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A certain ball team has an equal number of right- and

Expert replies
by showbiz » Fri Aug 03, 2012 4:04 am
A certain ball team has an equal number of right- and left-handed players. On a certain day, two-thirds of the players were absent from practice. Of the players at practice that day, one-third were left handed. What is the ratio of the number of right-handed players who were not at practice that day to the number of left-handed players who were not at practice?

1/3
2/3
5/7
7/5
3/2

C
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Source: — Problem Solving |

by eagleeye » Fri Aug 03, 2012 4:31 am
showbiz wrote:A certain ball team has an equal number of right- and left-handed players. On a certain day, two-thirds of the players were absent from practice. Of the players at practice that day, one-third were left handed. What is the ratio of the number of right-handed players who were not at practice that day to the number of left-handed players who were not at practice?

1/3
2/3
5/7
7/5
3/2

C
Let the total number of players be 18. Then we have 9 right handed and 9 left handed players.
On the particular day, 2/3 were absent => 12 absent, 6 present. Of the present ones, 1/3*6 = 2 are left handed. Therefore, 4 people are right-handed among those present.
Hence right handed players not at practice = 9-4 = 5
Left handed players not at practice = 9-2 = 7.
Ratio = 5/7.

C is correct.

(the choice of 18 here is strategic to make the math easier. I chose it because there were 2 fractions with 3 at the bottom and equal number of players on the team.)
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by Anurag@Gurome » Fri Aug 03, 2012 4:34 am
showbiz wrote:A certain ball team has an equal number of right- and left-handed players. On a certain day, two-thirds of the players were absent from practice. Of the players at practice that day, one-third were left handed. What is the ratio of the number of right-handed players who were not at practice that day to the number of left-handed players who were not at practice?

1/3
2/3
5/7
7/5
3/2

C
Let us assume that total no. of players = T
Then no. of right-handed players = T/2
No. of left-handed players = T/2
No. of players absent from practice = (2T/3)
No. of players present for practice = T - (2T/3) = T/3
No. of left-handed players present for practice = (1/3) * (T/3) = T/9
No. of left-handed players NOT present for practice = T/2 - T/9 = 7T/18
No. of right-handed players present for practice = T/3 - T/9 = 2T/9
No. of right-handed players NOT present for practice = T/2 - 2T/9 = (5T)/18

Therefore, # of right-handed players not present at practice : # of left-handed players not present at practice = 5T/18 : 7T/18 = [spoiler]5:7[/spoiler]

The correct answer is C.
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by GMATGuruNY » Fri Aug 03, 2012 9:44 am
showbiz wrote:A certain ball team has an equal number of right- and left-handed players. On a certain day, two-thirds of the players were absent from practice. Of the players at practice that day, one-third were left handed. What is the ratio of the number of right-handed players who were not at practice that day to the number of left-handed players who were not at practice?

1/3
2/3
5/7
7/5
3/2

C
This is an EITHER/OR group question.
Each player is EITHER left-handed OR right-handed.
Each player is EITHER present OR absent.
For an EITHER/OR group problem, use a GROUP GRID to organize the data.

Let L = left-handed, R = right-handed, P = present, A = absent.
To get a good number for the total, multiply all of the DENOMINATORS of the fractions described in the problem.
An equal number of left-handed and right-handed players implies 1/2.
2/3 were absent.
Of the players who attended practice, 1/3 were left-handed.
Let the total = 2*3*3 = 18.
Here's the grid:
_______________L_______R_______Total

P:

A:

total:____________________________18
Now let's fill in the grid step by step.
As soon as we know 2 entries in a row or a column, we can calculate the remaining entry in that row or column.

A certain ball team has an equal number of right- and left-handed players.
Two-thirds of the players were absent from practice.

_______________L_______R________Total

P:________________________________6

A:_______________________________12

total:__________9_______9_________18
Of the players at practice that day, one-third were left handed.
_______________L______R________Total

P:_____________2______4___________6

A:_______________________________12

total:__________9______9__________18
Complete the grid:
_______________L______R_________Total

P:_____________2______4__________6

A:_____________7______5_________12

total:__________9______9_________18
Answer the question:
(right-handed and absent)/(left-handed and absent) = 5/7.

The correct answer is C.
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by cypherskull » Sat Aug 04, 2012 5:30 am
Let x be the number of players on the team.

RH players = x/2
LH players = x/2

2x/3 - Absent
x/3 - Present

Of the players present, 1/3 LH.

Therefore, amongst the present players,

x/9 - LH
2x/9 - RH

Now pick a number for x so as to make it easier for calculations. Say x = 180. So, total number of LH players = total number of RH players = x/2 = 90.


Of the players present,

x/9 = 20 - LH
2x/9 = 40 - RH

Therefore, of the players absent,

LH = 90 - 20 = 70
RH = 90 - 40 = 50

Hence, the required ratio RH:LH = 50:70 = 5:7.
Regards,
Sunit

________________________________

Kill all my demons..And my angels might die too!
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by vinodsundaram » Mon Aug 06, 2012 1:31 am
Let n(LH) = x , n(RH) = x
therefore, Total players in the team = 2x

No of players present that day = (2x)/3

N(LH on that day) = (2x/3)*(1/3) = 2x/9
Of the player present on that day, 1/3 are left. implying 2/3 are right
N(RH on that day) = (2x/3)*(2/3)= 4x/9

Required ratio = ( n(RH) - n(RH on that day) ) / ( n(LH) - n(LH on that day) )
= ( (x - 4x/9) / (x-2x/9))
= (5x/9)/ (7x/9)
= (5/7)
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