numbers in the set and 2.' I got the answer wrong as a result of your phrasing which may also mean set a contains {5,7,9 , 9*7*9*2}.
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Re: For three consecutive odd integers, the product of seven a [#permalink] New post   03 Mar 2015, 13:09
Hey,

I used as an example 1,3,5 to create variables for the unknown odd consecutive numbers like this:
x, x+2, x+4

Then, based on the information in the stem:
(n+2)*7=n+4+n+35
7n + 14 = 2n + 39
5n = 25
n = 5

So, the numbers are 5,7,8

The whole set consists then of 5,7,9,10,14,18

Sum is: 63
Mean is: 63 / 6 = 10.5

I didn't find the stem confusing. It says the products (plural) of those integers and 2. So, it means the individual products of each one of the 3 integers and 2. To me it is perfectly clear and, so far, the veritas prep questions I have come across are excellent. So are the practice tests.
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Re: For three consecutive odd integers, the product of seven a [#permalink] New post   21 May 2015, 00:39
emmak wrote:
For three consecutive odd integers, the product of seven and the median of the integers is thirty-five greater than the sum of the other two integers. What is the average of Set A if Set A contains the original three consecutive odd integers as well as the products of those integers and 2?

A. 9.0
B. 10.5
C. 11.0
D. 12.5
E. 13.0

I think the last line should be "with 2" RATHER THAN "and 2"
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Re: For three consecutive odd integers, the product of seven a [#permalink] New post   30 Oct 2018, 18:26
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Re: For three consecutive odd integers, the product of seven a [#permalink]
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