GMAT Prep - DS Avg

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GMAT Prep - DS Avg

by oks » Fri Dec 26, 2008 4:24 pm
A certain list consists of five different integers. Is the average of the two greatest integers in the list greater than 70?

a. The median of the integers in the list is 70
b. The average of the integers in the list is 70.

My answer is the correct one - D, but it took me a long time to figure out that B could be a standalone solution as well.

Thanks!
Source: — Data Sufficiency |

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average

by nsharma » Fri Dec 26, 2008 6:35 pm
1. If median is 70 ( align them in increasing order) that means next two numbers are greater than 70 hence average of the two greatest integers will be more than 70.

2. I tried this by two trials of a set and figured out.

40 100 20 120 70 (100+120)/2 >70

60 60 90 60 80 (90+80)/2 > 70

i hope there is another short cut

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by vishubn » Fri Dec 26, 2008 6:56 pm
I think thats the only way :) ~!

i mean atleast to me !

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by cramya » Fri Dec 26, 2008 10:25 pm
Guys,
I have PM'd Stuart to provide some pointers if any on attacking these questions conceptually or in an easier way if it exists.

Regards,
Cramya

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by cramya » Fri Dec 26, 2008 10:36 pm
I am sure Stuart may have a more elegant solution for us.

My approach:

Stmt II

Average is 70 and the 5 integers are distinct.

If we try to prove that the average of the highest 2 numbers is less than 70 then we need to maxmize the first three numbers (in ascending order)

The best we can do is by highest possible integers below 70

68 69 70 71 72

(OR)

67 68 69 72 74

We see in the best cases possible the average of the the highest 2 numbers is always greater than the mean 70

Hance SUFF

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by scoobydooby » Fri Dec 26, 2008 11:01 pm
when deciding about statement 2 can we take some hint from the statement 1 that the median is 70? (since the statements never contradict each other in DS)
since the numbers are distinct integers, the highest 2 numbers will have to be greater than 70 and their average will be greater than 70 so B will be sufficient as well?

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by cramya » Fri Dec 26, 2008 11:13 pm
when deciding about statement 2 can we take some hint from the statement 1 that the median is 70?
The general rule of thumb is we should not use info from stmt I in stmt II and vice versa when evaluating statements independently. Its kind of a tricky/good question and I dont want to mislead anyone by saying something wrong. Once u have exhausted all possible considerations for each statement independently u can kind of take the other statment as a cue. May be I am confusing everyone heere but thats the best way I could put it. I am sure the experts may have a crystal clear answer

(since the statements never contradict each other in DS)
Another way to put this would be there would be something common between the 2 statements in general.

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by Ian Stewart » Sat Dec 27, 2008 10:43 am
scoobydooby wrote:when deciding about statement 2 can we take some hint from the statement 1 that the median is 70? (since the statements never contradict each other in DS)
As cramya points out, you can't use Statement 1 when you are considering Statement 2 alone - if you do, you're really using both Statements together. Indeed, while the median *might* be 70 when we use Statement 2, it might be different from 70 as well.
oks wrote:A certain list consists of five different integers. Is the average of the two greatest integers in the list greater than 70?

a. The median of the integers in the list is 70
b. The average of the integers in the list is 70.

My answer is the correct one - D, but it took me a long time to figure out that B could be a standalone solution as well.

Thanks!
Statement 2 is sufficient for quite a simple reason - if you remove the smallest element from a set, the average can't possibly go down (and unless all the elements are identical, the average will always go up). So if I have a five-element set with different elements and an average of 70, and then I take out the three smallest elements, the average must be larger than 70 afterwards.

If this were a real world problem, I think the answer would be clear. Say you're enrolled in an undergrad program where your courses are graded from 0 to 100, you're taking five courses, and you have an overall average of 70 on these five courses. If someone offers you the option to throw away your three lowest marks to calculate your overall average, would this be to your advantage? Well, of course it would, unless all five of your grades are 70, in which case your average would still be 70. That's the same idea as in this question.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by cramya » Sat Dec 27, 2008 11:10 am
Excellent analogy Stewart! Thank you!

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by vivek.kapoor83 » Mon Dec 29, 2008 11:16 am
i went it by somewht similar approach but dont know where fell short

stmt 1 is ok
in Stmt 2 ....
if avg is 70, then sum =350.So,largest 2 no. can be find out by subtracting 3 smallest from 350 and as all no are diff so, min possible values are 1,2,3 ...adding up all and sub from 350 - 6 = 344
So, 2 largest no sum is 344
..................now here i got trapped...pls help me any1...stuart...cramya.,,vittal

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by cramya » Mon Dec 29, 2008 2:35 pm
min possible values are 1,2,3 ...adding up all and sub from 350 - 6 = 344
So, 2 largest no sum is 344
..................now here i got trapped...pls help me any1...stuart...cramya.,,vittal

Here u have to maximize the first three numbers to prove insufficiency which can be best achieved with consecutive integers as I explained in my approach.

Even with the the first three numbers being a 68 69 70 or 67 68 69 there is no way the average of the last 2 be lesser than 70(since there ar 5 distinct integers).

Ian's approach would be the best but for mortals in math/quant :-) like us the consecutive integers may come to mind easier.

Hope this helps!

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by vivek.kapoor83 » Wed Dec 31, 2008 2:54 am
cramya,
even my approach i think was correct,
coz if u get 344 and u have to maintain the avg above 70 then those no hv to be greater than 70.
So, suff.

Ian - Would appriciate your comment on my solution.

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by Zipper » Wed Dec 31, 2008 4:28 am
vivek.kapoor83 wrote:cramya,
even my approach i think was correct,
coz if u get 344 and u have to maintain the avg above 70 then those no hv to be greater than 70.
So, suff.
You have to pick the greatest possible numbers in order to get the smallest possible numbers for the last two.And if their average is greater than 70 then you've proved it. You subtract 1, 2 and 3 however these are not the greatest possible integers, hell, they are not even the smallest possible integers :)

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by logitech » Wed Dec 31, 2008 1:56 pm
What IF the question was written as:

A certain list consists of FOUR different integers.

Would you still choose D ? :twisted:
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by cramya » Wed Dec 31, 2008 2:09 pm
Hypothetical problem:

LIST CONTAINS 4 DIFFERENT INTEGERS INSTEAD OF 5:

I would say yes but someone feel free to correct me if I am mistaken

Stmt I

I am sure we got that even for 4 integers
a<b<c<d

b+c/2 = 70
b+c = 140

b=69 c=71 is your best case to prove INSUFF not really since still not good enough to prove average of highest 2 is less than 70

Still d>71 so c+d/2 > 70

SUFF

Stmt II

a<b<c<d since they are different integers

The average of the integers in the list is 70

a+b+c+d / 4 = 70

a+b+c+d = 280

c+d = 280 - (a+b)

c+d/2 = 280 - (a+b) / 2

(a+b) cannot be greater than or equal to 140 since the average would then be greater than 70

So 280 - (a+b) / 2 will always be greater than 70

SUFF

Choose D)