Question Detail

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

1. A housewife saved Rs. 2.50 in buying an item on sale. If she spent Rs. 25 for the item, approximately how much percent she saved in the transaction

Answer: Option A

Explanation:

Actual Price = Rs.25 + Rs.2.50 = Rs.27.5

Saving = Rs.2.5

\begin{aligned}\text{Percentage Saving = }\dfrac{2.5}{27.5}\times 100 \\
= \dfrac{250}{27.5} = \dfrac{2500}{275}\\
= \dfrac{100}{11} = 9\dfrac{1}{11}\% \approx 9\%\end{aligned}

2. 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 %

3. If sales tax is reduced from 5% to 4%, then what difference it will make if you purchase an item of Rs. 1000

Answer: Option A

Explanation:

Clue: Answer will be 5% of 1000 - 4% of 1000

4. 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}

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 ?

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}