The weights of two iron balls are 3.5 kg and 7.5 kg. What is the percentage
weight of the 1st ball with respect to 2nd ball.
Answer: A
Weight of 1st ball \(= 3.5 kg\) and weight of second ball \(= 7.5 kg\)
⇒ Percentage weight of the 1st ball with respect to 2nd ball
\(= \frac{3.5}{7.5} \times 100 \)
\(= \frac{35}{3} \times 4 = \frac{140}{3} \)
\(= 46\frac{2}{3}\)
Please log in or register to report this question.
A student multiplied a number by \(\frac{5}{6}\) instead of \(\frac{6}{5}\). What is the percentage
error in the calculation?
Answer: B
The number be 30
When the student multiplied it by \(\frac{5}{6}\), ⇒ original result \(= \frac{5}{6} \times 30 = 25\)
When the student multiply it by \(\frac{6}{5}\), ⇒ New result \(= \frac{6}{5} \times 30 = 36\)
⇒ Percentage error in calculation \(= \frac{(36-25)}{36} \times 100 \)
\(\Rightarrow \frac{1100}{36} = 30.56\)
Please log in or register to report this question.
A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percentage of his total score did he make by running between the wickets?
Answer: B
Number of runs made by running \(= 110 - (3 \times 4 + 8 \times 6) \)
\(= 100 - (60) \)
\(= 50.\)
\(\therefore\) Required percentage \(= (\frac{50}{110} \times 100)\)% \(= 45\frac{5}{11}\)%
Please log in or register to report this question.
Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56 % of the sum of their marks. The marks obtained by them are:
Answer: C
Their marks be \((x + 9)\) and \(x\)
Then, \(x + 9 = \frac{56}{100}(x + 9 + x)\)
\(\Rightarrow 25(x + 9) = 14(2x + 9) \)
\(\Rightarrow 3x = 99 \)
\(\Rightarrow x = 33\)
So, their marks are 42 and 33.
Please log in or register to report this question.
A fruit seller had some apples. He sells 40 % apples and still has 420 apples. Originally, he had:
Answer: D
Suppose originally he had \(x\) apples.
Then, \((100 - 40)\)% of \(x = 420\)
\(\Rightarrow \frac{60}{100} \times x = 420 \)
\(\Rightarrow x = (\frac{420 \times 100}{60}) = 700\)
Please log in or register to report this question.
What percentage of numbers from 1 to 70 have 1 or 9 in the unit's digit?
Answer: C
Clearly, the numbers which have 1 or 9 in the unit's digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such number =14
Therefore, Required percentage \(=(\frac{14}{70} \times 100)\)% \(= 20\)%
Please log in or register to report this question.
If \(A = x\)% of \(y\) and \(B = y\)% of \(x\), then which of the following is true?
Answer: D
\(x\)% of \(y\) \(= (\frac{x}{100} \times y) = (\frac{y}{100} \times x) = y\)% of \(x\)
\(\therefore A = B\)
Please log in or register to report this question.
If 20 % of \(a = b\), then \(b\) % of 20 is the same as:
Answer: A
20 % of \(a = b \Rightarrow \frac{20}{100}a = b.\)
\(\therefore b\) % of \(20 = (\frac{b}{100} \times 20) = (\frac{20}{100}a \times \frac{1}{100} \times 20) = \frac{4}{100}a = 4\)% of \(a\)
Please log in or register to report this question.
In a certain school, 20 % of students are below 8 years of age. The number of students above 8 years of age is\(\frac{2}{3}\) of the number of students of 8 years of age which is 48. What is the total number of students in the school?
Answer: D
Let the number of students be \(x\).
Then, Number of students above 8 years of age \(= (100 -20)\)% of \(x = 80\)% of \(x\)
\(\therefore 80\)% of \(x = 40 + \frac{2}{3} \) of 48
\(\Rightarrow \frac{80}{100}x = 80\)
\(\Rightarrow x = 100.\)
Please log in or register to report this question.
Two numbers \(A\) and \(B\) are such that the sum of 5 % of \(A\) and 4 % of \(B \) is two-third of the sum of 6 % of \(A\) and 8 % of \(B. \) Find the ratio of \(A:B\)
Answer: D
5 % of \(A\) + 4% of \(B\) \(= \frac{2}{3}\) (6% of \(A + 8\)% of \(B\))
\(\Rightarrow \frac{5}{100}A + \frac{4}{100}B = \frac{2}{3}(\frac{6}{100}A + \frac{8}{100}B)\)
\(\Rightarrow \frac{1}{20}A + \frac{1}{25}B = \frac{1}{25}A + \frac{4}{75}B\)
\(\Rightarrow (\frac{1}{20} - \frac{1}{25})A = (\frac{4}{75} - \frac{1}{25})B\)
\(\Rightarrow \frac{1}{100}A = \frac{1}{75}B\)
\(\frac{A}{B} = \frac{100}{75} = \frac{4}{3}\)
\(\therefore Required ratio = 4: 3\)
Please log in or register to report this question.
