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# In a class, every student is either American or European. Every

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Manager
Joined: 07 Jun 2017
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Location: India
Concentration: Technology, General Management
GMAT 1: 660 Q46 V38
GPA: 3.6
WE: Information Technology (Computer Software)
In a class, every student is either American or European. Every [#permalink]

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12 Oct 2017, 21:34
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Question Stats:

59% (00:49) correct 41% (01:51) wrong based on 41 sessions

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In a class, every student is either American or European. Every student is wearing a blazer or a sweater. If a student is selected randomly, is the probability of the student being a European wearing a sweater greater than that of his/her being an American wearing a blazer?

(1) The probability that a randomly selected student is either an American or wearing a blazer is $$\frac{5}{6}$$

(2) The probability that a randomly selected student is either a European or wearing a sweater is $$\frac{3}{5}$$

Source: http://www.expertsglobal.com
[Reveal] Spoiler: OA

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Regards,
Naveen
email: nkmungila@gmail.com
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In a class, every student is either American or European. Every [#permalink]

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13 Oct 2017, 07:28
1
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nkmungila wrote:
In a class, every student is either American or European. Every student is wearing a blazer or a sweater. If a student is selected randomly, is the probability of the student being a European wearing a sweater greater than that of his/her being an American wearing a blazer?

(1) The probability that a randomly selected student is either an American or wearing a blazer is $$\frac{5}{6}$$

(2) The probability that a randomly selected student is either a European or wearing a sweater is $$\frac{3}{5}$$

Source: http://www.expertsglobal.com

We need to find European wearing a sweater and American wearing a blazer

Statement 1: this implies that probability of European wearing a sweater $$= 1-\frac{5}{6}=\frac{1}{6}$$

but we know nothing about American wearing a blazer. Hence Insufficient

Statement 2: this implies that probability of American wearing a blazer $$= 1-\frac{3}{5}=\frac{2}{5}$$

but we know nothing about European wearing a sweater. Hence Insufficient

Combing 1 & 2 we get our two required values and $$\frac{2}{5}>\frac{1}{6}$$. So we have a NO for our question stem. Hence Sufficient

Option C

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Re: In a class, every student is either American or European. Every [#permalink]

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14 Oct 2017, 06:36
3
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Please kindly imagine that. There are 4 types in the test:
- America + Blazer
- America + Sweater
- Europe + Blazer
- Europe + Sweater

1/ The proportion of student either an America or wearing blazer is 5/6 => It means that 5/6 of the number of students covers the students who are America and blazer => Only Europe + Sweater left => Proportion of Europe + Sweater = 1-5/6=1/6
=> cannot answer the question => insuff

2/ The proportion of student either an Europe or wearing sweater is 3/5 => It means that 3/5 of the number of students covers the students who are Europe and sweater => Only America + Blazer left => Proportion of America+ Blazer = 1-3/5=2/5
=> cannot answer the question => insuff

1 and 2 combine together => proportion of America + Blazer (2/5) > proportion of Europe + Sweater (1/6)

=> suff

Please kudo if it helps. Thanks.

Kudos [?]: 22 [3], given: 142

Manager
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Re: In a class, every student is either American or European. Every [#permalink]

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14 Oct 2017, 17:13
thingocanhnguyen wrote:
Please kindly imagine that. There are 4 types in the test:
- America + Blazer
- America + Sweater
- Europe + Blazer
- Europe + Sweater

1/ The proportion of student either an America or wearing blazer is 5/6 => It means that 5/6 of the number of students covers the students who are America and blazer => Only Europe + Sweater left => Proportion of Europe + Sweater = 1-5/6=1/6
=> cannot answer the question => insuff

2/ The proportion of student either an Europe or wearing sweater is 3/5 => It means that 3/5 of the number of students covers the students who are Europe and sweater => Only America + Blazer left => Proportion of America+ Blazer = 1-3/5=2/5
=> cannot answer the question => insuff

1 and 2 combine together => proportion of America + Blazer (2/5) > proportion of Europe + Sweater (1/6)

=> suff

Please kudo if it helps. Thanks.

the statement says about " OR" either an America or wearing a blazer

so we can not calculate in this way. am I wrong?

Kudos [?]: 2 [0], given: 74

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Re: In a class, every student is either American or European. Every [#permalink]

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14 Oct 2017, 19:20
1
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soodia wrote:
thingocanhnguyen wrote:
Please kindly imagine that. There are 4 types in the test:
- America + Blazer
- America + Sweater
- Europe + Blazer
- Europe + Sweater

1/ The proportion of student either an America or wearing blazer is 5/6 => It means that 5/6 of the number of students covers the students who are America and blazer => Only Europe + Sweater left => Proportion of Europe + Sweater = 1-5/6=1/6
=> cannot answer the question => insuff

2/ The proportion of student either an Europe or wearing sweater is 3/5 => It means that 3/5 of the number of students covers the students who are Europe and sweater => Only America + Blazer left => Proportion of America+ Blazer = 1-3/5=2/5
=> cannot answer the question => insuff

1 and 2 combine together => proportion of America + Blazer (2/5) > proportion of Europe + Sweater (1/6)

=> suff

Please kudo if it helps. Thanks.

the statement says about " OR" either an America or wearing a blazer

so we can not calculate in this way. am I wrong?

Hi soodia

Statement 1 gives probability of being an American +(or) probability of wearing a blazer. So when you remove this probability from the whole i.e. 1. you are left with Only probability of being an European or probability of wearing sweater. Since it is given that each student wears something, so the leftover European has to wear Sweater because probability of wearing blazer has been removed. Hence the statement when reduced from 1 gives probability of being European and wearing a sweater.

Similar logic can be applied for statement 2.

And by combining 1 & 2, you will get the desired result.

Hope this helps

Kudos [?]: 379 [1], given: 42

Manager
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Re: In a class, every student is either American or European. Every [#permalink]

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15 Oct 2017, 17:33
niks18 wrote:
soodia wrote:
thingocanhnguyen wrote:
Please kindly imagine that. There are 4 types in the test:
- America + Blazer
- America + Sweater
- Europe + Blazer
- Europe + Sweater

1/ The proportion of student either an America or wearing blazer is 5/6 => It means that 5/6 of the number of students covers the students who are America and blazer => Only Europe + Sweater left => Proportion of Europe + Sweater = 1-5/6=1/6
=> cannot answer the question => insuff

2/ The proportion of student either an Europe or wearing sweater is 3/5 => It means that 3/5 of the number of students covers the students who are Europe and sweater => Only America + Blazer left => Proportion of America+ Blazer = 1-3/5=2/5
=> cannot answer the question => insuff

1 and 2 combine together => proportion of America + Blazer (2/5) > proportion of Europe + Sweater (1/6)

=> suff

Please kudo if it helps. Thanks.

the statement says about " OR" either an America or wearing a blazer

so we can not calculate in this way. am I wrong?

Hi soodia

Statement 1 gives probability of being an American +(or) probability of wearing a blazer. So when you remove this probability from the whole i.e. 1. you are left with Only probability of being an European or probability of wearing sweater. Since it is given that each student wears something, so the leftover European has to wear Sweater because probability of wearing blazer has been removed. Hence the statement when reduced from 1 gives probability of being European and wearing a sweater.

Similar logic can be applied for statement 2.

And by combining 1 & 2, you will get the desired result.

Hope this helps

YES. nice explanation
thank you niks18

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Re: In a class, every student is either American or European. Every   [#permalink] 15 Oct 2017, 17:33
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