Question Detail

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

1. In expressing a length of 81.472 km as nearly as possible with the three significant digits, find the percentage error

0.35%
0.34%
0.034%
0.035%

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}

2. 88% of 370 + 24% of 210 - x = 118

Answer: Option D

3. Evaluate 28% of 450 + 45% of 280

Answer: Option C

Explanation:

= (28/100) * 450 + (45/100) * 280
= 126 + 126 = 252

4. 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}

5. What is 15 percent of 34

5.10
4.10
3.10
2.10

Answer: Option A

Explanation:

It will be 15% of 34
= (15/100) * 34 = 5.10

Rajneesh Singh 9 years ago

the last one question solve method is long.
method should be 60%+30%-15%=75%
25%=96
then ans 24

mastguru 9 years ago replied

thanks Rajneesh :)
Shan Jutt 9 years ago

Hy... I prepared that quiz for nts test..hope they will help..pray for my test..

mastguru 9 years ago replied

Wish you all the good luck !