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# At an elite baseball camp, 60% of players can bat both right-handed

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Current Student
Joined: 21 Jul 2013
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At an elite baseball camp, 60% of players can bat both right-handed  [#permalink]

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03 Jan 2015, 08:22
3
15
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Difficulty:

65% (hard)

Question Stats:

66% (02:18) correct 34% (01:41) wrong based on 227 sessions

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At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% of the players who bat left-handed do not bat right-handed, what is the probability that a player selected at random does not bat left-handed?

A.15%
B. 20%
C. 25%
D. 30%
E. 40%

This appeared in a free test by Veritas Prep. I do not really understand the solution provided. Can someone explain please?
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Re: At an elite baseball camp, 60% of players can bat both right-handed  [#permalink]

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03 Jan 2015, 12:09
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Hi All,

This question can be considered an Overlapping Sets question (and solved with the Tic-Tac-Toe Board), but you don't need to organize your information in that way to get to the correct answer.

The key to this question is in realizing that some Left-handed batters ALSO bat with their Right-hands. Here's how you can answer the question by TESTing VALUES:

To start, let's choose 100 for the TOTAL number of players:

60% can bat with BOTH hands = 60 players

**Note: the TOTAL number of players who can bat with their LEFT hands include these 60 players AND the players who can bat with the Left hands ONLY.**

Next, we're told that 25% of players who can bat with their LEFT hands DO NOT bat with their Right hands.

X = Total who can bat Left-handed
25% of X DO NOT use their Right hands
75% of X CAN ALSO use their Right hands

Early on, we had 60 players who could bat with BOTH hands; THAT group is the 75% mentioned above...

.75X = 60
X = 80

There are 80 players who are Left-handed (20 are JUST Left-handed; 60 can bat with BOTH hands).

We're asked for the probability of randomly selecting a NON-Left-handed player.

Total = 100
NON-Left-handed = 20

20/100 = 20%

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At an elite baseball camp, 60% of players can bat both right-handed  [#permalink]

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03 Jan 2015, 11:32
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whitehalo wrote:
At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% of the players who bat left-handed do not bat right-handed, what is the probability that a player selected at random does not bat left-handed?

A.15%
B. 20%
C. 25%
D. 30%
E. 40%

This appeared in a free test by Veritas Prep. I do not really understand the solution provided. Can someone explain please?

Let the total number of people who batted left handed = x
total number of people who bat left-handed, but do not bat right handed = x/4
thus x= 60 + (x/4)
or x=80

thus total number of people who batted right handed = 100-80 =20

thus required probability = (20/100)*100 = 20%
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Re: At an elite baseball camp, 60% of players can bat both right-handed  [#permalink]

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03 Jan 2015, 19:54
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I felt the explanation in the file attached might make the task easy ..Correct me If Im wrong

whitehalo wrote:
At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% of the players who bat left-handed do not bat right-handed, what is the probability that a player selected at random does not bat left-handed?

A.15%
B. 20%
C. 25%
D. 30%
E. 40%

This appeared in a free test by Veritas Prep. I do not really understand the solution provided. Can someone explain please?

Attachments

Untitled.jpg [ 22.93 KiB | Viewed 8096 times ]

VP
Joined: 07 Dec 2014
Posts: 1230
Re: At an elite baseball camp, 60% of players can bat both right-handed  [#permalink]

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02 Apr 2016, 16:16
1
let L=total left handed batters
L/4=left handed batters who do not bat right handed
L-L/4=3L/4=left handed batters who also bat right handed
let P=total players
3L/4=.6P
P=5L/4
L/(5L/4)=4/5=80% probability that player selected at random will bat left handed
100%-80%=20% probability that player selected at random will not bat left handed
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Re: At an elite baseball camp, 60% of players can bat both right-handed  [#permalink]

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03 Jan 2015, 10:05
Apparently, I don't understand it either.

Three options:

Both = .6
Right only = ?
Left only = .25

We are looking for Right only

1 - Both - Left only = .15

Right only = .15

Seem like the logical choice to me....

whitehalo wrote:
At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% of the players who bat left-handed do not bat right-handed, what is the probability that a player selected at random does not bat left-handed?

A.15%
B. 20%
C. 25%
D. 30%
E. 40%

This appeared in a free test by Veritas Prep. I do not really understand the solution provided. Can someone explain please?
Manager
Joined: 27 Dec 2013
Posts: 191
Re: At an elite baseball camp, 60% of players can bat both right-handed  [#permalink]

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03 Jan 2015, 10:53
Me too failed to undstnd this ques.

am i missing something.

aeropower wrote:
Apparently, I don't understand it either.

Three options:

Both = .6
Right only = ?
Left only = .25

We are looking for Right only

1 - Both - Left only = .15

Right only = .15

Seem like the logical choice to me....

whitehalo wrote:
At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% of the players who bat left-handed do not bat right-handed, what is the probability that a player selected at random does not bat left-handed?

A.15%
B. 20%
C. 25%
D. 30%
E. 40%

This appeared in a free test by Veritas Prep. I do not really understand the solution provided. Can someone explain please?
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Posts: 107
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Re: At an elite baseball camp, 60% of players can bat both right-handed  [#permalink]

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01 Apr 2016, 23:07
1
whitehalo wrote:
At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% of the players who bat left-handed do not bat right-handed, what is the probability that a player selected at random does not bat left-handed?

A.15%
B. 20%
C. 25%
D. 30%
E. 40%

This appeared in a free test by Veritas Prep. I do not really understand the solution provided. Can someone explain please?

I guess the tricky part is "25% of the players who bat left-handed do not bat right-handed". This also means that 75% of the players who bat left handed also bat right handed.
=> 75%of left handed players = 60.
Therefore no. of left handed players = 80
Therefore number of players who bat only right handed = 20
Hence probability = 20/100 = 20%

Please suggest if this approach is correct.
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Posts: 8336
Re: At an elite baseball camp, 60% of players can bat both right-handed  [#permalink]

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01 Apr 2016, 23:42
MeghaP wrote:
whitehalo wrote:
At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% of the players who bat left-handed do not bat right-handed, what is the probability that a player selected at random does not bat left-handed?

A.15%
B. 20%
C. 25%
D. 30%
E. 40%

This appeared in a free test by Veritas Prep. I do not really understand the solution provided. Can someone explain please?

I guess the tricky part is "25% of the players who bat left-handed do not bat right-handed". This also means that 75% of the players who bat left handed also bat right handed.
=> 75%of left handed players = 60.
Therefore no. of left handed players = 80
Therefore number of players who bat only right handed = 20
Hence probability = 20/100 = 20%

Please suggest if this approach is correct.

Hi,
Mention that you are starting with the total number as 100
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Re: At an elite baseball camp, 60% of players can bat both right-handed  [#permalink]

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01 Aug 2017, 20:06
The trick is that it's a set problem and not a venn diagram.

So if 25% of the batters who bat left don't bat right that means it's 20%. As if 20 people don't bat right and 60% do then that's 2/8 or 25%
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Posts: 4229
Re: At an elite baseball camp, 60% of players can bat both right-handed  [#permalink]

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25 Nov 2019, 11:31
Top Contributor
whitehalo wrote:
At an elite baseball camp, 60% of players can bat both right-handed and left-handed. If 25% of the players who bat left-handed do not bat right-handed, what is the probability that a player selected at random does not bat left-handed?

A.15%
B. 20%
C. 25%
D. 30%
E. 40%

We can solve using the Double Matrix Method.

The Double Matrix Method can be used for most questions featuring a population in which each member has two characteristics associated with it.
Here, we have a population of baseball players, and the two characteristics are:
- bats left-handed or DOESN'T bat left-handed
- bats right-handed or DOESN'T bat right-handed

NOTICE that the question does not ask us to find an actual number. It asks us to find a probability. This means we can assign whatever value we wish to the total number of couples.
So, let's say there are 100 players, which we'll add to our diagram:

60% of players can bat both right-handed and left-handed
60% of 100 = 60, so 60 players can bat both right-handed AND left-handed .
Add that to the diagram to get:

25% of the players who bat left-handed do not bat right-handed
Hmmm, we don't know the number of left-handed players, so we can't find 25% of that value.
So, let's assign a variable.
Let's let x = left-handed batters, and add it to our diagram:

So, x of the 100 players bat left handed.

25% of the players who bat left-handed do not bat right-handed
If x players bat left-handed, then 25% of x do not bat right-handed.
In other words, 0.25x = number of players who do not bat right-handed

At this point, we see that the two left-hand boxes add to x.
So, we can write the equation: 60 + 0.25x = x
Rearrange to get 60 = 0.75x
Rewrite 0.75 as fraction to get: 60 = (3/4)x
Multiply both sides by 4/3 to get: 80 = x
If x = 80, then we know that 80 of the 100 players bat left-handed.
This means that the remaining 20 players DO NOT bat left handed.

So, P(player doesn't bat left-handed) = 20/100 = 20%

Cheers,
Brent

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Re: At an elite baseball camp, 60% of players can bat both right-handed   [#permalink] 25 Nov 2019, 11:31
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