In a set of positive integers {j,k,l,m,n,o} in which j

Question

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 :-)

desaichinmay22 · Answer

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

mikemcgarry · Answer

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 :-)

GMATGuruNY · Answer

Whereas Statement 1 describes an inequality, Statement 2 indicates an actual equation.
Since Statement 2 seems more constrained, start with Statement 2.

Statement 2:
All of the values must be positive integers.
Since  k+o = \frac{4}{3}(l + m) ,  l+m  must be a MULTIPLE OF 3.

Let  l=3  and  m=12 , with the result that  k+o = \frac{4}{3}(3+12) = 20 
If  k=2  and  o=18 , the following case is possible:
{1, 2, 3, 12, 13, 18}
In this case:
Mean  = \frac{1+2+3+12+13+18}{6} = \frac{49}{6} =  more than 8
Median  = \frac{3+12}{2} = 7.5 
Since mean > median, the answer to the question stem is YES.

To reduce the mean, try to bring  l  and  m  closer to each other.

Let  l=7  and  m=8 , while keeping  k=2  and  o=18 .
The following case is possible:
{1, 2, 7, 8, 9, 18}
In this case:
Mean =  \frac{1+2+7+8+9+18}{6} = \frac{45}{6} = \frac{15}{2} = 7.5 
Median  = \frac{7+8}{2} = 7.5 
Since mean = median, the answer to the question stem is NO.

Since the answer is YES in the first case but NO in the second case, INSUFFICIENT.

Statement 1:
The two cases tested in Statement 2 also satisfy Statement 1.
In the first case,  n+o=13+18=31  and  j+k=1+2=3 , with the result that  n+o  is more than twice  j+k .
In the second case,  n+o=9+18=27  and  j+k=1+2=3 , with the result that  n+o  is more than twice  j+k .

Since each case satisfies both statements -- and the answer is YES in the first case but NO in the second case -- the two statements combined are INSUFFICIENT.

E

bumpbot · Answer

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In a set of positive integers {j,k,l,m,n,o} in which j<k<l<m

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