Question Detail

Out of 450 students of a school, 325 play football, 175 play cricket and 50 neither play football nor cricket. How many students play both football and cricket ?

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}

1. The ratio 5:20 expressed as percent equals to

50 %
125 %
25 %
None of above

Answer: Option C

Explanation:

Actually it means 5 is what percent of 20, which can be calculated as,
(5/20)*100 = 5 * 5 = 25

2. 2 is what percent of 50

Answer: Option B

Explanation:

2/50 * 100 = 1/25 * 100 = 4 %

3. An inspector rejects 0.08% of the meters as defective, How many meters he examine to reject 2 meteres

1200
2400
1400
2500

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}

4. Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate got

Answer: Option C

Explanation:

Total number of votes polled = (1136 + 7636 + 11628) = 20400

So, Required percentage = 11628/20400 * 100 = 57%

5. In a hotel, 60% had vegetarian lunch while 30% had non-vegetarian lunch and 15% had both type of lunch. If 96 people were present, how many did not eat either type of lunch ?

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