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maaverick
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This can be solved using Venn diagrams.

35% failed in one subject and 42% in another. 15% failed in both.

Below is the venn diagram representing the failures, in a sample of 100.

____________
______|_________
20% | 15% | 27% |
|_____ |________|
|
____________|
Obviously, those failing in one of the subjects = 20 + 27 = 47%

Total = 47% of 2500 = 1175.
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I think it is easier to use Venn diagrams...At least for me...
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47% of 2500
Venn Diagram are a good idea. One will not commit any mistake.
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Thanks guys, the OA is C
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Ans is c i.e. 1175

since in A subject 35% of 2500 failed i.e. 875
in B subject 42% of 2500 failed i.e. 1050
Failed in both the subject 15%of 2500 i.e. 375

student failed in A alone=875-375=500
student failed in B alone=1050-375=675
Student failed in either of the two subjects=500+675=1175
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I dont understand the answer. The question asks how many passed in either subject but not in both?

Aren't you guys calculating how many failed in either subject but not in both?
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OA is not C but D.

The question asks for the number passed, not for failed.

I cross verified in another forums and the OA is those forums is D.

https://books.google.com/books?id=wOpOVK ... ct&f=false
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maaverick
In an examination, 35% candidates failed in one subject and 42% failed in another subject while 15% failed in both the subjects. If 2500 candidates appeared in the examination, how many passed in either subject but not in both?

a) 325
b) 1075
c) 1175
d) 2125
e) 2250

We can use the following formula:

100 = Percent failed one subject + Percent failed another subject - Percent failed both subjects + Percent failed neither subject

100 = 35 + 42 - 15 + P

100 = 62 + P

P = 38

We see that 38 percent of the candidates failed neither subject, i.e., 38 percent passed both subjects. Now we can use the following formula to find the number who passed either subject but not both:

Number who passed either subject but not both = number who passed only one subject + number who passed only another subject = (number who passed one subject - number who passed both subjects) + (number who passed another subject - number who passed both subjects)

N = 2500 x 0.65 - 2500 x 0.38 + 2500 x 0.58 - 2500 x 0.38

N = 2500 x (0.65 - 0.38 + 0.58 - 0.38)

N = 2500 x 0.47

N = 1175

Alternate Solution:

We can use the following formula:

100 = Percent failed one subject + Percent failed another subject - Percent failed both subjects + Percent failed neither subject

100 = 35 + 42 - 15 + P

100 = 62 + P

P = 38

We know that the percentage that passed both exams is 38% and the percentage that failed both exams is 15%. Therefore, the percentage that passed exactly one exam is (100% - 38% - 15% = 47%). Thus, the number of individuals who passed exactly one exam is (0.47)(2500) = 1175.

Answer: C
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Rupstar
maaverick
In an examination, 35% candidates failed in one subject and 42% failed in another subject while 15% failed in both the subjects. If 2500 candidates appeared in the examination, how many passed in either subject but not in both?

a) 325
b) 1075
c) 1175
d) 2125
e) 2250

c
if 35 failed in A and 42 in B and 15 in both
those failing in A or B but not in A and B is
p(Au(notB))+p((notA)uB)=p(a)+p(b)-2p(AnB)
=42+35-30
=47%
1175

Rupstar, your logic is correct to me but the question asked for those "PASSING" in A or B but not in A and B.

It does not look like OP typed the question correctly because it is not possible to find out how many students passed in A or B and not in A and B. It is difficult to see in a venn diagram but very easy to see in a table:




We know that 62% of students failed at least one test. Because of this, we know that 38% of students passed at least one test. However, we have no information about what % of users passed A, or B, or A&B.
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joondez
Rupstar
maaverick
In an examination, 35% candidates failed in one subject and 42% failed in another subject while 15% failed in both the subjects. If 2500 candidates appeared in the examination, how many passed in either subject but not in both?

a) 325
b) 1075
c) 1175
d) 2125
e) 2250

c
if 35 failed in A and 42 in B and 15 in both
those failing in A or B but not in A and B is
p(Au(notB))+p((notA)uB)=p(a)+p(b)-2p(AnB)
=42+35-30
=47%
1175

your logic is correct to me but the question asked for those "PASSING" in A or B but not in A and B.

It does not look like OP typed the question correctly because it is not possible to find out how many students passed in A or B and not in A and B. It is difficult to see in a venn diagram but very easy to see in a table:



We know that 62% of students failed at least one test. Because of this, we know that 38% of students passed at least one test. However, we have no information about what % of users passed A, or B, or A&B.

Hi,

The question asks for the no. of students who passed either A or B, not both the subjects A and B.

A Pass + B Pass = 500 + 675 = 1175

The diagram needs to be drawn as below:
Attachments

Answer.png
Answer.png [ 3.85 KiB | Viewed 73137 times ]

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hi,

n(A)= fail
n(B)= Fail

then question is asking , n(A')+n(B')-n(A'nB') not n(A) only + n(b) only.
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Hi
can please someone explain why the phrase below can't mean 3 different groups of student exactly:
35% failed in 1st subject and passed in another
42% failed in 2nd and failed the 1st
15% failed the 1st and the 2nd
total 92%
and 8% passed both subjects
Quote:
In an examination, 35% candidates failed in one subject and 42% failed in another subject while 15% failed in both the subjects.
so 100 = Percent failed one subject + Percent failed another subject + Percent failed both subjects + Percent failed neither subject

how do you define from the problem definition that it's indeed 100 = Percent failed one subject + Percent failed another subject - Percent failed both subjects + Percent failed neither subject
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