2. 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}
3. If x% of y is 100 and y% of z is 200, then find the relation between x and z.
z = x
2z = x
z = 2x
None of above
Answer: Option C
Explanation:
It is , y% of z = 2(x% of y)
=> yz/100 = 2xy/100
=> z = 2x
4. 2.09 can be expressed in terms of percentage as
2.09%
20.9%
209%
0.209%
Answer: Option C
Explanation:
While calculation in terms of percentage we need to multiply by 100, so
2.09 * 100 = 209.
5. 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 ?
75
100
125
150
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}