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selim
In an exam 62% of the students were declared as passed. However, due to complication error, 20% of the students who have actually passed were shown as failed and 20% of the students who have actually failed were declared as passed. what % of the students actually passed?

a) 68
b) 70
c) 72
d) 75
e) 75.5

Let there be p passed in 100 students, so 100-p failed..
So 62 consists of 80%pass and 20% fail..
0.8p+0.2(100-p)=62......0.8p+20-0.2p=62..
0.6p=42.....p=70

B
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Whats wrong with.this approach
62*80% +30*20%?

Posted from my mobile device
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chetan2u
selim
In an exam 62% of the students were declared as passed. However, due to complication error, 20% of the students who have actually passed were shown as failed and 20% of the students who have actually failed were declared as passed. what % of the students actually passed?

a) 68
b) 70
c) 72
d) 75
e) 75.5

Let there be p passed in 100 students, so 100-p failed..
So 62 consists of 80%pass and 20% fail..
0.8p+0.2(100-p)=62......0.8p+20-0.2p=62..
0.6p=42.....p=70

B

chetan2u I don't know if you will see this after so long but can you please explain why you did "62% consists of 80% passed & 20% failed"? That part completely went over my head...
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As someone else asked above why does this not work:
62*8/10 + 38*2/10

The more i look at it, the more confused i get. How can the % of passed people increase when the portion of false positives is larger than false negatives?

I think the problem is worded incorrectly, or at least portrays the wrong meaning. As written, the problem reads like they are of two separate quantities "Of the 62% marked passed, only 80% passed, and of the 38% marked failed, 20% passed". Which is completely different from how it's being solved here "62% is the mixture of 80% pass and 20% fail" or "Of the 62% that passed, 20% of the passed were removed and 20% failed was added"
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VeritasKarishma
does this make sense?

actual passed is x and actual failed is 100-x

0.2x needs to be added to the 62, since 20% of the actual passed had been classified as failed
and
0.2(100-x) need to be taken out of 62 since they had failed but shown as passed.

thus:

0.62 + 0.2x - 0.2(100-x) = x (that is, adding the actual passed and taking out the actul failed, shd give us the actual passed, x)

simplifying we get:

0.62 - 20 = 0.6x

---- this is where the arithmetic fails......that 20 is a problem.....if i turn it into a 0.2, i get 70!

which makes me think if my logic is incrrect.?
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the answer shouldn't be 70, the error I think is including 80% of passed students in 62, where it should be 100%
so the equation is:
62= p+ 20%(100-p)
p= 52.5%
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Rule of alligation

80....... 20
..... 62
42........ 18

Actually passed : actually failed = 42:18 = 7:3

In percentage actually passed
= 7/10 * 100
= 70%
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