Two numbers are less than third number by 30% and 37% respectively. How much percent is the second number less than by the first

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 C

Explanation:

Let the third number is x.
then first number = (100-30)% of x
= 70% of x = 7x/10

Second number is (63x/100)

Difference = 7x/10 - 63x/100 = 7x/10

So required percentage is, difference is what percent of first number

=> (7x/100 * 10/7x * 100 )% = 10%

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 D

Explanation:

It means that 0.08% of x = 2
\begin{aligned}
=> ( \frac{8}{100 \times 100} \times x) = 2 \\
=> x = \frac{2 \times 100 \times 100}{8} \\
=> x = 2500
\end{aligned}

Answer: Option B

Explanation:

Clearly,
x% of 150 = 150 - x [as x is number of boys]

\begin{aligned}
=> x + \frac{x}{100} * 150 = 150 \\

=> \frac{5}{2}x = 150 \\

=> x = 60 \\
\end{aligned}