In an exam 62% of the students were declared as passed. However, due t

Question

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

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

chetan2u · Accepted Answer

Hi,

Overall 62% have been declared pass... 
say total is T..
Thus 0.62T are declared passed

This, 0.62T consists of two groups.
1) Actual passed- 20% of these were erroneously declared FAIL.. So 80% are part of the declared passed, that is 62% of total. If P actually passed, then 0.80 are part of 0.62T
2) Actual failed- 20% of these were erroneously declared PASS.. So 20% are part of the declared passed, that is 62% of total. If P% actually passed, then (100-p)% actually failed. 20% of (100-P)%=0.20(100-P), who are part of 0.62T
 
Let us take T as 100..
Thus  0.8p+0.2(100-p)=62 ......
 0.8p+20-0.2p=62 ..
 0.6p=42.....p=70

KarishmaB · Answer

Say there were 100 students. p passed and (100 - p) failed.

62 = (80/100)*p + (20/100)*(100 - p)

42 = 3p/5

p = 70

Answer (B)

chetan2u · Answer

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

juthi223 · Answer

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

Posted from my mobile device

DazednConfused · Answer

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...

fireagablast · Answer

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"

Mansoor50 · Answer

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.?

saif39571 · Answer

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%

Devil90 · Answer

Rule of alligation

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

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

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

In an exam 62% of the students were declared as passed. However, due t

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In an exam 62% of the students were declared as passed. However, due t

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