Question Detail

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.

Answer: Option C

Explanation:

Failed in mathematics, n(A) = 34
Failed in English, n(B) = 42
$\begin{aligned} n(A\cup B) = n(A)+n(B)-n(A\cap B) \\ = 34+42-20 = 56 \\ \text{Failed in either or both subjects are 56} \\ \text{Percentage passed = }(100-56)\% \\ = 44\% \end{aligned}$

1. Teacher took exam for English, average for the entire class was 80 marks. If we say that 10% of the students scored 95 marks and 20% scored 90 marks then calcualte average marks of the remaining students of the class.

Answer: Option C

Explanation:

Lets assume that total number of students in class is 100 and required average be x.
Then from the given statement we can calculate :
(10 * 95) + (20 * 90) + (70 * x) = (100 * 80)

=> 70x = 8000 - (950 + 1800) = 5250

=> x = 75.

2. How many litres of pure acid are there in 8 litres of a 20% solution

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

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
$\begin{aligned} => ( \frac{8}{100 \times 100} \times x) = 2 \\ => x = \frac{2 \times 100 \times 100}{8} \\ => x = 2500 \end{aligned}$