In an examination, 34% of the students failed in mathematics and 42% failed in English. If 20% of the students failed in both the subjects, then find the percentage of students who passed in both the subjects.

Answer: Option B

Explanation:

Students who play cricket, n(A) = 325
Students who play football, n(B) = 175
Total students who play either or both games,
\begin{aligned}
= n(A\cup B) = 450-50 = 400\\
\text{Required Number}, n(A \cap B) \\
= n(A)+n(B)-n(A\cup B) \\
= 325 + 175 - 400 = 100
\end{aligned}

Answer: Option B

Explanation:

Question of this type looks a bit typical, but it is too simple, as below...

It will be 8 * 20/100 = 1.6

Answer: Option B

Explanation:

Let the total number is x,
then,

(100-25)% of (100 - 10)% x = 4050
=> 75% of 90% of x = 4050
=> 75/100 * 90/100 * x = 4050
=> x = (4050*50)/27 = 6000

Answer: Option C

Explanation:

It is , y% of z = 2(x% of y)
=> yz/100 = 2xy/100
=> z = 2x

Answer: Option C

Explanation:

Failed in mathematics, n(A) = 34
Failed in English, n(B) = 42
\begin{aligned}
n(A\cup B) = n(A)+n(B)-n(A\cap B) \\
= 34+42-20 = 56 \\
\text{Failed in either or both subjects are 56} \\
\text{Percentage passed = }(100-56)\% \\
= 44\%
\end{aligned}