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nipukumar
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Post subject: Average problem  Posted: Thu Nov 25, 2010 6:08 pm |
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Posts: 11
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If the average (arithmetic mean) of 5 positive temperatures is x degrees Fahrenheit, then the sum of the 3 greatest of these temperatures, in degrees Fahrenheit, could be
(A) 6x
(B) 4x
(C) 35x
(D) 23x
(E) 53x
I could not find the exact answer for this Q. Could some through some light on it with explanation.
Thanks,
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nipukumar
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Post subject: Re: Average problem Posted: Thu Nov 25, 2010 6:16 pm |
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Posts: 11
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Errata to the above posted problem. Correct answer choices are below. There was a copy-paste error on the earlier ones.
(A) 6x
(B) 4x
(C) 5x/3
(D) 3x/2
(E) 3x/5
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nipukumar
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Post subject: Re: Average problem Posted: Thu Nov 25, 2010 6:24 pm |
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My interpretation to the problem is as follows.
Say 5 Temp in Farenh. are a,b,c,d,e
Question stem says: a+b+c+d+e/5 = x
Now say c,d,e are the greatest. For these 3 to be greatest i assumed a and b to be 0 a=b=0 ,
then the above equation becomes,
0+0+c+d+e/5 =x
=> c+d+e=5x
But this is not the answer.
I am thinking if i can assume temperature a and b as 0 or not.
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atul.prasad
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Post subject: Re: Average problem Posted: Fri Nov 26, 2010 9:29 am |
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Hi,
The question says average of 5 POSITIVE numbers.
0 is not a positive number, so you cant choose 0 as one of the temperatures.
This problem can be better solved by elimination since the question asks "what could be". So there is no definite value for this question
Given from the question that the sum of the temperatures is 5*x
Couple of observations:
a) The sum of the largest 3 temperatures cannot be > 5x
b) The sum of the smaller 2 numbers cannot be greater than the sum of the largest 3.
(A) 6x
[Ruled out since it is >5x]
(B) 4x
[Correct Answer]
(C) 5x/3
[Sum of smaller 2 numbers would then be 5x -5x/3 = 10x/3 which contradicts statement b above]
(D) 3x/2
[Sum of smaller 2 numbers would be 5x - 3x/2 = 7x/2 , which contradicts the statement b above]
(E) 3x/5
[Same reasoning as C and D]
Hope this helps.
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jnelson0612
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Post subject: Re: Average problem Posted: Fri Nov 26, 2010 11:28 am |
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Posts: 1857
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Atul, as always, excellent work!
The problem can also be done using the number picking method. I chose to set x=10. If the average of five numbers is 10, then the total of all five added together is 50. Which of the answer choices could be the sum of the largest three?
A) 6x = 6(10) = 60 NO. This is larger than the sum of all five of the numbers
B) 4x = 4(10) = 40 YES. This could be the sum of the three largest, leaving 10 as the sum of the two smallest.
C) 5x/3 = 5(10)/3 = 16 2/3 NO. The largest three cannot have a sum this small, as that would leave 33 1/3 as the sum of the two smallest. Illogical.
D) 3x/2 = 3(10)/2 = 15. NO, for same reason stated in C
E) 3x/5 = 3(10)/5 = 6. NO for same reason stated in C.
The only possible answer is B.
_________________
Jamie Nelson
ManhattanGMAT Instructor
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nipukumar
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Post subject: Re: Average problem Posted: Fri Nov 26, 2010 7:04 pm |
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Excellent analysis. B is infact the right answer.
'0' can be counted as integer but not a positive or negative integer. Key takeaway for me.
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ChrisB
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Post subject: Re: Average problem Posted: Sat Nov 27, 2010 7:37 pm |
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Posts: 91
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Hi,
You're welcome and good takeaway re: positive integers.
Thanks,
Chris
_________________
Chris Brusznicki
MGMAT Instructor
Chicago, IL
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agha79
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Post subject: Re: Average problem Posted: Tue Feb 08, 2011 6:35 pm |
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I also used the number picking approach. I picked numbers 1 – 5
1 + 2 + 3 + 4 + 5 = 3
5
This gives us average of 3 and makes X = 3
Now the sum of three largest values (3, 4, & 5) is 12 and this is the magic number that’s what we need to look for in answer choices.
Now in answer choices when you replace x with 3 only choice B works so “B” is the answer.
I hope this helps
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RonPurewal
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Post subject: Re: Average problem Posted: Wed Feb 09, 2011 9:12 am |
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agha79 wrote: I also used the number picking approach. I picked numbers 1 – 5
1 + 2 + 3 + 4 + 5 = 3
5
This gives us average of 3 and makes X = 3
Now the sum of three largest values (3, 4, & 5) is 12 and this is the magic number that’s what we need to look for in answer choices.
Now in answer choices when you replace x with 3 only choice B works so “B” is the answer.
I hope this helps eeeehhh, yeah, but this only worked because you got really lucky. plug-ins are dangerous on "could be" problems, because a significant number of plug-ins won't make any of the answers work. (try other sets of numbers; the vast majority of those sets won't give an answer of 4x. you won the lottery with these choices.)
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