In a class 80% have passed english,70% passed Hindi 10%
didnot passed either. If 144 students passed both. What is
the total strength of the class.
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let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) - n(e intersection h)
=> t = 0.8t + 0.7t - 144
=> t = 1.5t - 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t - 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240
| Is This Answer Correct ? | 98 Yes | 6 No |
Answer / binz
let no of students passed eng be .8t
" " " " " hind be .7t
did not pass either be .1t
then total no of students =.8t+.7t+.1t-144=t
1.6t-t=144
t=144/.6
=240
| Is This Answer Correct ? | 60 Yes | 4 No |
Answer / karthik
Another way:
let t = students who passed one or both subjects,
and x = total students
n(e U h) = n(e) + n(h) - n(e intersection h)
=> x = 0.8t + 0.7t - 144
=> x = 1.5t - 144
=>1.5t - x = 144 --------(1)
students who failed both subjects is 10%
=>x-t = 0.1x
=>0.9x - t = ---------(2)
Solving (1) & (2), we get x = 240.
| Is This Answer Correct ? | 33 Yes | 0 No |
Answer / pravin
the above method s th easiest way to get the ans....
| Is This Answer Correct ? | 12 Yes | 1 No |
Answer / aprameya shyam
let t = total no of students.. then
(e U h) = n(e) + n(h) - n(e intersection h)
=> t = 8t/10 + 7t/10 - 144
=> t = 288 students
| Is This Answer Correct ? | 10 Yes | 41 No |
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