Raman's salary was decreased by 50% and subsequently increased by 50%. How much percent does he loss.

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}

Answer: Option A

Explanation:

Actual Price = Rs.25 + Rs.2.50 = Rs.27.5

Saving = Rs.2.5

\begin{aligned}\text{Percentage Saving = }\dfrac{2.5}{27.5}\times 100 \\
= \dfrac{250}{27.5} = \dfrac{2500}{275}\\
= \dfrac{100}{11} = 9\dfrac{1}{11}\% \approx 9\%\end{aligned}

Answer: Option A

Explanation:

Let the number be x,
then,
\begin{aligned}
\frac{5}{3} - \frac{3}{5} = \frac{16}{15}x
\end{aligned}

Error% = \begin{aligned}
(\frac{16}{15}x * \frac{3}{5} * 100)% = 64%
\end{aligned}

Answer: Option C

Explanation:

Lets assume that total number of students in class is 100 and required average be x.
Then from the given statement we can calculate :
(10 * 95) + (20 * 90) + (70 * x) = (100 * 80)

=> 70x = 8000 - (950 + 1800) = 5250

=> x = 75.

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