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Set Q consists of 4 consecutive integers, while set T

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Set Q consists of 4 consecutive integers, while set T [#permalink] New post 27 Aug 2006, 07:38
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Set Q consists of 4 consecutive integers, while set T consists of 7 consecutive integers. If the sum of the integers in Q is equal to the sum of the integers in T, is the median of the elements in T a positive number?

(1) Q is a subset of T.
(2) The largest integer in set Q is also in set T.

Last edited by kevincan on 27 Aug 2006, 08:19, edited 2 times in total.
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Re: DS: Sets of Consecutive Integers [#permalink] New post 27 Aug 2006, 07:51
kevincan wrote:
Set Q consists of 4 consecutive integers, while set T consists of 7 consecutive integers. What is the sum of the elements in Q?

(1) Q is a subset of S.
(2) The largest integer in set Q is also in set S.


Are u sure that the question is correct? Seems like it should be T instead of S. well, whatever it is, do let me know!

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 [#permalink] New post 27 Aug 2006, 08:20
GMATT73 wrote:
E?


You're right, but I left out part of the question. Now corrected!
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 [#permalink] New post 28 Aug 2006, 05:59
Question has been corrected. Let's try now!
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 [#permalink] New post 28 Aug 2006, 10:03
Let Q = {n, n+1, n+2, n+3}
Let T = {k, k+1, k+2, .., k+6}

Given
Sum Q = Sum T
4n+(1+2+3) = 7k+(1+2+3..6)
4n = 7k+(4+5+6)

4n=7k+15...........(1)

Q: Is median of T > 0?
i.e. k+3 > 0 i.e. k >-3

Is k > -3?


S1: Q subset of T
Therefore, Q is part of T
n can be {k, k+1, k+2, k+3}


Solving (1) with values for n:
a. n=k:
-3k = 15 => k = -5 < -3.

b.n=k+1
4k+4 = 7k+15
-3k = 11; k is integer. Not possible.

c. n=k+2
4k+8 = 7k+15
or -3k=7 Not possible

d. n=k+3
4k+12 = 7k+15
-3k=3=> k=-1 > -3

Not sufficient as two conflicting answers from (a) and (d)

S2: n+3 belongs to T
n+3 can be anyone of the elements of the set T


a. n+3 = k
4k-12 = 7k+15
-3k = 27, k=-9 < -3

b. n+3 = k+1
4k-8 =7k+15
-3k=23 Not possible

c. n+3 = k+2
4k-4 = 7k+15
-3k = 19 Not possible

d. n+3 = k+3
-3k = 15, k = -5 < -3

e. n+3 = k+4
4k+4 = 7k+15
-3k = 11 Not possible

f. n+3 = k+5
4k+8 = 7k+15
-3k = 7 Not possible

g. n+3 = k+6
4k+12 = 7k+15
-3k = 3, k = -1 > -3

From (a), (d), (g) conflicting answers.

Not sufficient.


S1 & S2:
n+3 = k+3 => k = -5 < -3
n+3 = k+4 => Not possible
n+3 = k+5 => Not possible
n+3 = k+6 => k = -1 > -3


Conflicting answers.

Not sufficient.


Answer: E.
  [#permalink] 28 Aug 2006, 10:03
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