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

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Two different teams, X and Y, which have no members in  [#permalink]

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22 Mar 2011, 06:55
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25% (medium)

Question Stats:

73% (01:07) correct 27% (01:47) wrong based on 130 sessions

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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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22 Mar 2011, 07:04

(1) gives us how a weighed average, that can be derived from the height of players in X and Y teams, is nearer to Y, so Y has more players.

(2) does not give any information about number of players in team Y, so not sufficient.
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Two different teams, X and Y, which have no members in  [#permalink]

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25 Oct 2017, 12:43
1

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 !

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Re: Two different teams, X and Y, which have no members in  [#permalink]

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26 Nov 2017, 00:33
hellosanthosh2k2 wrote:

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 !

Hi Santhosh

When nothing else is mentioned, I think its quite safe to assume that all members will merge. Though I believe actual GMAT will take utmost care of such things, as a question is revised and approved after many interventions by qualified people (I believe). Still when not mentioned, I think we can take this assumption.
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Re: Two different teams, X and Y, which have no members in  [#permalink]

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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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Joined: 02 Sep 2009
Posts: 58427
Re: Two different teams, X and Y, which have no members in  [#permalink]

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28 Feb 2018, 01:00
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.

Hope it's clear.
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Two different teams, X and Y, which have no members in  [#permalink]

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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
Two different teams, X and Y, which have no members in   [#permalink] 28 Feb 2018, 01:08
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