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# 100 students appeared for two examinations. 60 passed the first

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Joined: 27 May 2014
Posts: 502
GMAT 1: 730 Q49 V41
100 students appeared for two examinations. 60 passed the first  [#permalink]

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15 Feb 2017, 00:47
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Difficulty:

5% (low)

Question Stats:

88% (01:34) correct 12% (02:13) wrong based on 118 sessions

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100 students appeared for two examinations. 60 passed the first, 50 passed the second and 30 passed both. Find the probability that a student selected at random has failed in both the examinations?

(A) 1/5
(B) 1/7
(C) 5/7
(D) 5/6
(E) 6/7

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Re: 100 students appeared for two examinations. 60 passed the first  [#permalink]

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16 Feb 2017, 12:51
1
Hi saswata4s,

What is the source of this question? I ask because while it's meant to be an Overlapping Sets question, the answer choices are written in such a way that you can avoid almost of the "implied" math and still answer the question. Since there are exactly 100 students, the fraction that failed both Exams MUST be something "out of 100." Four of the five answer choices CANNOT be reduced from X/100, so they cannot be the correct answer. Only one of them can....

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Joined: 08 Feb 2017
Posts: 14
Re: 100 students appeared for two examinations. 60 passed the first  [#permalink]

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11 Mar 2017, 21:13
1
Can solve this quest by applying set theory
Let A= PPL who passed in 1st test =60
Let C= PPL who passed in 2nd test =50
Let B= overlap i.e. PPL who passed in both= 30
Then A+B=60
Therefore A=30
B+C= 50
Therefore C=20
A+B+C= 80
Remaining= 20 (PPL who did not pass at all)
=20/100= 1/5
Hence option A
Intern
Joined: 06 Nov 2016
Posts: 25
Re: 100 students appeared for two examinations. 60 passed the first  [#permalink]

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15 Feb 2017, 01:07
Lets calculate the number of students failed in each of the two exams.

A- 40 , B - 50. Now, the next thing to figure out is if these two events are dependent or independent events. These two events are independent events as the outcome of one does not influence the outcome of other. In this case the following formula hold true. p(A and B) = p(A) * p(B).

p(A) = 2/5 * 1/2 = 1/5.
Manager
Joined: 31 Jul 2017
Posts: 195
Location: Tajikistan
Re: 100 students appeared for two examinations. 60 passed the first  [#permalink]

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15 Jun 2018, 08:14
I used Venn Diagram to solve this question.
1st circle gives us 30 passed
2nd circle gives us 20 passed
inner part of both circles 1 and 2 give us 30 students. In total we have 80,but since total number of students is 100, we conclude that 20 students didn't pass the test. So, 20/100=1/5
Intern
Joined: 01 Aug 2019
Posts: 1
100 students appeared for two examinations. 60 passed the first  [#permalink]

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16 Dec 2019, 22:55
Please correct me if i am wrong in any of the following steps-
(i am not using Venn Diagram)

As per the question, it states that 30 students pass both the tests (i.e 30 out of 100 passed both examinations)

this means that rest of 70 students will be grouped as "students who didn't passed the both tests"

This itself implies the following-
students failed in 1st exam + students failed in 2nd exam + students failed in both exams= 70
=> 40 + 50 + students failed in both = 70
=> students failed in both = -20 (ie 20 students failed in both)

thus probability is 20/100 => 1/5

please correct me if i am wrong.
100 students appeared for two examinations. 60 passed the first   [#permalink] 16 Dec 2019, 22:55

# 100 students appeared for two examinations. 60 passed the first

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