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Two different teams, X and Y, which have no members in

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rxs0005
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Two different teams, X and Y, which have no members in [#permalink] New post   22 Mar 2011, 06:55
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Two different teams, X and Y, which have no members in common, will merge to form a new team. If the average (arithmetic mean) height of the players on team X is 5 feet 7 inches and the average height of the players on team Y is 5 feet 10 inches, which team has more players?

(1) If a new team, team Z, is formed from teams X and Y it will have an average height of 5 feet 9 inches.
(2) There are 12 players on team X.
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Two different teams, X and Y, which have no members in [#permalink] New post   25 Oct 2017, 12:43
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Hi the answer is A.

But i have a question, though it may look silly.

See, statement 1 says team z is formed from team X and team Y, but it is never mentioned that all players of team X and team Y will be included in team Z.
Please note new team (formed out of merge) given in the question and team Z mentioned in statement 1 may be different.

Say only 1 out of 100 players from team X and 24 out of 25 players from Y are included in Z, then, weighted average will be definitely towards the average of Y, proving Y has more players, but actually X has more players.

Confused !

Please help.
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lor12345
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Re: Two different teams, X and Y, which have no members in [#permalink] New post   28 Feb 2018, 00:47
rxs0005 wrote:
Two different teams, X and Y, which have no members in common, will merge to form a new team. If the average (arithmetic mean) height of the players on team X is 5 feet 7 inches and the average height of the players on team Y is 5 feet 10 inches, which team has more players?

(1) If a new team, team Z, is formed from teams X and Y it will have an average height of 5 feet 9 inches.
(2) There are 12 players on team X.


Hi,
Can anybody explain me how to calculate answer 1?
Thanks
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Bunuel
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Re: Two different teams, X and Y, which have no members in [#permalink] New post   28 Feb 2018, 01:00
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lor12345 wrote:
rxs0005 wrote:
Two different teams, X and Y, which have no members in common, will merge to form a new team. If the average (arithmetic mean) height of the players on team X is 5 feet 7 inches and the average height of the players on team Y is 5 feet 10 inches, which team has more players?

(1) If a new team, team Z, is formed from teams X and Y it will have an average height of 5 feet 9 inches.
(2) There are 12 players on team X.


Hi,
Can anybody explain me how to calculate answer 1?
Thanks


Two different teams, X and Y, which have no members in common, will merge to form a new team. If the average (arithmetic mean) height of the players on team X is 5 feet 7 inches and the average height of the players on team Y is 5 feet 10 inches, which team has more players?

The average (arithmetic mean) height of the players on team X = 5 feet 7 = 67 inches;
The average (arithmetic mean) height of the players on team Y = 5 feet 10 inches = 70 inches.

(1) If a new team, team Z, is formed from teams X and Y it will have an average height of 5 feet 9 inches (69 inches). Since the overall average is closer to the average of team Y than it is to the average of team X, then team Y must have more members than team X. Sufficient.

Algebraically: \(69 =\frac{67x + 70y}{x + y}\). After simplifying we'll get 2x = y. Hecen, y > x.

(2) There are 12 players on team X. Clearly insufficient.

Answer: A.

Hope it's clear.
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Two different teams, X and Y, which have no members in [#permalink] New post   28 Feb 2018, 01:08
lor12345 wrote:
rxs0005 wrote:
Two different teams, X and Y, which have no members in common, will merge to form a new team. If the average (arithmetic mean) height of the players on team X is 5 feet 7 inches and the average height of the players on team Y is 5 feet 10 inches, which team has more players?

(1) If a new team, team Z, is formed from teams X and Y it will have an average height of 5 feet 9 inches.
(2) There are 12 players on team X.


Hi,
Can anybody explain me how to calculate answer 1?
Thanks


Ignore the 5ft and only look at the inches to save time (since both have 5 fts).

Team X = 7 inch Avg
Team Y = 10 inch Avg
Team Z = 9 inch Avg

From here you can already guess that team Y has more players than X (Seeing that the combined Team Z has an average skewed closer to Team Y).

But if you want to calculate the exact ratio (which is not needed for this question:

\(\frac{Y-Z}{Z-X} = \frac{10-9}{9-7} = \frac{1}{2}\)

Team Y has twice as many players than team X.

Hopes this helps :-)
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Re: Two different teams, X and Y, which have no members in [#permalink] New post   16 Apr 2020, 09:52
I have 1 doubt
Statement 1 gives only about the avg height
What if only tall players are selected from Team X(for ex lets say 6ft) and Short players are selected from Team Y.
Will it make a difference in choosing answer?
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GlobalNomad13
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Re: Two different teams, X and Y, which have no members in [#permalink] New post   17 Apr 2020, 15:54
Answer A - I used the weighed average method.

Started with statement 2 - clearly insuff
Statement 1: the new average is closer to the average in team Y, it was enough to tell me that team Y must carry "more weight".
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Re: Two different teams, X and Y, which have no members in [#permalink] New post   08 Nov 2023, 01:19
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Re: Two different teams, X and Y, which have no members in [#permalink]
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