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Question Detail
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%
1. If 15% of 40 is greater than 25% of a number by 2, the number is
- 14
- 16
- 18
- 20
Answer: Option B
Explanation:
15/100 * 40 - 25/100 * x = 2 or x/4 = 4 so x = 16
2. 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%
3. In expressing a length of 81.472 km as nearly as possible with the three significant digits, find the percentage error
- 0.35%
- 0.34%
- 0.034%
- 0.035%
Answer: Option C
Explanation:
Error = (81.5 - 81.472) = 0.028
Required percentage = \begin{aligned}
\frac{0.028}{81.472} \times 100 = 0.034 %
\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. If number x is 10% less than another number y and y is 10% more than 125, then find out the value of x
- 123.55
- 123.65
- 123.75
- 123.85
Answer: Option C
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