Quantitative Aptitude
PERCENTAGE MCQs
Percentages
Number of runs made by running = 110 - (3 x 4 + 8 x 6)
= 110 - (60)
= 50.
So, Required percentage = \(\left(\frac{50}{110}\times100\right)\) % = \(45\frac{5}{11}\) %
Let their marks be (x + 9) and x.
Then, x + 9 = \(\frac{56}{100}\left(x+9+x\right)\)
25(x + 9) = 14(2x + 9)
3x = 99
x = 33
So, their marks are 42 and 33.
Suppose originally he had x apples.
Then, (100 - 40)% of x = 420.
\(\frac{60}{100}\times x = 420\)
x = \(\left(\frac{420\times100}{60}\right)= 700.\)
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
So, Required percentage = \(\left(\frac{14}{70}\times100\right)\) % = 20%
x% of y = \(\left(\frac{x}{100}\times y\right) = \left(\frac{y}{100}\times x\right)\) = y% of x
So, A=B.
20% of a = b \(\Rightarrow\frac{20}{100}a = b\)
So, b% of 20 = \(\left(\frac{b}{100}\times 20\right) = \left(\frac{20}{100}a\times\frac{1}{100}\times20\right)\) = 4% of a.
Let the number of students be x. Then,
Number of students above 8 years of age = (100 - 20)% of x = 80% of x.
So, 80% of x = 48 + \(\frac{2}{3} of 48\)
\(\frac{80x}{100} = 80\)
x = 100
5% of A + 4% of B = (6% of A + 8% of B)\(\frac{2}{3}\)
= \(\frac{5}{100}A + \frac{5}{100}B = \frac{2}{3}\left(\frac{6}{100}A + \frac{8}{100}B\right)\)
= \(\frac{1}{100}A +\frac{1}{100}B =\frac{1}{25}A + \frac{1}{75}B\)
= \(\left(\frac{1}{20}-\frac{1}{25}\right)A =\left(\frac{4}{75}- \frac{1}{25}\right)B\)
= \(\frac{1}{100}A =\frac{1}{75}B\)
\(\frac{A}{B} =\frac{100}{75}=\frac{4}{3 }.\)
So, Required ratio = 4:3.
Let the number be x.
Then, error = \(\frac{3}{5}x-\frac{5}{3}x=\frac{16}{15}x.\)
Error% = \(\left(\frac{16x}{15}\times\frac{3}{5}x\times100\right)\) % = 64%
Number of valid votes = 80% of 7500 = 6000.
Valid votes polled by other candidate = 45% of 6000
= \(\left(\frac{45}{100}\times6000\right) = 2700.\)