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Question Detail
How many litres of pure acid are there in 8 litres of a 20% solution
- 1.5
- 1.6
- 1.7
- 1.8
Answer: Option B
Explanation:
Question of this type looks a bit typical, but it is too simple, as below...
It will be 8 * 20/100 = 1.6
1. 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 %
2. 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 ?
- 27
- 26
- 25
- 24
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
3. Two numbers are less than third number by 30% and 37% respectively. How much percent is the second number less than by the first
- 8%
- 9%
- 10%
- 11%
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%
4. A student multiplied a number by 3/5 instead of 5/3. What is the percentage error.
- 64%
- 65%
- 66%
- 67%
Answer: Option A
Explanation:
Let the number be x,
then,
\begin{aligned}
\frac{5}{3} - \frac{3}{5} = \frac{16}{15}x
\end{aligned}
Error% = \begin{aligned}
(\frac{16}{15}x * \frac{3}{5} * 100)% = 64%
\end{aligned}
5. In an examination, 34% of the students failed in mathematics and 42% failed in English. If 20% of the students failed in both the subjects, then find the percentage of students who passed in both the subjects.
- 40%
- 42%
- 44%
- 46%
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}
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