Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are

Answer: Option C

Explanation:

Error = (81.5 - 81.472) = 0.028
Required percentage = \begin{aligned}
\frac{0.028}{81.472} \times 100 = 0.034 %
\end{aligned}

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}

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 D

Explanation:

\begin{aligned}
n(A) = \left(\frac{60}{100}*96\right) = \frac{288}{5} \\
n(B) = \left(\frac{30}{100}*96\right) = \frac{144}{5} \\
n(A\cap B) = \left(\frac{15}{100}*96\right) = \frac{72}{5} \\
\text{People who have either or both lunch} \\
n(A\cup B) = \frac{288}{5}+\frac{144}{5}-\frac{72}{5} \\
= \frac{360}{5} = 72
\end{aligned}

So People who do no have either lunch were = 96 -72
= 24