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# In a football club, 1/4th of the players are capped players and ......

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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3078
In a football club, 1/4th of the players are capped players and ......  [#permalink]

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Updated on: 23 Jan 2019, 05:34
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Difficulty:

15% (low)

Question Stats:

86% (02:00) correct 14% (02:42) wrong based on 39 sessions

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Everything you need to know about Tree Structures - Exercise Question #2

In a football club, $$\frac{1}{4}^{th}$$ of the players are capped players and the remaining are uncapped players. Half of the capped players and all the uncapped players are defenders. If there are 36 players, in the club, who are uncapped, then what is the number of players, in the club, who are not defenders?

A. 3
B. 6
C. 9
D. 12
E. 18

To read the article: Everything you need to know about Tree Structures

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Originally posted by EgmatQuantExpert on 23 Jan 2019, 04:18.
Last edited by EgmatQuantExpert on 23 Jan 2019, 05:34, edited 1 time in total.
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Re: In a football club, 1/4th of the players are capped players and ......  [#permalink]

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23 Jan 2019, 04:59
total uncapped players = 36
3*x/4 = 36
x= 48
total players in club = 48
so
capped = 48 * 1/4 = 12

defenders = 6+ 36 = 42

non defenders = 48-42 = 6
IMO B

we can solve question using 2x2 matrix as well..

[
quote="EgmatQuantExpert"]Everything you need to know about Tree Structures - Exercise Question #2

In a football club, $$\frac{1}{4}^{th}$$ of the players are capped players and the remaining are uncapped players. Half of the capped players and all the uncapped players are defenders. If there are 36 players, in the club, who are uncapped, then what is the number of players, in the club, who are not defenders?

A. 3
B. 6
C. 9
D. 12
E. 18

To read the article: Everything you need to know about Tree Structures

[/quote]
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3078
In a football club, 1/4th of the players are capped players and ......  [#permalink]

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01 Feb 2019, 02:21

Solution

Given:
We are given that,
• $$\frac{1}{4}^{th}$$ of the players in a football club are capped players, and the remaining players are uncapped players
o Half of the capped players are defenders

• The number of uncapped players in the club = 36

To find:
• The number of players, in the club, who are not defenders

Approach and Working:
• Total number of players are divided as capped and uncapped, and all the players of these two groups are either defenders are non-defenders.

Now, let’s represent the above structure in the form a tree diagram.

Now, let’ see what we need to find : The number of players, in the club, who are not defenders = $$C_O + U_O$$
• Thus, we need to find the values of $$U_O$$ and $$C_O$$, to answer this question

Let us assume that the total number of players in the club = T
• From the given information , we can say that the number of capped players = 25% of T = $$\frac{T}{4}$$
• Which implies, that the number of uncapped players = $$T – \frac{T}{4} = \frac{3T}{4}$$

Now, from the diagram we can see that, total capped players = $$C_D + C_O$$
• Thus, the equation will be $$C_D + C_O = \frac{T}{4}$$
• And, from the given information, we get, $$C_D = (\frac{1}{2}) * (\frac{T}{4}) = \frac{T}{8}$$
o Thus, $$C_O = \frac{T}{4} – \frac{T}{8} = \frac{T}{8}$$ ……………. (1)

We are also given that all uncapped players are defenders
• Thus, $$U_D = \frac{3T}{4}$$ and $$U_O = 0$$ ………………….…… (2)

So, from (1) and (2), we get, $$U_O + C_O = 0 + \frac{T}{8} = \frac{T}{8}$$ ………………. (3)
• To find T, we have information that, number of uncapped players, $$U = \frac{3T}{4} = 36$$
o This implies, $$T = 36 * \frac{4}{3} = 48$$

• Substituting this value of T in (3), we get
o $$U_O + C_O = \frac{48}{8} = 6$$

Therefore, the number of players, in the club, who are not defenders = 6

Hence, the correct answer is option B.

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In a football club, 1/4th of the players are capped players and ......   [#permalink] 01 Feb 2019, 02:21
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