Question

In an examination 70% of student pass in c-programming and 60% pass in computer architecture, while 25% students failed and 220 students pass in both the subjects. find total number of students.

Answer 1

Let the total number of students be 'x'.

The number of students passed in C-programming  = 70% of x = $\frac{70}{100} \times x$ = 0.7x

So, n(Programming) = n(P)= 0.7x

The number of students passed in Computer architecture = 60% of x = $\frac{60}{100} \times x$ = 0.6x

n(Architecture) = n(A)= 0.6x

The number of students failed in both the subjects

=$n(P^c \cap A^c)=25 of x = \frac{25}{100} \times x = 0.25x$

The number of students passed in both the subjects

= $n(P \cup A)=220$

Now, using the formula,

$n(P \cup A)=n(P)+n(A)-n(P\cap A)$

x - $n(P^c \cap A^c) = 0.7x+0.6x-220$

x - 0.25x = 1.3x - 220

0.75x = 1.3x - 220

0.75x - 1.3x = -220

-0.55x = -220

x = 400

Therefore, the total number of students are 400.