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

Answer: Option D

Explanation:

Let the origianl salary = Rs. 100

It will be 150% of (50% of 100)
= (150/100) * (50/100) * 100 = 75

So New salary is 75, It means his loss is 25%

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

Answer: Option D

3. What will be the fraction of 20%

1/4
1/5
1/10
None of above

Answer: Option B

Explanation:

It will 20*1/100 = 1/5

4. 1/2 is what percent of 1/3

150%
200%
250%
300%

Answer: Option A

Explanation:

1/2/1/3 * 100 = 1/2 * 3/1 * 100 = 150 %

5. 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}