Question Detail
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 %
1. 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
2. Total number of boys and girls in a school is 150. If the number of boys is x, then girls become x% of the total number of students. The number of boys is
- 50
- 60
- 70
- 80
Answer: Option B
Explanation:
Clearly,
x% of 150 = 150 - x [as x is number of boys]
\begin{aligned}
=> x + \frac{x}{100} * 150 = 150 \\
=> \frac{5}{2}x = 150 \\
=> x = 60 \\
\end{aligned}
3. 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}
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 ?
- 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}
5. 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%