Question

in a cetain examination, 40%of the students passed in mathematics, 50% passes in english and 25% passed in both subjects..If 120 students failed in both the subjects, find the total number of students who sat the test

Answer 1

$\large\underline{\sf{Solution-}}$

Given that, in a cetain examination,

• 40%of the students passed in mathematics.

• 50% passes in english.

• 25% passed in both subjects.

• 120 students failed in both the subjects.

Let assume that

• Total number of students appeared in the test be x.

Let further assume that

• A represents the set of students who passed in Mathematics.

• B represents the set of students who passed in English.

So,

$\rm \: n(A) = \frac{40}{100} \times x = 0.4x$

$\rm \: n(B) = \frac{50}{100} \times x = 0.5x$

$\rm \: n(A\cap B) = \frac{25}{100} \times x = 0.25x$

So, number of students who passed in at least one of the subject is

$\rm \: n(A\cup B)$

$\rm \: = n(A) + n(B) - n(A\cap B)$

$\rm \: = 0.4x + 0.5x - 0.25x$

$\rm \: = 0.65x$

Now, it is given that number of students who failed in both the subjects is 120.

$\rm \: n(A'\cap B') = 120$

$\rm \: x - n(A\cup B) = 120$

$\rm \: x - 0.65x = 120$

$\rm \: 0.35x = 120$

$\rm \: x = \dfrac{120}{0.35}$

$\rm\implies \:x = 342.85 \: \approx \: 343$

So, 343 students sat in the test.

Answer 2

Step-by-step explanation:

\large\underline{\sf{Solution-}}

Solution−

Given that, in a cetain examination,

40%of the students passed in mathematics.

50% passes in english.

25% passed in both subjects.

120 students failed in both the subjects.

Let assume that

Total number of students appeared in the test be x.

Let further assume that

A represents the set of students who passed in Mathematics.

B represents the set of students who passed in English.

So,

\begin{gathered}\rm \: n(A) = \frac{40}{100} \times x = 0.4x \\ \end{gathered}

n(A)=

100

40

×x=0.4x

\begin{gathered}\rm \: n(B) = \frac{50}{100} \times x = 0.5x \\ \end{gathered}

n(B)=

100

50

×x=0.5x

\begin{gathered}\rm \: n(A\cap B) = \frac{25}{100} \times x = 0.25x \\ \end{gathered}

n(A∩B)=

100

25

×x=0.25x

So, number of students who passed in at least one of the subject is

\begin{gathered}\rm \: n(A\cup B) \\ \end{gathered}

n(A∪B)

\begin{gathered}\rm \: = n(A) + n(B) - n(A\cap B) \\ \end{gathered}

=n(A)+n(B)−n(A∩B)

\begin{gathered}\rm \: = 0.4x + 0.5x - 0.25x \\ \end{gathered}

=0.4x+0.5x−0.25x

\begin{gathered}\rm \: = 0.65x \\ \end{gathered}

=0.65x

Now, it is given that number of students who failed in both the subjects is 120.

\begin{gathered}\rm \: n(A'\cap B') = 120 \\ \end{gathered}

n(A

′

∩B

′

)=120

\begin{gathered}\rm \: x - n(A\cup B) = 120 \\ \end{gathered}

x−n(A∪B)=120

\begin{gathered}\rm \: x - 0.65x = 120 \\ \end{gathered}

x−0.65x=120

\begin{gathered}\rm \: 0.35x = 120 \\ \end{gathered}

0.35x=120

\begin{gathered}\rm \: x = \dfrac{120}{0.35} \\ \end{gathered}

x=

0.35

120

\rm\implies \:x = 342.85 \: \approx \: 343⟹x=342.85≈343

So, 343 students sat in the test.