If the distance between two stations \(A\) and \(B\) is \(200 \ km\). A train covers the distance from \(A\) to \(B\) At the average speed of \(100 \ km/hr\) and returns from \(B\) to \(A\) at the average speed of \(x \ km/hr\). The average speed of the train during the whole journey is \(70 \ km/hr\), find the value of \(x \ ?\)
\(53 \ km/hr\)
\(52.84 \ km/hr\)
\(53.84 \ km/hr\)
\(55.23 \ km/hr\)
Answer: (c) \(53.84 \ km/hr\)Solution: Given values, \(x_1 = 100 \ km/hr\), \(x_2 = x \ km/hr\), and the average speed \(= 70 \ km/hr\), then $$ Average \ speed = \frac{2 \ x_1 \ x_2}{x_1 + x_2} $$ $$ 70 = \frac{2 \times 100 \times x}{100 + x} $$ $$ 7000 + 70 \ x = 200 \ x $$ $$ 130 \ x = 7000 $$ $$ x = \frac{7000}{130} = 53.84 \ km/hr $$
A man travels first \(60 \ km\) in \(1 / hr\) and second \(50 \ km\) in \(x \ hr\). If the average speed of the man is \(40 \ km/hr\) then find the value of \(x \ ?\)
Find the average of numbers between \(10\) to \(30\) divisible by \(3 \ ?\)
\(18.5\)
\(19.5\)
\(20.5\)
\(16.5\)
Answer: (b) \(19.5\).Solution: First number divisible by \(3\) after \(10\) is \(12\).last number divisible by \(3\) before \(30\) is \(27\). Then $$ Average = \frac{12 + 27}{2} $$ $$ = \frac{39}{2} = 19.5 $$
The average of five numbers is \(70\), if one number is excluded then Average becomes \(80\). Find the excluded number?
\(30\)
\(32\)
\(35\)
\(36\)
Answer: (a) \(30\)Solution: Average increases = \(80 - 70 = 10\)Total increased number = \(10 \times 4 = 40\)It means is \(40\) less the average, then excluded number = \(70 - 40 = 30\)
A bus covers \(100 \ km\) distance, first \(30 \ km\) at the speed of \(50 \ km/hr\), second \(30 \ km\) at the speed of \(60 \ km/hr\) and last \(40 \ km\) at the speed of \(x \ km/hr\). If the average speed of the bus during the whole journey is \(65 \ km/hr\), then find the value of \(x \ ?\)
Find the average of first four even natural numbers after multiplying by \(5\) and divide by by \(2 \ ?\)
\(13.5\)
\(14.5\)
\(12.5\)
\(10.5\)
Answer: (c) \(12.5\)Solution: First four even natural numbers = \(2, 4, 6, 8\)First four even natural numbers after multiplying by \(5\) and divide by \(2\) = \(5, 10, 15, 20\) then, $$ Average = \frac{5 + 10 + 15 + 20}{4} $$ $$ = \frac{50}{4} = 12.5 $$
If the average of \(8\) numbers is \(30\) then find the average after multiplying the numbers by \(5 \ ?\)
\(150\)
\(160\)
\(158\)
\(156\)
Answer: (a) \(150\)Solution: Average of \(8\) numbers is \(30\).total value of \(8\) numbers = \(8 \times 30 = 240\)total value of \(8\) numbers after multiplying the numbers by \(5\) = \(240 \times 5 = 1200\) then, $$ Average = \frac{1200}{8} = 150 $$
If the average weight of four men is \(65 \ kg\) and the ratio of their weight is \(2 : 4 : 3 : 5\), then find the weight of lowest weight man?
\(35.50 \ kg\)
\(37.25 \ kg\)
\(37.14 \ kg\)
\(41.26 \ kg\)
Answer: (c) \(37.14 \ kg\)Solution: Let the weight of men = \(2x, 4x, 3x, 5x\) then, $$ 65 = \frac{2x + 4x + 3x + 5x}{4} $$ $$ 14x = 260 $$ $$ x = 18.57 $$ the weight of lowest weight man = \(2 \times 18.57 = 37.14 \ kg\)
Find the average of first three natural numbers of multiplying by \(10\) and divide by \(5 \ ?\)
\(2\)
\(4\)
\(6\)
\(8\)
Answer: (b) \(4\)Solution: First three natural numbers = \(1, 2, 3\)First three natural numbers after multiplying by \(10\) and divide by \(5\) = \(2 , 4, 6\) then, $$ Average = \frac{2 + 4 + 6}{3} = \frac{12}{3} = 4 $$
If a man travels three consecutive distances \(100 \ km\), \(150 \ km\), and \(200 \ km\), at the speed of \(45 \ km/hr\), \(55 \ km/hr\), and \(x \ km/hr\). If the average speed of the man during the whole journey is \(60 \ km/hr\), then find the value of \(x \ ?\)