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In a set of positive integers {j,k,l,m,n,o} in which j<k<l<m [#permalink]
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15 Aug 2014, 13:04
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In a set of positive integers {j,k,l,m,n,o} in which j<k<l<m<n<o, is the mean larger than the median? 1. The sum of n and o is more than twice the sum of j and k. 2. The sum of k and o is 4/3 the sum of l and m This question was quite hard IMO. How should I do this effectively. It took me more than 2 minutes and I had to guess and move on.
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Re: In a set of positive integers {j,k,l,m,n,o} in which j<k<l<m [#permalink]
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15 Aug 2014, 16:19
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mahendru1992 wrote: In a set of positive integers {j,k,l,m,n,o} in which j<k<l<m<n<o, is the mean larger than the median?
1. The sum of n and o is more than twice the sum of j and k. 2. The sum of k and o is 4/3 the sum of l and m
This question was quite hard IMO. How should I do this effectively. It took me more than 2 minutes and I had to guess and move on. Dear mahendru1992, I'm happy to respond. I don't particularly like this question. I think it is very picayune. A good GMAT question that is very challenging will almost always have something of elegance about it  it may be hard to see, but there is some magically elegant solution. That is not apparent in this problem. The two statements individually do not give enough information, and are insufficient. The real question is: if we combine everything, do we have enough information. I would solve this by picking appropriate numbers. To satisfy the second statement, the more precise constraint, I will make the middle two equal to 3, with a sum of six, and create a sum of 8 for k & o: {_, 2, 3, 3, _ 6} As long as the first blank is 2 or less, and the second blank is between 3 & 6, this will also satisfy the first statement. Now, let's fill in some numbers that produce sums divisible by six. For example: {1, 2, 3, 3, 3, 6}  mean = 3, median = 3 both equal. That's a good starting point. Suppose we make that fifth number bigger {1, 2, 3, 3, 6, 6}  sum = 21, mean = 3.5 median = 3 mean is bigger. Now, go back to the first set of six, and make the first number smaller, even negative: {5, 2, 3, 3, 3, 6}  sum = 12, mean = 2, median = 3 now, the median is bigger So, given the constraints, we can pick examples that make the mean either bigger than, smaller than, or equal to the median. Clearly, we do not have sufficient information to give a definitive answer to the prompt question. Answer = (E). Does all this make sense? Mike
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Re: In a set of positive integers {j,k,l,m,n,o} in which j<k<l<m [#permalink]
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16 Aug 2014, 00:01
mikemcgarry wrote: mahendru1992 wrote: In a set of positive integers {j,k,l,m,n,o} in which j<k<l<m<n<o, is the mean larger than the median?
1. The sum of n and o is more than twice the sum of j and k. 2. The sum of k and o is 4/3 the sum of l and m
This question was quite hard IMO. How should I do this effectively. It took me more than 2 minutes and I had to guess and move on. Dear mahendru1992, I'm happy to respond. I don't particularly like this question. I think it is very picayune. A good GMAT question that is very challenging will almost always have something of elegance about it  it may be hard to see, but there is some magically elegant solution. That is not apparent in this problem. The two statements individually do not give enough information, and are insufficient. The real question is: if we combine everything, do we have enough information. I would solve this by picking appropriate numbers. To satisfy the second statement, the more precise constraint, I will make the middle two equal to 3, with a sum of six, and create a sum of 8 for k & o: {_, 2, 3, 3, _ 6} As long as the first blank is 2 or less, and the second blank is between 3 & 6, this will also satisfy the first statement. Now, let's fill in some numbers that produce sums divisible by six. For example: {1, 2, 3, 3, 3, 6}  mean = 3, median = 3 both equal. That's a good starting point. Suppose we make that fifth number bigger {1, 2, 3, 3, 6, 6}  sum = 21, mean = 3.5 median = 3 mean is bigger. Now, go back to the first set of six, and make the first number smaller, even negative: {5, 2, 3, 3, 3, 6}  sum = 12, mean = 2, median = 3 now, the median is bigger So, given the constraints, we can pick examples that make the mean either bigger than, smaller than, or equal to the median. Clearly, we do not have sufficient information to give a definitive answer to the prompt question. Answer = (E). Does all this make sense? Mike Hi Mike, thank you for answering the question but how're you taking values such as 3,3 and 5? The question states that the values have to be positive and not equal to each other



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Re: In a set of positive integers {j,k,l,m,n,o} in which j<k<l<m [#permalink]
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16 Aug 2014, 01:46
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In a set of positive integers {j,k,l,m,n,o} in which j<k<l<m<n<o, is the mean larger than the median? Mean= (j+k+l+m+n+o)/6 Median=(l+m)/2 (j+k+l+m+n+o)/6>(l+m)/2 (j+k+l+m+n+o)>3(l+m) (j+k+n+o)>2(l+m)
statement 1 : (n+o)>2(j+k) doesn't help to answer whether (j+k+n+o)>2(l+m) statement insufficient
statement 2: (k+o)=4/3(l+m) 3(k+o)=4(l+m) 3/2(k+o)=2(l+m) Therefore,(j+k+n+o)>3/2(k+o) solving above, we get 2(j+n)>(k+o)..? no sufficient information to state that.
1+2 combined are also insufficient to answer whether 2(j+n)>(k+o) Ans  E



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Re: In a set of positive integers {j,k,l,m,n,o} in which j<k<l<m [#permalink]
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16 Aug 2014, 13:41
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mahendru1992 wrote: Hi Mike, thank you for answering the question but how're you taking values such as 3,3 and 5? The question states that the values have to be positive and not equal to each other Dear mahendru1992, Yes, but it's very easy to get around that. Think about it this way. Start with two middle numbers with a sum of 24. {j, k, 11, 13, n, o} Now, make k & o have of a sum of 32. This satisfies the second statement. {j, 10, 11, 13, n, 22} So far, sum = 56. Now, make j = 4, which brings the sum to an even 60, and then k = 18. {4, 10, 11, 13, 18, 22} Sum = 78, mean = 13, and of course median = 12. Mean > median. If we could bring the sum below 72, then the mean would be less than 12. Well, we have to keep k, m, l, and o as is, because they have the fixed ratio of the second statement. We can lower both j and o, as long as (n + o) > 2*(j + k)  that will be no problem to maintain. Subtract 4 from n and 3 from j. {1, 10, 11, 13, 14, 22} Now, the sum is 71, and 71/6 < 72/6 = 12, so the mean is less than 12, and median still equals 12. Mean < median. Often, a powerful way of working with averages is to think about the sum and control the sum. Does all this make sense? Mike
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